5 Double Integrals
Multivariable calculus is a critical tool for solving problems in physics, engineering, and other sciences. However, before delving into the higher-dimensional world of multivariable calculus, it’s important to have a strong foundation in the basics.
In this book, we focus on the fundamentals of multiple integration-the processes of integrating functions of several variables. We’ll discuss both Cartesian and polar coordinate systems, along with the various techniques for evaluating integrals. With a solid understanding of these basic concepts, you’ll be ready to tackle the more challenging problems in multivariable calculus.
Multivariable calculus is the study of functions of multiple variables. It allows for the determination of maxima and minima, as well as other important information about how a function behaves. In many cases, multiple variables can be treated as a single variable, allowing for the use of traditional calculus methods. However, in other cases, multiple variables require a more sophisticated approach.
This is where multivariable calculus comes in. Multivariable calculus is particularly important in physics and engineering, where multiple physical quantities often need to be considered simultaneously. It is also useful in economics, where multiple factors can affect the behavior of a market or individual consumers. In short, multivariable calculus is a powerful tool that can be used to analyze complicated situations involving multiple variables.
Double integrals are a type of multiple integration, meaning they allow you to integrate over multiple variables simultaneously. In other words, double integrals let you find the total amount of something (like area or volume) that is spread out over multiple dimensions.
To calculate a double integral, you first need to choose an appropriate coordinate system. Once you’ve done that, you can divide the region into smaller pieces and then add up the integrals over each of those pieces. Double integrals can be used to calculate all sorts of things, from the volume of an irregular solid to the probability of two random variables taking on certain values. So whatever it is you’re trying to integrate, a double integral is probably the way to go!
Integrals are a fundamental tool in calculus, and iterated integrals are a powerful extension of this concept. With multiple integration, we can calculate the volume of a solid region bounded by multiple surfaces. This technique can also be used to calculate the moments of a 3D object, or to find the center of mass of an irregular shape.
In short, iterated integrals are a versatile tool that can be used in a wide range of applications. So if you’re ever feeling lost in a sea of multiple integrals, just remember that each one is just an extension of the basic concept of integration. And with a little practice, you’ll be iterating your way to success in no time!
Double integrals in polar coordinates can be a bit of a pain to wrap your head around. After all, multiple integration is already confusing enough - now you have to do it in a different coordinate system? But once you understand the basics, it’s not so bad. And who knows, you might even find it fun.
To understand multiple integrals in polar coordinates, it helps to think about a two-dimensional coordinate system like a cake. The x-axis is like the diameter of the cake, and the y-axis is like the height. If we want to find the volume of the cake, we need to integrate both the diameter and the height. That’s where multiple integrals come in.
With multiple integrals in polar coordinates, we’re essentially doing the same thing - but with a twist. Instead of using a rectangular coordinate system, we’re using a polar coordinate system. This means that our x-axis is now the angle, and our y-axis is now the radius. And just like before, if we want to find the volume (or more accurately, the area) of our cake, we need to integrate both the angle and the radius.
Integrals are a fundamental tool in calculus, and the double integral is a powerful extension of the single integral. Double integrals can be used to calculate the area of irregular shapes, the volume of solids with curved surfaces, and the amount of flow through an orientable surface. In fact, multiple integrals can be applied in countless ways to solve problems in physics, engineering, and other fields.
In many ways, multiple integrals are the natural extension of single integrals. Just as a single integral can be thought of as the area under a curve, a double integral can be thought of as the volume under a surface.
More generally, multiple integrals can be used to calculate the measure of any region in space. This makes them a powerful tool for solving problems in physics and engineering. In addition, multiple integrals can be used to calculate more complex quantities than their single-variable counterparts.
For example, double integrals can be used to calculate moments and centroids. As a result, multiple integrals are an essential tool for anyone looking to tackle problems in mathematics or the sciences.
So next time you’re struggling with a difficult integral, remember that you’re not alone - and that there’s probably a multiple integral out there that can help.
Triple integrals are multiple integrals where the domain of integration is three-dimensional. In other words, they’re integrals over three variables. These integrals can be difficult to calculate, but using cylindrical or spherical coordinates can make the process much simpler. Cylindrical coordinates are particularly well-suited for problems involving cylindrical objects, such as cylinders and cones. Spherical coordinates are best for problems involving spheres or other round objects.
We’ve all been there. We’re given a multiple integral to solve, and we just don’t know where to start. Should we use cylindrical coordinates? Or spherical coordinates? How do we even set up the integral in those coordinate systems? And what if the region of integration isn’t nice and simple like a box or a cylinder?
Thanks to the magic of triple integrals, we can solve multiple integrals no matter what the coordinate system or the shape of the region. In this article, we’ll look at how to set up triple integrals in both cylindrical and spherical coordinate systems.
With a little practice, you’ll be able to tackle any multiple integral, no matter how complicated it might seem at first. Either way, these multiple integrals can be tricky, but with a little practice, you’ll be a pro in no time!
Multiple integration is a powerful tool that can be used to calculate integrals of many different functions. However, it is sometimes necessary to change the variables in a multiple integral in order to make the calculation easier. This process is known as a change of variables.
There are many different ways to change the variables in a multiple integral, but the most common method is to use a substitution. This involves replacing the original variables with new ones that are more suitable for the particular problem.
For example, if we want to integrate a function that is only defined on a certain region of space, we can use a change of variables to transform the integral into one that is defined on all of space. This can be a very useful technique when dealing with multiple integrals.
Multivariable calculus may sound like a mouthful, but it’s just a fancy way of saying “calculus with more than one variable.” And as anyone who’s taken calculus knows, one variable is already plenty. So what’s the big deal with multiple variables? Well, for starters, multiple integration can be a real pain. You’ve got multiple integrals, double integrals, triple integrals… it can all get pretty confusing pretty quickly.
And then there’s the issue of finding critical points. With multiple variables comes multiple partial derivatives, and keeping track of them all can be a challenge. But don’t despair! These challenges can be overcome with practice and patience. After all, there’s nothing wrong with taking things one step at a time. Just remember: take a deep breath, relax, and trust in the process. You’ll get there eventually.
I hope that this tour has given you a better understanding of how I teach calculus in this book. I take a step-by-step approach, breaking down each concept and providing plenty of examples along the way. My goal is to make the process as easy and painless as possible, so you can focus on learning the material and not worrying about the mechanics of the math. With a little practice, you’ll be able to tackle any calculus problem that comes your way!
5.1 The Volume Under a Surface
Consider the rectangle given by \[ R=[a,b]\times[c,d]=\{(x,y)\mid a\leq x\leq b, c\leq x\leq d\}. \] We wish to construct a (regular) partition of \(R\). To do so, let \[ a=x_0 < x_1 < x_2 < \cdots < x_{i-1} < x_i < \cdots < x_n=b \] be a partition of \([a,b]\) into subintervals each of width \(\Delta x=(b-a)/n\). Similarly let \[ c=y_0 < y_1 < y_2 < \cdots < y_{j-1} < y_j < \cdots < y_m=d \] be a partition of \([c,d]\) into subintervals each of width \(\Delta y=(d-c)/m\). The vertical lines \(x=x_i\) for \(0\leq i \leq n\) and the horizontal lines \(y=y_j\) for \(0\leq j \leq m\) form subrectangles which partition \(R\) as shown in \(\ref{2dpartition}\). Since each of the subrectangles have the same area, namely \(\Delta A=\Delta x \, \Delta y\) we call this type of partition \(\mathcal{P}\) a regular partition of the rectangle \(R\).
For example, let \(R=[-1,2]\times[-2,2]\). In \(\ref{fig:part1}\) we have sketched the graph of the regular partition of \(R\) with \(n=6\) subintervals of \([-1,2]\) and \(m=8\) subintervals of \([-2,2]\). For this partition we have \(\Delta x=1/2\) and \(\Delta y=1/2\), making the area for each subrectangle \(\Delta A =1/4\).
If we increase the number of subintervals of both \([a,b]\) and \([c,d]\), then the number of subrectangles of \(R\) increases. This process is illustrated in \(\ref{fig:part234}\).
Now we assume that we have a function \(f\) of two variables whose domain contains a rectangle \(R\) and that we have a regular partition \(\mathcal{P}\) of \(R\) as described above. From each subrectangle in \(\mathcal{P}\) we choose a representative point \(\left(x_{ij}^*,y_{ij}^*\right)\) and form the sum,
\[\begin{equation} \label{dbrmsum} \sum _{i=1}^n \sum _{j=1}^m f\left(x_{ij}^*,y_{ij}^*\right)\Delta A \end{equation}\] where \(\Delta A\) is the area of each subrectangle. A summation of this form is called a Riemann sum of \(f\) with respect to the partition \(\mathcal{P}\) and the subrectangle representatives \((x_{ij}^*,y_{ij}^*).\) Notice that a regular partition of \(R\) is completely determined by \(n\) and \(m\).
The next three examples are an illustration of how Riemann sums can be used to estimate the volume of the 3-dimensional solid region over the \(xy\)-plane bound below by \(R\) and above by the surface \(z=f(x,y)\) (see \(\ref{fig:surboundvolfunc}\)).
Example 5.18 Consider the surface and the rectangle \[\begin{equation} z=5 - \frac{1}{4} x^2 - \frac{1}{5} y^2 \qquad R=\{(x,y)\mid 0\leq x \leq 2, 0\leq y\leq 3\}\end{equation}\] Using the lower left corners as subrectangle representatives, determine the Riemann sums when \(n=2\) and \(m=3\). :::
Solution. When \(n=2\) and \(m=3\) we have \[ \Delta x=\frac{2-0}{2}=1, \qquad \Delta y=\frac{3-0}{2}=1, \qquad \text{and} \qquad \Delta A=\Delta x \Delta y =1. \] We are choosing the lower left corners as subrectangle representatives (using \(0<1<2\) and \(0<1<2<3\)), that is, \[\begin{align*} \left(x_{11}^*,y_{11}^*\right)=(0,0) \qquad \left(x_{12}^*,y_{12}^*\right)=(0,1) \qquad \left(x_{13}^*,y_{13}^*\right)=(0,2) \\[5pt] \left(x_{21}^*,y_{21}^*\right)=(1,0) \qquad \left(x_{22}^*,y_{22}^*\right)=(1,1) \qquad \left(x_{24}^*,y_{23}^*\right)=(1,2) \end{align*}\] as depicted in \(\ref{figex:surboundvolone}\). Now we find the Riemann sum for this partition and chosen representatives: \[\begin{align*} & \sum _{i=1}^2 \sum _{j=1}^3 f\left(x_{ij}^*,y_{ij}^*\right)\Delta A = \sum _{i=1}^2 \sum _{j=1}^3 f\left(x_{ij}^*,y_{ij}^*\right) \\[5pt] & = \sum _{i=1}^2 \left[ f\left(x_{i1}^*,y_{i1}^*\right) + f\left(x_{i2}^*,y_{i2}^*\right)+ f\left(x_{i3}^*,y_{i3}^*\right)\right] \\[5pt] & = f\left(x_{11}^*,y_{11}^*\right) + f\left(x_{12}^*,y_{12}^*\right)+ f\left(x_{13}^*,y_{13}^*\right) + f\left(x_{21}^*,y_{21}^*\right) + f\left(x_{22}^*,y_{22}^*\right) + f\left(x_{23}^*,y_{23}^*\right) \\[5pt] & = f(0,0) + f(0,1)+ f(0,2)+f(1,0) + f(1,1) + f(1,2) \\[5pt] & = 5+\frac{24}{5}+\frac{21}{5}+\frac{19}{4}+\frac{91}{20}+\frac{79}{20} = \frac{109}{4}=27.25. \end{align*}\]
Example 5.10 Consider the surface and the rectangle \[\begin{equation} z=5 - \frac{1}{4} x^2 - \frac{1}{5} y^2 \qquad R=\{(x,y)\mid 0\leq x \leq 2, 0\leq y\leq 3\}\end{equation}\] Using the lower left corners as subrectangle representatives, determine the Riemann sums when \(n=4\) and \(m=6\). :::
Solution. When \(n=4\) and \(m=6\) we have \[ \Delta x=\frac{2-0}{4}=\frac{1}{2}, \qquad \Delta y=\frac{3-0}{6}=\frac{1}{2}, \qquad \text{and} \qquad \Delta A=\Delta x \Delta y =\frac{1}{4} \] Again using lower left corners as subrectangle representatives (see \(\ref{figex:surboundvoltwo}\)) we find the Riemann sum for this partition to be: \[\begin{align*} & \sum _{i=1}^4 \sum _{j=1}^6 f\left(x_{ij}^*,y_{ij}^*\right)\Delta A = \frac{1}{4}\left[\sum _{i=1}^4 \sum _{j=1}^6 f\left(x_{ij}^*,y_{ij}^*\right)\right] \\[5pt] & \qquad = \frac{1}{4} \left[ f\left(0,0\right) + f\left(0,\frac{1}{2}\right) + f\left(0,1\right) + f\left(0,\frac{3}{2}\right) + f\left(0,2\right) + f\left(0,\frac{5}{2}\right) \right. \\[5pt] &\qquad\qquad + f\left(\frac{1}{2},0\right) + f\left(\frac{1}{2},\frac{1}{2}\right) + f\left(\frac{1}{2},1\right) + f\left(\frac{1}{2},\frac{3}{2}\right) + f\left(\frac{1}{2},2\right) + f\left(\frac{1}{2},\frac{5}{2}\right) \\[5pt] &\qquad\qquad\qquad + f\left(0,1\right) + f\left(1,\frac{1}{2}\right) + f\left(1,1\right) + f\left(1,\frac{3}{2}\right) + f\left(1,2\right) + f\left(1,\frac{5}{2}\right) \\[5pt] & \qquad\qquad\qquad\qquad \left. +f\left(\frac{3}{2},0\right) +f\left(\frac{3}{2},\frac{1}{2}\right) +f\left(\frac{3}{2},1\right) +f\left(\frac{3}{2},\frac{3}{2}\right) +f\left(\frac{3}{2},2\right) +f\left(\frac{3}{2},\frac{5}{2}\right) \right] \\[5pt] & \qquad = \left(\frac{1}{4}\right)\frac{415}{4}=25.9375. \end{align*}\]
Example 5.2 Consider the surface and the rectangle \[\begin{equation} z=5 - \frac{1}{4} x^2 - \frac{1}{5} y^2 \qquad R=\{(x,y)\mid 0\leq x \leq 2, 0\leq y\leq 3\}\end{equation}\] Using the lower left corners as subrectangle representatives, determine the Riemann sums when \(n=6\) and \(m=12\). :::
Solution. When \(n=6\) and \(m=12\) we have \[ \Delta x=\frac{2-0}{6}=\frac{1}{3}, \qquad \Delta y=\frac{3-0}{12}=\frac{1}{4}, \qquad \text{and} \qquad \Delta A=\Delta x \Delta y =\frac{1}{12} \] Using lower left corners as subrectangle representatives we find the Riemann sum for this partition to be \[\begin{align*} & \sum _{i=1}^6 \sum _{j=1}^{12} f\left(x_{ij}^*,y_{ij}^*\right)\Delta A = \frac{1}{12}\left[\sum _{i=1}^6 \sum _{j=1}^{12} f\left(x_{ij}^*,y_{ij}^*\right)\right] \\[5pt] &\qquad = \frac{1}{12} \left[ f\left(0,0\right) + \cdots +f\left(0,\frac{11}{4}\right) + f\left(\frac{1}{3},0\right) + \cdots +f\left(\frac{1}{3},\frac{11}{4}\right) \right. \\[5pt] & \qquad \qquad\qquad\left. + \cdots + f\left(\frac{4}{3},0\right) + \cdots +f\left(\frac{4}{3},\frac{11}{4}\right) + f\left(\frac{5}{3},0\right) + \cdots +f\left(\frac{5}{3},\frac{11}{4}\right) \right] \\[5pt] & \qquad= \frac{1}{12}\left(\frac{18223}{60}\right)=\frac{14923}{60} =25.3097. \end{align*}\] As depicted in \(\ref{figex:surboundvolthree}\), the volume of the solid region that lies directly above \(R\) and below the surface \(z=f(x,y)\) is approximately \(25.3097\).
Definition 5.1 Let \(f\) be defined on the rectangle \(R\) and suppose that \(f(x,y)\geq 0\) on \(R\). Then the volume \(V\) of the solid region that lies directly above \(R\) and below the surface \(z=f(x,y)\) is \[\begin{equation} \label{defvolurie} V=\lim_{m,n \to \infty} \sum_{i=1}^n \sum_{j=1}^m f(x_{ij}^*,y_{ij}^*) \, \Delta A \end{equation}\] if this limit exists.
Midpoint Rule
It can be proven that if \(f\) is a continuous function on \(R\), then the limit in \(\ref{defvolurie}\) always exists, no matter how the subrectangle representatives \((x_{ij}^*, y_{ij}^*)\) are chosen. For example, if \(a=x_0 < x_1 < x_2 < \cdots < x_{i-1} < x_i < \cdots < x_n=b\) is a regular partition of \([a,b]\) and \(c=y_0 < y_1 < y_2 < \cdots < y_{j-1} < y_j < \cdots < y_m=d\) is a regular partition of \([c,d]\), then we can use midpoints, namely, from each subrectangle \([x_i,x_{i+1}]\times [y_j, y_{j+1}]\) we choose \[ (x_{ij}^*, y_{ij}^*) = \left( \frac{\overbrace{ (a+\Delta x (i-1))}^{\text{left endpoint}}\, + \overbrace{(a+(\Delta x) i)}^{\text{right endpoint}}}{2} , \frac{ \overbrace{(c+\Delta y (j-1))}^{\text{lower endpoint}}+\overbrace{(c+(\Delta y) j)}^{\text{upper endpoint}}}{2} \right) \] as subrectangle representatives. This is called the midpoint rule .
Example 5.1 Consider the surface and the rectangle \[\begin{equation} z=2+\frac{1}{2}x^2+\frac{1}{3}y^2 \qquad R=\{(x,y)\mid -1\leq x \leq 2, -3\leq y\leq 4\}\end{equation}\] Use the midpoint rule to find the Riemann sum determined by the regular partition of \(R\) with \(n=12\) and \(m=28\) to estimate the volume of the solid that lies under the graph of the surface \(z=f(x,y)\) and directly above the rectangle \(R\).
Solution. When \(n=12\) and \(m=28\) we have \[ \Delta x=\frac{2-(-1)}{12}=\frac{1}{4}, \quad \Delta y=\frac{4-(-3)}{28}=\frac{1}{4}, \quad \text{and} \quad \Delta A=\Delta x \Delta y =\frac{1}{16}. \] The subrectangle representations are: \[\begin{align*} (x_{ij}^*, y_{ij}^*) & = \left( \frac{(-1+\frac{1}{4}(i-1))+(-1+\frac{1}{4} i)}{2}, \frac{-3+\frac{1}{4}(j-1))+(-3+\frac{1}{4} j)}{2} \right) \\ & =\left( -\frac{9}{8}+\frac{1}{4}i, -\frac{25}{8}+\frac{1}{4} j \right) \end{align*}\] Using midpoints we find the Riemann sum for this partition to be: \[\begin{align*} & \sum _{i=1}^{12} \sum _{j=1}^{28} f\left(x_{ij}^*,y_{ij}^*\right)\Delta A = \frac{1}{16}\left[\sum _{i=1}^{12} \sum _{j=1}^{28} f\left(-\frac{9}{8}+\frac{1}{4}i, -\frac{25}{8}+\frac{1}{4} j \right)\right] \\[5pt] & \qquad = \frac{1}{16} \left[ f\left(-\frac{7}{8},-\frac{23}{8}\right) +f\left(-\frac{7}{8},-\frac{21}{8}\right) + f\left(-\frac{7}{8},-\frac{19}{8}\right) \right. \\[5pt] & \qquad \qquad \qquad \qquad \left.+\cdots +f\left(\frac{15}{8},\frac{27}{8}\right) + f\left(\frac{15}{8},\frac{29}{8}\right) +f\left(\frac{15}{8},\frac{31}{8}\right) \right] \\[5pt] & \qquad = \frac{1}{16}\left(\frac{10591}{8}\right)=82.7421875. \end{align*}\]
Definition of Double Integral
In our previous examples, we used regular partitions to form Riemann sums for functions defined over rectangular regions. This is not necessity. To measure the size of the rectangles in the partition \(P,\) we define the norm \(\|P\|\) of the partition to be the length of the longest diagonal of any of the subrectangles in the partition. Refine a partition \(P\) by subdividing the cells in such a way that the norm decreases. When this process is applied to the Riemann sum and the norm decreases to zero we have the double integral of \(f\) over \(R\).
Definition 5.2 If a function \(f\) is defined on a closed, bounded rectangular region \(R\) in the \(x y\)-plane, then the double integral of \(f\) over \(R\) is defined by \[ \iint_R f(x,y)d A=\lim _{\|P\|\to 0}\sum _{k=1}^N f\left(x_k^*,y_k^*\right)\triangle A_k \] provided this limit exists, in which case, \(f\) is said to be integrable over \(R.\)
Let \(f(x,y)\) be a function that is continuous on the region \(D\) that can be contained in a rectangle \(R.\) Define the function \(F(x,y)\) on \(R\) as \(f(x,y)\) if \((x,y)\) is in \(D\) and 0 otherwise. If \(F\) is integrable over \(R\), we say that \(f\) is integrable over \(D\), and the double integral of \(f\) over \(D\) is defined as \[\begin{equation} \label{leftrightint} \iint_D f(x,y)d A=\iint_R F(x,y)d A. \end{equation}\]
The function \(F(x,y)\) may have discontinuities on the boundary of \(D,\) but if \(f(x,y)\) is continuous on \(D\) and the boundary of \(D\) is fairly ``well behaved”, then it can be shown that the double integral of the right side of \(\ref{leftrightint}\) exists and hence that the double integral of the left side of \(\ref{leftrightint}\) exists.
Properties of Double Integrals
Theorem 5.1 Assume that all the given integrals exist on a rectangular region \(R.\)
- For constants \(a\) and \(b\), \[ \iint_R [a f(x,y)+b g(x,y)]\, dA =a \iint_R f(x,y) \, d A+b\iint_R g(x,y) \, dA. \]
- If \(f(x,y)\geq g(x,y)\) throughout a rectangular region \(R,\) then \[ \iint_R f(x,y)d A\geq \iint_R g(x,y)d A. \]
- If the rectangular region of integration \(R\) is subdivided into two (disjoint) subrectangles \(R_1\)and \(R_2,\) then \[ \iint_R f(x,y)d A= \iint_{R_1} f(x,y)d A+\iint_{R_2} f(x,y)d A. \]
Exercises
Exercise 5.1 If \(f\) is a constant function, say \(f(x,y)=k\) and \(R = [a,b] \times [c,d]\) show that \[ \iint_R k dA=k(b-a)(d-c). \]
Exercise 5.2 Approximate the given double integral by dividing the rectangle \(R\) with vertices \((0,0)\), \((4,0)\), \((4,2)\), and \((0,2)\) into eight squares and find the sum.
- \(\iint_R (x+y) \, dA\)
- \(\iint_R x y \, dA\)
- \(\iint_R (x^2+y^2) \, dA\)
- \(\iint_R\frac{1}{(x+1)(y+1)} \, dA\)
Exercise 5.3 Find an approximation for the volume \(V\) of the solid lying under the graph of the elliptic paraboloid \(z=8-2x^2-y^2\) and above the rectangular region \(R=[0,1]\times [0,2]\). Use a regular partition \(\mathcal{P}\) of \(R\) with \(m=n=2\), and choose the subrectangular representation point \((x_{ij}^*,y_{ij}^*)\) as indicated.
- The lower left-hand corner of \(R_{ij}\).
- The upper left-hand corner of \(R_{ij}\).
- The lower right-hand corner of \(R_{ij}\).
- The upper right-hand corner of \(R_{ij}\).
- The center of \(R_{ij}\).
Exercise 5.4 Calculate the double Riemann sum of \(f\) for the partition of \(R\) given by the indicated lines and the given choice of \((x_{ij}^*,y_{ij}^*)\). Also calculate the norm of the partition.
- \(f(x,y)=x^2+4y\), \(R=[0,2]\times[0,3]\), \(x=1, y=1, y=2\); \((x_{ij}^*,y_{ij}^*)\) is the upper right corner of \(R_{ij}\)
- \(f(x,y)=x^2+4y\), \(R=[0,2]\times[0,3]\), \(x=1, y=1, y=2\); \((x_{ij}^*,y_{ij}^*)\) is the center of \(R_{ij}\)
- \(f(x,y)=xy-y^2\), \(R=[0,5]\times[0,4]\), \(x=1, x=2, x=3, x=4, y=2\); \((x_{ij}^*,y_{ij}^*)\) is the center of \(R_{ij}\)
- \(f(x,y)=2x+x^2y\), \(R=[-2,2]\times[-1,1]\), \(x=-1, x=0, x=1, y=-1/2, y=0, y=1/2\); \((x_{ij}^*,y_{ij}^*)\) is the lower left corner of \(R_{ij}\)
- \(f(x,y)=x^2-y^2\), \(R=[0,5]\times[0,2]\), \(x=1, x=3, x=4, y=1/2, y=1\); \((x_{ij}^*,y_{ij}^*)\) is the upper left corner of \(R_{ij}\)
- \(f(x,y)=5xy^2\), \(R=[1,3]\times[1,4]\), \(x=1.8, x=2.5, y=2, y=3\); \((x_{ij}^*,y_{ij}^*)\) is the lower right corner of \(R_{ij}\)
Exercise 5.5 Find the Riemann sum \[ \sum_{i=1}^m\sum_{j=1}^n f(x_{ij}^*,y_{ij}^*) \Delta A \] of \(f\) over the region \(R\) with respect to the regular partition \(\mathcal{P}\) with the indicated values of \(m\) and \(n\).
- \(f(x,y)=2x+3y\), \(R=[0,1]\times[0,3]\), \(m=2, n=3\); \((x_{ij}^*,y_{ij}^*)\) is the lower left-hand corner of \(R_{ij}\)
- \(f(x,y)=x^2-2y\), \(R=[1,5]\times[1,3]\), \(m=4, n=2\); \((x_{ij}^*,y_{ij}^*)\) is the upper right-hand corner of \(R_{ij}\)
- \(f(x,y)=x^2+2y^2\), \(R=[-1,3]\times[0,4]\), \(m=4, n=2\); \((x_{ij}^*,y_{ij}^*)\) is the upper right-hand corner of \(R_{ij}\)
- \(f(x,y)=2xy\), \(R=[-1,1]\times[-2,2]\), \(m=4, n=4\); \((x_{ij}^*,y_{ij}^*)\) is the center of \(R_{ij}\)
Exercise 5.6 Evaluate the integral \(\iint_R (4-2y)\, dA\), where \(R=[0,1]\times [0,1]\), by identifying it as the volume of a solid.
Exercise 5.7 The integral \[ \iint_R \sqrt{9-y^2} \, dA, \] where \(R=[0,4]\times[0,2]\), represents the volume of a solid. Sketch the solid.
Exercise 5.8 If \(f\) is a constant function, \(f(x,y)=k\), and \(R=[a,b]\times [c,d]\), show that \[ \iint_R k \, dA =k(b-a)(d-c). \]
Exercise 5.9 If \(R=[0,1]\times [0,1]\), show that \[ 0\leq \iint_R \sin(x+y) \, dA \leq 1. \]
Exercise 5.10 Evaluate \(\iint_{R} y e^{xy} \, dA\) where \(R=[0,2]\times [0,1]\).
Iterate Integrals Over Rectangular Regions
::: {#thm- } [Fubini’s Theorem for Rectangular Regions] If \(f\) is a continuous function of \(x\) and \(y\) over the rectangle \(R:\) \(a\leq x\leq b,\) \(c\leq y\leq d,\) then the double integral of \(f\) over \(R\) may be evaluated by either iterated integral: \[ \iint_R f(x,y)d A=\int _a^b \left[\int_c^d f(x,y) \, dy\right]dx=\int _c^d \left[\int_a^b f(x,y) \, dx\right]dy. \]
Example 5.3 Evaluate the double integral \[\begin{equation*} \iint_R x \sin x y \,dA \end{equation*}\] over the region \(R=\{ (x,y) \mid 0\leq x\leq \pi ,\) \(0\leq y\leq 1\}\).
Solution. Since the integrand is continuous over \(R\) we use \(\ref{fubrect}\) to find \[\begin{equation} \iint_R x \sin x y \, dA =\int _0^{\pi }\int _0^1 x \sin x y \, dy \, dx =-\int_0^{\pi } (\cos x-1) \, dx=\pi . \end{equation}\] See \(\ref{figfub:surboundvolone}\) for a visualization.
Example 5.4 Evaluate the double integral \[\begin{equation*} \iint_R \left(\frac{4+x^2}{1+y^2}\right)d A \end{equation*}\] over the region \(R=\{ (x,y) \mid 0\leq x\leq 1 ,\) \(0\leq y\leq 1\}\).
Solution. Since the integrand is continuous over \(R\) we use \(\ref{fubrect}\) to find \[\begin{equation} \iint_R \left(\frac{4+x^2}{1+y^2}\right)d A =\int _0^1\int _0^1\left(\frac{4+x^2}{1+y^2}\right)dx dy =\int_0^1 \frac{13}{3 \left(y^2+1\right)} \, dy =\frac{13 \pi }{12}. \end{equation}\] See \(\ref{figfub:surboundvoltwo}\) for a visualization.
Example 5.5 Evaluate the double integral \[\begin{equation*} \iint_R \frac{2 x y}{x^2+1}dA \end{equation*}\] over the region \(R=\{(x,y) \mid 0\leq x\leq 1, 1\leq y\leq 3\}\).
Solution. Since the integrand is continuous over \(R\) we use \(\ref{fubrect}\) to find \[\begin{equation} \iint_R \frac{2 x y}{x^2+1} \, dA =\int _0^1\int _1^3\frac{2x y}{x^2+1} \, dy \, dx =\int_0^1 \frac{8 x}{x^2+1} \, dx=4 \ln (2). \end{equation}\] See \(\ref{figfub:surboundvolthree}\) for a visualization.
Example 5.6 Evaluate the double integral \[\begin{equation*} \iint_R \left(x^2e^{x y}\right)d A \end{equation*}\] over the region \(R=\{ (x,y) \mid 0\leq x\leq 1 ,\) \(0\leq y\leq 1\}\).
Solution. Since the integrand is continuous over \(R\) we use \(\ref{fubrect}\) to find \[ \iint_R \left(x^2e^{x y}\right)d A =\int _0^1\int _0^1x^2e ^{x y}dy dx =\int_0^1 \left(x e^x-x\right) \, dx=\frac{1}{2}. \] See \(\ref{figfub:surboundvol}\) for a visualization.
Example 5.7 Evaluate the double integral \[\begin{equation*} \int _3^4\int _1^2\frac{x}{x-y}dy\, dx. \end{equation*}\]
Solution. Since the integrand is continuous over \(R\) we use \(\ref{fubrect}\) to find \[ \int _3^4\int _1^2\frac{x}{x-y} \, dy \, dx =\int_3^4 x (\ln (1-x)-\ln (2-x)) \, dx =\frac{1}{2}-10 \ln (2)+\frac{15}{2} \ln (3). \] See \(\ref{figfub:surboundvolfive}\) for a visualization.
Example 5.8 Evaluate the double integral \[\begin{equation*} \int _2^3\int _{-1}^2\frac{1}{(x+y)^2}dy\, dx. \end{equation*}\]
Solution. Since the integrand is continuous over \(R\) we use \(\ref{fubrect}\) to find \[ \int _2^3\int _{-1}^2\frac{1}{(x+y)^2} \, dy \, dx =\int_2^3 \left(\frac{1}{-x-2}-\frac{1}{1-x}\right) \, dx =\ln (2)+\ln (4)-\ln (5) =\ln \left(\frac{8}{5}\right). \] See \(\ref{figfub:surboundvolsix}\) for a visualization.
How does this example not contradiction \(\ref{fubrect}\)?
Example 5.9 Show that the iterated integrals \[\begin{equation*} \int _0^1\int _0^1\frac{y-x}{(x+y)^3} \, dy \, dx \quad \text{ and } \quad \int _0^1\int _0^1\frac{y-x}{(x+y)^3} \, dx \, dy \end{equation*}\] have different values.
Solution. The first integral \[\begin{equation} \int _0^1\int _0^1\frac{y-x}{(x+y)^3} \, dy \, dx =\int_0^1 -\frac{1}{(x+1)^2} \, dx =-\frac{1}{2}. \end{equation}\] and the second integral \[\begin{equation} \int _0^1\int _0^1\frac{y-x}{(x+y)^3} \, dx \, dy =\int_0^1 \frac{1}{(y+1)^2} \, dy =\frac{1}{2}. \end{equation}\] do indeed have different values.
Iterated Integrals Over Non-Rectangular Regions
Let \(R\) be a region in the \(xy\)-plane. Then \(R\) is called a
- vertically simple region if \(R\) can be described by the inequalities \[\begin{equation*} D_1: \quad a\leq x\leq b, \quad g_1(x)\leq y\leq g_2(x) \end{equation*}\] where \(g_1(x)\) and \(g_2(x)\) are continuous functions of \(x\) on \([a,b].\)
- horizontally simple region if \(R\) can be described by the inequalities \[\begin{equation*} D_2: \quad c\leq y\leq d, \quad h_1(y)\leq x\leq h_2(y) \end{equation*}\] where \(h_1(x)\) and \(h_2(x)\) are continuous functions of \(y\) on \([c,d].\)
::: {#thm- } [Fubini’s Theorem for Non-Rectangular Regions] If \(D_1\) is a vertically simple region, then \[ \iint_{D_1} f(x,y) \, d A=\int _a^b \int _{g_1(x)}^{g_2(x)}f(x,y) \, dy \, dx \] whenever both integrals exist. Similarly, for a horizontally region \(D_2,\) \[ \iint_{D_2} f(x,y) \, d A=\int _c^d \int _{h_1(y)}^{h_2(y)}f(x,y)\, dx \, dy\] whenever both integrals exist.
Example 5.11 Evaluate the double integral \[\begin{equation*} \iint_D x y \, d A \end{equation*}\] where \(D\) is the region bounded by the line \(y=x-1\) and the parabola \(y^2=2x+6\).
Solution. The region \(D\) is both a vertically simple and a horizontally simple region, but the description of \(D\) as a vertically simple region is more complicated because the lower boundary consists of two parts. Therefore we express \(D\) as a horizontally simple region: \[
D=\left\{(x,y)\mid-2\leq y\leq 4,\frac{y^2}{2}-3\leq x\leq y+1\right\}.
\] as shown below. Then the double integral becomes
\[
\iint_D x y \, d A
=\int _{-2}^4\int _{(1/2)y^2-3}^{y+1}x y \, dx \,dy
=\frac{1}{2}\int_{-2}^4 \left(-\frac{y^5}{4}+4 y^3+2 y^2-8y\right) \, dy
=36.
\] If we had expressed \(D\) as a vertically simple region, then we would have obtained
\[
\iint_D x y \, d A=\int _{-3}^{-1}\int _{-\sqrt{2x+6}}^{\sqrt{2x+6}}x y \, dy \, dx+\int _{-1}^5\int _{x-1}^{\sqrt{2x+6}}x y \, dy \, dx=36.
\] See \(\ref{figreg:surboundvol}\) for a visualization.
Example 5.12 Evaluate the iterated integral \[ \int _0^1\int _{x^2}^{\sqrt{x}}x y^2 \, dy \, dx. \]
Solution. We express the region of integration \(D\) as a vertically simple region: \[ D=\left\{(x,y)\mid 0\leq x\leq 1, x^2 \leq y \leq \sqrt{x} \right\}. \] We use \(\ref{fubthmrect}\) to obtain \[ \int _0^1\int _{x^2}^{\sqrt{x}}x y^2 \, dy \, dx =\int _0^1x \left(\frac{x^{3/2}}{3}-\frac{x^6}{3}\right) \, dx =\frac{3}{56}. \] See \(\ref{figreg:surboundvoltwo}\) for a visualization.
Example 5.13 Evaluate the iterated integral \[ \int _0^{2\sqrt{3}}\int _{\left.y^2\right/6}^{\sqrt{16-y^2}} \, dx \, dy. \]
Solution. We express the region of integration \(D\) as a horizontally simple region: \[ D=\left\{(x,y)\mid 0\leq y\leq 2\sqrt{3}, \frac{y^2}{6} \leq x \leq \sqrt{16-y^2} \right\}. \] We use \(\ref{fubthmrect}\) to obtain \[ \int _0^{2\sqrt{3}}\int _{\left.y^2\right/6}^{\sqrt{16-y^2}} \, dx \, dy =\int _0^{2\sqrt{3}}\left(\sqrt{16-y^2}-\frac{y^2}{6}\right) \, dy =\left(\frac{1}{18} \left(12 \sqrt{3}+48 \pi \right) \right). \] See \(\ref{figreg:surboundvolthree}\) for a visualization.
Example 5.14 Evaluate the iterated integral \[ \int _0^1\int _{-x^2}^{x^2} \, dy \, dx. \]
Solution. We express the region of integration \(D\) as a vertically simple region: \[ D=\left\{(x,y)\mid 0\leq x\leq 1, -x^2 \leq y \leq x^2 \right\}. \] The region of integration is sketched in \(\ref{fig:surboundvolfour}\). We find \[\begin{align*} & \int _0^1\int _{-x^2}^{x^2} \, dy \, dx =\int _0^12 x^2 \, dx =\frac{2}{3}. \end{align*}\]
Example 5.15 Sketch the region of integration and evaluate \[\begin{equation*} \int _{-2}^1\int _{y^2+4y}^{3y+2} \, dx \, dy. \end{equation*}\]
Solution. The region is horizontally simple region as \(-2\leq y\leq 1\) and \(y^2+4y\leq x\leq 3y+2.\) We find \[ \int _{-2}^1\int_{y^2+4y}^{3y+2} dx \, dy=\int_{-2}^1 \left(-y^2-y+2\right) \, dy=\frac{9}{2}. \] This region could also be considered as a vertically simple region as \(-4\leq x\leq 5\) and \(\frac{x-2}{3}\leq y\leq -2+\sqrt{x+4}.\) We find \[ \int _{-4}^5\int_{(x-2)/3}^{-2+\sqrt{x+4}} dy \, dx=\int_{-4}^5 \left(\left(-2+\sqrt{x+4}\right)-\left(\frac{x-2}{3}\right)\right) \, dx =\frac{9}{2}. \] See \(\ref{figreg:surboundvolfive}\) for a visualization.
Example 5.16 Sketch the region of integration and evaluate \[\begin{equation*} \int _0^{\pi /2}\int _0^{\sin x}e^y \cos x \, dy \, dx. \end{equation*}\]
Solution. The region is vertically simple region as \(0\leq x\leq \frac{\pi }{2}\) and \(0\leq y\leq \sin x.\) We find
\[
\int _0^{\pi /2}\int _0^{\sin x}e^y \cos x \, dy \, dx=\int_0^{\pi /2} \left(e^{\sin x}\cos x-\cos x\right) \, dx=e-2.\] The region is also horizontally simple region as \(0\leq y\leq 1\) and \(\text{Arcsin}(y)\leq x\leq \frac{\pi }{2}.\) We find \[
\int _0^1\int _{\arcsin y }^{\pi /2}e^y \cos x \, dx \, dy=e-2.
\] See \(\ref{figreg:surboundvolsix}\) for a visualization.
Example 5.17 Sketch the region of integration and evaluate the iterated integral \[\begin{equation*} \int _0^1\int _x^1\sin \left(y^2\right) \, dy \, dx. \end{equation*}\]
Solution. If we try to evaluate the integral as it stands, we are faced with the task of first evaluating \[\begin{equation}
\int \sin y^2 \,dy.
\end{equation}\] We elect to change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. We have
\[
\int _0^1\int _x^1\sin y^2 \, dy \, dx=\iint_D \sin y^2 \, d A
\] where \(D=\{(x,y) \mid 0\leq x\leq 1,x\leq y\leq 1\}.\) This region has an alternate description: \[
D=\{(x,y) \mid 0\leq y\leq 1,0\leq x\leq y\}.
\] Thus we can express the double integral as an iterated integral in the reverse order: \[
%\int _0^1\int _x^1\sin y^2 \, dy \, dx
%=
\iint_D \sin y^2 \, d A
=\int _0^1\int _0^y\sin y^2 \, dx \, dy
=\int _0^1\left.\left[x \sin y^2\right]\right|_{0}^{y}dy
=\frac{1}{2}(1-\cos 1).
\]
Volume as a Double Integral
::: {#thm- } Volume as a Double Integral If \(f(x,y)\geq 0\) on the rectangular region \(R,\) then the product \(f\left(x_k^*,y_k^*\right)\text{$\triangle $A}_k\) is the volume of a parallelepiped (a box) with height \(f\left(x_k^*,y_k^*\right)\) and base area \(\text{$\triangle $A}_k.\) The Riemann sum \[\begin{equation}
\sum _{k=1}^N f\left(x_k^*,y_k^*\right)\text{$\triangle $A}_k
\end{equation}\] provides an estimate of the total volume under the surface \(z=f(x,y)\) over \(R,\) and if \(f\) is continuous, we expect the approximation to improve by using more refined partitions. That is, the volume under \(z=f(x,y)\) over the domain \(R\) is given by
\[\begin{equation}
\lim _{\|P\|\to 0}\sum _{k=1}^N f\left(x_k^*,y_k^*\right)\triangle A_k=\iint_R f(x,y)d A
\end{equation}\] when \(f(x,y)\geq 0\) on the rectangular region \(R.\)
Example 5.19 Find the volume of the bounded solid that lies inside both the cylinder \(x^2+y^2=3\) and the sphere \(x^2+y^2+z^2=7\).
Solution. The volume is given by
\[
V=4\int _0^{\sqrt{3}}\int _0^{\sqrt{3-y^2}}\sqrt{7-x^2-y^2} \, dx \, dy\] or also \[
V=4\int _0^{\sqrt{3}}\int _0^{\sqrt{3-x^2}}\sqrt{7-x^2-y^2} \, dy \, dx\approx 22.0336.
\]
Example 5.20 Find the volume of the solid region bounded below by the given rectangle in the \(x y\)-plane and above by the graph of the given surface \[ f(x,y)=\frac{x y}{\sqrt{x^2+y^2+1}} \] on \(0\leq x\leq 1, 0\leq y\leq 1\)
Solution. The volume is given by
\[\begin{align*}
& \int _0^1\int _0^1\frac{x y}{\sqrt{x^2+y^2+1}} \, dy \, dx
=\int_0^1 x \left(\sqrt{x^2+2}-\sqrt{x^2+1}\right) \, dx \\
& \qquad =\frac{1-2 \sqrt{2}}{3}+\frac{-2 \sqrt{2}+3 \sqrt{3}}{3}
=\frac{1-4 \sqrt{2}+3 \sqrt{3}}{3}
\approx 0.179766
\end{align*}\]
Example 5.21 Find the volume of the solid region bounded below by the given rectangle in the \(x y\)-plane and above by the graph of the surface \[ f(x,y)=(x+y)^5 \] on \(0\leq x\leq 1, 0\leq y\leq 1\).
Solution. The volume is \[\begin{equation} \int _0^1\int _0^1(x+y)^5dy\, dx =\int_0^1 \left(\frac{1}{6} (x+1)^6-\frac{x^6}{6}\right) \, dx =3. \end{equation}\]
Example 5.22 Find the volume of the bounded solid between the two elliptic paraboloids \[ z=\left.x^2\right/(9+y^2-4) \quad \text{and}\quad z=\left.-x^2\right/(9-y^2+4). \]
Solution. Using symmetry, the volume is given by
\[\begin{align*}
V& =8\int _0^6\int _0^{\sqrt{4-\left.x^2\right/9}}\int _0^{4-\left.x^2\right/9-y^2} \, dz \, dy \, dx
\\
& =8\int _0^6\int _0^{\sqrt{4-\left.x^2\right/9}}\left(-\frac{x^2}{9}-y^2+4\right) \, dy \, dx
\\
& =8\int _0^6\left(\frac{8}{3} \sqrt{4-\frac{x^2}{9}}-\frac{2}{27} x^2 \sqrt{4-\frac{x^2}{9}}\right) \, dx
= 48
\approx 150.796.
\end{align*}\]
Example 5.23 Find the volume of the bounded solid bounded below by the rectangle \(R: 1\leq x\leq 2, 1\leq y\leq 2\)
in the \(x y\)-plane and above by the graph of \[\begin{equation}
z=f(x,y)=\frac{x}{y}+\frac{y}{x}.
\end{equation}\]
Solution. The volume is given by the following double integral and can be compute using Fubini’s theorem.
\[\begin{align*}
\iint_R \left(\frac{x}{y}+\frac{y}{x}\right) \, dA
& =\int _0^2\int _1^2\left(\frac{x}{y}+\frac{y}{x}\right) \, dy \, dx
\\
& =\int _0^2 \left.\left[x \ln |y|+\frac{y^2}{2x}\right]\, \right|^2_1 dx
\\
& =\left.\left[\frac{3}{2}\ln |x|\frac{+1}{2}(\ln 2)x^2\right]\, \right|^2_1
= \text{3 ln 2}.
\end{align*}\]
5.2 Exercises
Exercise 5.11 Sketch the region of integration and evaluate \(\int _{-2}^1\int _{y^2+4y}^{3y+2} \, dx \, dy\).
Exercise 5.12 Evaluate \(\iint_R (x^2+y^2)^{3/2} \, dA\) where \(R\) is the unit circle centered at \((0,0)\).
Exercise 5.13 Evaluate the following iterated integrals.
- \(\int _0^2\int _0^1\left(x^2+x y+y^2\right)dy\, dx.\)
- \(\int _1^2\int _0^{\pi }x \cos y \, dy\, dx.\)
- \(\int _0^{\ln (2)}\int _0^1e^{x+2y}dx\, dy.\)
- \(\int _3^4\int _1^2\frac{x}{x-y}dy\, dx\)
- \(\int _2^3\int _{-1}^2\frac{1}{(x+y)^2}dy\, dx\)
Exercise 5.14 Use iterated integration to compute the double integral of the given rectangular region for each of the following.
- \(R=\{(x,y)\mid1\leq x\leq 2,0\leq y\leq 1\}\)
\[ \iint_R x^2y\, dA \] - \(R=\{(x,y)\mid-1\leq x\leq 0,0\leq y\leq \ln (2)\}\) \[ \iint_R2x e^y\, dA \]
- \(R=\left\{(x,y)\mid 0\leq x\leq \frac{\pi }{4}, \, 0\leq y\leq \frac{\pi }{2}\right\}\)
\[ \iint_R\sin (x+y)\, dA \] - \(R=\{(x,y)\mid0\leq x\leq \pi ,0\leq y\leq 1\}\) \[ \iint_Rx \sin (x y)\, dA \]
- \(R=\{(x,y)\mid0\leq x\leq 1,0\leq y\leq 2\}\) \[ \iint_Rx\sqrt{1-x^2}e^{3y}\, dA \]
- \(R=\{(x,y)\mid0\leq x\leq 1,1\leq y\leq 2\}\) \[ \iint_Rx e^{x y}\, dA \]
Exercise 5.15 Find the volume of the solid bounded below by the given rectangular region in the \(x y\)-plane and above the graph of the given function.
- \(z=2x+3y\),
$ R={(x,y)x,0y} $ - \(z=x \ln ( x y)\), $ R={(x,y)x,1ye} $
- \(z=x \cos y+y \sin x.\) \[ R=\left\{(x,y)\mid0\leq x\leq \frac{\pi }{2}\, ,0\leq y\leq \frac{\pi}{2}\right\} \]
Exercise 5.16 Find the volume of the solid region bounded below by the given rectangle in the \(x y\)-plane and above by the graph of the given surface.
- \(f(x,y)=\frac{x y}{\sqrt{x^2+y^2+1}}\) on \(0\leq x\leq 1, 0\leq y\leq 1\)
- \(f(x,y)=(x+y)^5\) on \(0\leq x\leq 1, 0\leq y\leq 1\)
Exercise 5.17 Sketch the region and evaluate the iterated integral over the non-rectangular region.
- \(\int _0^4\int _0^{4-x}x ydy\, dx.\)
- \(\int _1^e\int _0^{\ln (x)}x ydy\, dx.\)
- \(\int _0^2\int _0^{\sin (x)}y \cos xdy\, dx.\)
- \(\int _{-2}^1\int _{y^2+4y}^{3y+2} dx\, dy.\)
- \(\int _0^{2\sqrt{3}}\int _{\left.y^2\right/6}^{\sqrt{16-y^2}} dx\, dy.\)
- \(\int _0^1\int _{-x^2}^{x^2} dy\, dx.\)
- \(\int _0^4\int _{x^2}^{4x} dy\, dx.\)
- \(\int _0^1\int _{x^2}^{\sqrt{x}}x y^2dy\, dx\)
- \(\int _{-1}^3\int _{\tan ^{-1}x}^{\pi /4}x ydy\, dx.\)
- \(\int _0^4\int _0^{4-x}x ydy\, dx.\)
- \(\int _0^7\int _{x^2-6x}^x \sin x dy\, dx\)
Exercise 5.18 Find the volume of the solid bounded by \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1. \] (Setup but do not evaluate the double integral)
Exercise 5.19 Find the volume of the solid that lies inside both the cylinder \(x^2+y^2=3\) and the sphere \(x^2+y^2+z^2=7\). (Setup but do not evaluate the double integral).
Exercise 5.20 For each of the following sketch the region of integration and write an equivalent integral with the order of integration reversed for
- \(\int _0^1\int _0^{2y}f(x,y)\, dx\, dy.\)
- \(\int _0^1\int _{x^2}^{\sqrt{x}}f(x,y)\, dy\, dx.\)
- \(\int _0^4\int _{y/2}^{\sqrt{y}}f(x,y)\, dx\, dy.\)
- \(\int _1^2\int _{\ln (x)}^2f(x,y)\, dy\, dx.\)
- \(\int _0^3\int _{y/3}^{\sqrt{4-y}}f(x,y)\, dx\, dy.\)
Exercise 5.21 Evaluate the double integral \[ \iint_D x y \, dA \] where \(D\) is the triangular region in the \(x y\)-plane with vertices \((0,0),\) \((1,0),\) and \((4,1).\)
Exercise 5.22 Evaluate the double integral \[ \iint_D \left(x^2-x y-1\right)\, dA \] where \(D\) is the triangular region in the \(x y\)-plane bounded by the lines \(x-2y+2=0,\) \(x+3y-3=0,\) and \(y=0.\)
Exercise 5.23
- Write the iterated integral \(I=\int_0^1 \int_y^1 \sin ( x^2 ) \, dx \, dy\) as a double integral over a domain \(R\).
- Sketch the domain \(R\).
- Evaluate the double integral by reversing the order of integration.
5.3 Double Integrals In Polar Coordinates
The polar conversion formulas are used to convert from rectangular to polar coordinates:
\[\begin{equation} x=r \cos \theta, \quad y=r \sin \theta, \quad r=\sqrt{x^2+y^2}, \quad \tan \theta =\frac{y}{x}. \end{equation}\]
::: {#thm- } [Fubini’s Theorem in Polar Coordinates] If \(f\) is continuous in the polar region \(R\) described by \[ 0\leq r_1(\theta )\leq r\leq r_2(\theta ) \qquad \qquad \alpha \leq \theta \leq \beta \] (with $0-$), then \[\begin{equation} \iint_R f(r,\theta )d A=\int _{\alpha }^{\beta } \int_{r_1(\theta )}^{r_2(\theta )}f(r,\theta )r \, dr \, d\theta . \end{equation}\] :::
Example 5.24 Evaluate \[\begin{equation} \iint_R \left(3x+4y^2\right)\, dA \end{equation}\] where \(R\) is the upper-half plane bounded by the circles \(x^2+y^2=1\) and \(x^2+y^2=4\).
Solution. The region of integration \(R\) is described as \[ R=\left\{(x,y) \mid y\geq 0,1\leq x^2+y^2\leq 4\right\}. \] It is an upper ring with polar coordinates given by \(1\leq r\leq 2,\) \(0\leq \theta \leq \pi .\) Therefore \[\begin{align*} \iint_R\left(3x+4y^2\right) \, d A & =\int_0^{\pi }\int_1^2\left(3r \cos \theta +4r^2\sin ^2 \theta \right) r \, dr \, d\theta \\ & =\int_0^{\pi } \left(7 \cos \theta +15\sin ^2\theta \right)\, d\theta =\frac{15\pi }{2}\end{align*}\] by using the trigonometric identity \(\sin^2 \theta=\frac{1-\cos (2\theta)}{2}\).
Example 5.25 Evaluate \[\begin{equation} \int_0^{\pi /2}\int_1^3 r e^{-r^2} \, dr \, d\theta. \end{equation}\]
Solution. In polar coordinates the region of integration is described as \[ R=\{(r,\theta) \mid 0\leq \theta \leq \frac{\pi}{2}, 1\leq r \leq 3\}. \] Using Fubini’s Theorem in polar coordinates the iterated integral is evaluated as \[\int_0^{\pi /2}\int_1^3 r e^{-r^2}drd\theta =\int_0^{\pi /2} \frac{e^8-1}{2 e^9 } \, d\theta =\frac{\left(e^8-1\right) \pi }{4 e^9}. \]
Example 5.26 Evaluate \[\begin{equation} \int_0^{\pi /2}\int_1^2 \sqrt{4-r^2}r \,dr \, d\theta. \end{equation}\]
Solution. In polar coordinates the region of integration is described as \[ R=\{(r,\theta) \mid 0\leq \theta \leq \frac{\pi}{2}, 1\leq r \leq 2\}.\] Using polar coordinates the iterated integral is evaluated as \[ \int_0^{\pi /2}\int_1^2 \sqrt{4-r^2}r drd\theta =\int_0^{\pi /2} \sqrt{3} \, d\theta =\frac{\sqrt{3} \pi }{2}. \]
Example 5.27 Evaluate \[\begin{equation} \int_0^{\pi }\int_0^4 r^2 \sin ^2\theta \, dr \, d\theta. \end{equation}\]
Solution. In polar coordinates the region of integration is described as \[ R=\{(r,\theta) \mid 0\leq \theta \leq \pi, 0\leq r \leq 4\}. \] Using polar coordinates the iterated integral is evaluated as \[ \int_0^{\pi }\int_0^4 r^2 \sin ^2\theta dr \, d\theta =\int_0^{\pi } \frac{64 \sin ^2\theta }{3} \, d\theta =\frac{32 \pi }{3} \] by using the trigonometric identity \(\sin^2 \theta=\frac{1-\cos (2\theta)}{2}\).
Example 5.28 Evaluate \[\begin{equation} \int_0^{\pi /2}\int_1^3 r^2 \cos ^2\theta \, dr \, d\theta . \end{equation}\]
Solution. In polar coordinates the region is described as \[
R=\{(r,\theta) \mid 0\leq \theta \leq \frac{\pi}{2}, 1\leq r \leq 3\}.
\]
Using polar coordinates the iterated integral is evaluated as \[
\int_0^{\pi /2}\int_1^3 r^2 \cos ^2\theta \, dr \, d\theta =\int_0^{\pi /2} \frac{26 \cos ^2\theta }{3} \, d\theta =\frac{13 \pi }{6}
\] by using the trigonometric identity \(\cos^2 \theta=\frac{1+\cos (2\theta)}{2}\).
Example 5.29 Evaluate \[\begin{equation} \int_0^{\pi }\int_0^{1+\sin \theta } \, dr \, d\theta . \end{equation}\]
Solution. In polar coordinates the region is described as \[ R=\{(r,\theta) \mid 0\leq \theta \leq \pi, 0\leq r \leq 1+\sin \theta\} \] Using polar coordinates the iterated integral is evaluated as \[ \int_0^{\pi }\int_0^{1+\sin \theta } drd\theta =\int_0^{\pi } (\sin \theta +1) \, d\theta =2+\pi. \]
Example 5.30 Evaluate \[\begin{equation} \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\ln \left(x^2+y^2+9\right) \, dy \, dx. \end{equation}\]
Solution. In polar coordinates \[\begin{align*} \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\ln \left(x^2+y^2+9\right) \, dy \, dx & =\int_0^{2\pi }\int_0^3\ln \left(r^2+9\right) rdrd\theta \\ & =\int_0^{2\pi }\left[-\frac{9}{2} (1+\ln (9)-2 \ln (18))\right]d\theta \\ = (-9 \pi (1+\ln (9)-2 \ln (18))). %\approx 73.0473. \end{align*}\]
Example 5.31 Evaluate \[\begin{equation} \int_0^2\int_0^{\sqrt{2x-x^2}}\frac{x-y}{x^2+y^2} \, dy \, dx \end{equation}\]
Solution. In polar coordinates \[\begin{align*} \int_0^2\int_0^{\sqrt{2x-x^2}}\frac{x-y}{x^2+y^2} \, dy \, dx & =\int_0^{\pi /2}\int_0^{2\cos \theta }(\sin \theta -\cos \theta ) \, dr \, d\theta \\ & =\int_0^{\pi /2}2 \cos \theta (\sin \theta -\cos (\theta )) \, d\theta \\ & =\frac{2-\pi }{2}. \approx -0.570796 \end{align*}\]
Example 5.32 Use a double integral to find the area enclosed by one loop of the four leaved rose \(r=\cos 2\theta .\)
Solution. From the sketch of the curve we see that a loop is given by the region \[ R=\left\{(r,\theta )\mid\frac{-\pi }{4}\leq \theta \leq \frac{\pi }{4}\, , 0\leq r\leq \cos 2\theta \right\}. \] So the area is \[ \iint_R \, d A=\int_{-\pi /4}^{\pi /4} \int_0^{\cos 2\theta }r dr \, d\theta =\frac{1}{4}\int_{-\pi /4}^{\pi /4} (1+\cos 4\theta ) \, d\theta =\frac{\pi }{8}. \]
Example 5.33 Find the volume of the bounded solid region bounded below by the rectangle \(R:\) \(0\leq x\leq 1,\) \(0\leq y\leq 1\) in the \(x y\)-plane and above by the graph of \[ z=f(x,y)=\sqrt{x+y}. \]
Solution. The volume is \[\begin{align*} \int_0^1\int_0^1\sqrt{x+y}dy\, dx & =\int_0^1 \left(\frac{2}{3} \left(-x^{3/2}+(1+x)^{3/2}\right)\right) \, dx \\ & =\frac{8}{15} \left(-1+2 \sqrt{2}\right). \end{align*}\]
Example 5.34 Find the volume of the bounded solid region below by the rectangle \(R:\) \(1\leq x\leq 2,\) \(1\leq y\leq e\) in the \(x y-\text{plane}\) and above by the graph of \[ z=f(x,y)=x \ln (x y). \]
Solution. The volume is \[\begin{align*} \int_1^2\int_1^ex \ln (x y)dy\, dx & =\int_1^2 (x (1+(-1+e) \ln x)) \, dx \\ & =-\frac{3}{4} (-3+e)+2(-1+e) \ln 2. \end{align*}\] Note in this example we used the formulas \[ \int \ln udu=u \ln u-u+C, u>0 \quad \text{and} \quad \int u \ln udu=\frac{-1}{4}u^2+\frac{1}{2}u^2\ln u. \]
Example 5.35 Find the volume of the bounded solid region bounded by the plane \(z=0\) and the paraboloid \(z=1-x^2-y^2.\)
Solution. If we put \(z=0\) in the equation of the paraboloid, we get \(x^2+y^2=1.\) This means that the plane intersects the paraboloid in the circle \(x^2+y^2=1,\) so the solid lies under the paraboloid and above the circular disk \(R\) given by \(x^2+y^2\leq 1.\) In polar coordinates the region of integration \(R\) is given by \(0\leq r\leq 1,\) \(0\leq \theta \leq 2\pi\). Since \(1-x^2-y^2=1-r^2,\) the volume is \[ V=\iint_R \left(1-x^2-y^2\right)d A=\int_0^{2\pi}\int_0^1\left(1-r^2\right)r \, dr \, d\theta = \frac{\pi}{4} \int_0^{2\pi} \, d\theta = \frac{\pi }{2}. \]
Example 5.36 Find the volume of the bounded solid region under the paraboloid \(z=x^2+y^2,\) above the \(x y-\text{plane},\) and inside the cylinder \(x^2+y^2=2x\)
Solution. The solid lies above the disk \(R\) whose boundary circle has equation \(x^2+y^2=2x\) or \((x-1)^2+y^2=1.\) In polar coordinates the boundary is \(r^2=2 r \cos \theta\) or \(r=2 \cos \theta .\) Thus the region of integration is \[ R=\left\{(r,\theta )\mid -\frac{\pi }{2}\leq \theta \leq \frac{\pi }{2}, \, 0\leq r\leq 2 \cos \theta \right\} \] and so the volume is \[ V=\iint_R \left(x^2+y^2\right)d A=\int_{-\pi /2}^{\pi /2}\int_0^{2 \cos \theta }r^2rdrd\theta =4\int_{-\pi /2}^{\pi /2}\cos ^4\theta d\theta =\frac{3\pi }{2}\] by using the trigonometric identity \(\cos^4 \theta=\left(\frac{1+\cos(2\theta)}{2}\right)^2\).
Example 5.37 Find the volume of the bounded solid common to the cylinder \(x^2+y^2=2\) and the ellipsoid \(3x^2+3y^2+z^2=7.\)
Solution. In polar coordinates \[\begin{align*} V & =2\int_0^{2\pi }\int_0^{\sqrt{2}}\sqrt{7-3r^2}r \, dr \, d\theta =2\int_0^{2\pi }\left(-\frac{1}{9}+\frac{7 \sqrt{7}}{9}\right)\, d\theta \\ & =\frac{4\pi}{9} \left(-1+7 \sqrt{7}\right). \approx 24.4629. \end{align*}\]
Example 5.38 Find the volume of the solid bounded above by the cone \(z=x^2+y^2,\) below by the plane \(z=0,\) and on both sides by the cylinder \(x^2+y^2=y.\)
Solution. In polar coordinates
\[
V=\int_0^{\pi }\int_0^{\sin \theta }r^3 \, dr \, d\theta
=\int_0^{\pi }\frac{\sin ^4 \theta }{4} \, d\theta
=\frac{3 \pi }{32}
\approx 0.294524.
\]
5.4 Exercises
Exercise 5.24 Evaluate \(\int_0^{\pi }\int_0^{1+\sin \theta } \, dr \, d\theta\).
Exercise 5.25 Sketch the region of integration and then evaluate the iterated integral in polar coordinates.
- \(\int_0^{\pi /2}\int_1^3r e^{-r^2} drd\theta\)
- \(\int_0^{\pi /2}\int_1^2\sqrt{4-r^2} drd\theta\)
- \(\int_0^{\pi }\int_0^4r^2\sin ^2\theta dr \, d\theta\)
- \(\int_0^{\pi /2}\int_1^3r ^2\cos ^2\theta drd\theta\)
- \(\int_0^{\pi}\int_0^{1+\sin \theta } dr \, d\theta\)
Exercise 5.26 Sketch the enclosed region and then use an iterated integral to find the area of the region.
- \(r=2 \cos \theta\)
- \(r=4 \cos 3\theta\)
- \(r=1\) and \(r=2\sin\theta\)
- \(r=1\) and \(r=1+\cos\theta\)
Exercise 5.27 Use polar coordinates to evaluate the iterated integrals.
- \(\int_0^1\int_x^{\sqrt{x}}\left(x^2+y^2\right)^{3/2} dy dx\)
- \(\int_0^2\int_0^{\sqrt{2x-x^2}}\frac{x-y}{x^2+y^2} dy dx.\)
Exercise 5.28 Evaluate the following iterated integrals by converting to polar coordinates.
- \(\int_0^3\int_0^{\sqrt{9-x^2}}x dy dx\)
- \(\int_0^2\int_0^{\sqrt{4-y^2}}e^{x^2+y^2} dx dy\)
- \(\int_0^3\int_0^{\sqrt{9-x^2}}x dy dx\)
- \(\int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\ln \left(x^2+y^2+9\right) dy dx\)
- \(\int_0^2\int_y^{\sqrt{8-y^2}}\frac{1}{\sqrt{1+x^2+y^2}}dx dy\)
- \(\int_0^3\int_0^{\sqrt{9-x^2}}\cos \left(x^2+y^2\right) dy dx\)
- \(\int_0^4\int_0^{\sqrt{4y-y^2}}\frac{1}{\sqrt{x^2+y^2}} dx dy\)
Exercise 5.29 Find the volume of the solid region common to the cylinder \(x^2+y^2=2\) and the ellipsoid \(3x^2+3y^2+z^2=7.\)
Exercise 5.30 Find the volume of the ice cream cone bounded by the hemisphere \(z=\sqrt{8-x^2-y^2}\) and the cone \(z=\sqrt{x^2+y^2}.\)
Exercise 5.31 Find the volume of the solid region bounded above by the cone \(z=x^2+y^2,\) below by the plane \(z=0,\) and on both sides by the cylinder \(x^2+y^2=y.\)
Exercise 5.32 Find the volume of the given solid region common to the cylinder \(x^2+y^2=2\) and the ellipsoid \(3x^2+3y^2+z^2=7.\)
Exercise 5.33 Find the volume of the given solid region bounded above by the cone \(z=x^2+y^2,\) below by the plane \(z=0,\) and on both sides by the cylinder \(x^2+y^2=y.\)
Exercise 5.34 For a constant \(a\) where \(0\leq a\leq R,\) the plane \(z=R-a\) cuts off a \(\texttt{"}\)cap\(\texttt{"}\) from the hemisphere \(z=\sqrt{R^2-x^2-y^2}.\) Use a double integral in polar coordinates to find the volume of the cap. For what values of \(a\) is the volume of the cap half the volume of the hemisphere?
5.5 Applications of Double Integrals
We will consider the following applications: average value of a function over a region, mass of a lamina, electric charge, moments and center of mass, moments of inertia, and probability density functions.
5.6 Average Value
Recall the average value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length if the interval. For an integrable function of two variables defined on a bounded region in the plane, the average value is the integral over the region divided by the area of the region.
Example 5.39 Which do you think will be larger, the average value of \(f(x,y)=xy\) over the square \(0\leq x\leq 1\), \(0\leq y \leq 1\), or over the quarter circle \(x^2+y^2\leq 1\) in the first quadrant?
Solution. The average value of \(f\) over the square is \[ \int_0^1\int_0^1 xy dy dx=\int_0^1\frac{1}{2}x dx=\frac{1}{4}. \] Using polar coordinates, the average value of \(f\) over the quarter circle is \[ \frac{1}{\pi/4}\int_0^{\pi/2}\int_0^1 r^3\cos \theta \sin \theta dr \, d\theta =\frac{4}{\pi}\int_0^{\pi/2} \frac{1}{4}\cos \theta\sin\theta d\theta =\frac{1}{2\pi}. \] Therefore the average value of \(xy\) is larger over the square.
5.7 Mass of a Lamina
A planar lamina is a flat plate that occupies a region \(R\) in the plane that is so thin it can be regarded as two dimensional. If \(m\) is the lamina’s mass and \(A\) is the area of the region \(R,\) then \(\delta =m/A\) is the density of the lamina (in units of mass per unit area). A lamina is called homogeneous if its density \(\delta (x,y)\) is constant over \(R\) and nonhomogeneous if \(\delta (x,y)\) varies from point to point.
Definition 5.3 If \(\delta\) is a continuous density function of the lamina corresponding to a plane region \(R\) then the (total) mass \(m\) of the planar lamina is given by
\[
m=\iint_R\delta (x,y) \, dA.
\]
Example 5.40 Find the mass of the planar lamina occupying the region \(R\) bounded by the parabola \(y=2-x^2\) and the line \(y=x\) if \(\delta (x,y)=x^2.\)
Solution. Begin by drawing the parabola and the line, and by finding their points of intersection \((-2,-2)\) and \((1,1)\). Considering the region \(R\) as vertically simple, the mass of the lamina is \[ m =\iint_R \int_x^{2-x^2}x^2 \, dA =\int^1_{-2}\int_x^{2-x^2}x^2 \, dy \, dx =\int^1_{-2} x^2(2-x^2-x) \, dx =\frac{63}{20}. \]
5.8 Electric Charge
Physicists also consider other types of density that can be treated in the same manner. For example, if an electric charge is distributed over a region \(R\) and the charge density (in units of charge per unit area) is given by \(\delta (x,y)\) at a point \((x,y)\) in \(R,\) then the total charge \(Q\) is given by
\[
Q=\iint_R\delta (x,y) d A.
\]
Example 5.41 Charge is distributed over the triangular region \(R\) described by \((0,1),\) \((1,1),\) and \((1,0)\) so that the charge density at \((x,y)\) is \(\delta (x,y)= x y,\) measured in coulombs per square meter. Find the total charge.
Solution. Considering the region \(R\) as a vertically simple region, we have
\[
Q =\iint_R\delta (x,y) \, d A
=\int _0^1\int _{1-x}^1x y \, dy \, dx
=\frac{1}{2}\int_0^1 \left(2x^2-x^3\right) \, dx
=\frac{5}{24}.
\] Thus the total charge is \(5/24\) C.
5.9 Moments and Center of Mass of a Lamina
The moment of an object about an axis measures the tendency of the object to rotate about that axis. It is defined as the product of the object’s mass and the signed distance from the axis. Let \(\delta\) denote a continuous density function of a body. Then \(M_x\) and \(M_y,\) the first moments about the \(x\)-axis and \(y\)-axis, respectively are \[ M_x=\iint_Ry \delta (x,y) \, dx \, dy \qquad \text{and} \qquad M_y=\iint_Rx \delta (x,y) \, dx \, dy \] The center of mass of the lamina covering \(R\) is the point \((\overline{x},\overline{y})\) where the mass \(m\) can be concentrated without affecting the moments \(M_x\) and \(M_y\); that is \(m \overline{x}=M_y\) and \(m \overline{y}\). If the density \(\delta\) is constant, the point \((\overline{x}, \overline{y})\) is called the centroid of the region.
Example 5.42 A lamina occupies a region \(R\) in the \(xy\)-plane bounded by the parabola \(y=x^2\) and the line \(y=1\). Find the center of mass of the lamina if its mass density at a point \((x,y)\) is directly proportional to the distance between the point and the \(x\)-axis.
Solution. The mass density of the lamina is \(\delta(x,y)=ky\) where \(k\) is a constant. Since the region \(R\) is symmetric with respect to the \(y\)-axis and the density of the lamina is directly proportional to the distance from the \(x\)-axis, we see that the center of mass is located on the \(y\)-axis and so \(\overline{x}=0\). To find \(\overline{y}\) we view \(R\) as a vertically simple region and compute \[ m=\underset{R}{\int \int} \delta d A =\int^1_{-1} \int_{x^2}^1 k y \, dy \, dx =\frac{k}{2}\int_{-1}^1 (1-x^4)\, dx = \frac{4k}{5}. \] We compute \[ \overline{y} =\frac{1}{m} \int^1_{-1} \int_{x^2}^1 k y^2 \, dy \, dx =\frac{5}{12} \int^1_{-1} (1-x^6) \, dx =\frac{5}{7} \] Therefore, the center of mass is \(\left(0,\frac{5}{7}\right)\).
5.10 Probability Density Functions
Quantities that range continuously over an interval of real numbers are called continuous random variables. Every continuous random variable \(X\) has a probability density function \(f\) with the property that the probability of \(X\) lying between the numbers \(a\) and \(b\) is given by the integral
\[
P(a\leq X\leq b)=\int _a^bf(x) \, dx.
\]
In general, \(f(x)\geq 0\) for all \(x,\) and since the value of \(X\) is always some real number, it follows that \(P(-\infty <X<\infty )=1,\) so \(\int _{-\infty }^{\infty } f(x)dx=1.\) In geometric terms, the probability \(P(a\leq X\leq b)\) is the area under the graph of \(f\) over the interval \(a\leq x\leq b.\)
If \(X\) and \(Y\) are both continuous random variables, then the joint probability density function for two random variables \(X\) and \(Y\) is a function of two variables \(f(x,y)\) such that \(f(x,y)\geq 0\) for all \((x,y)\) and \[
P[(X,Y) \text{ in } D]=\underset{D}{\int \int }f(x,y) \, dA
\]
where \(P[(X,Y) \text{ in } D]\) denotes the probability that \((X,Y)\) is in the region \(D.\) Note that \[
P\left[(X,Y) \text{ in } \mathbb{R}^2\right]=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(x,y) \, dx \, dy =1.
\] Geometrically \(P[(X,Y) \text{ in } D]\) may be thought of as the volume under the surface \(z=f(x,y)\) above the region \(D\).
Example 5.43 Suppose the joint probability density function for the random variable \(X\) and \(Y\) is modeled by
\[
f(x,y)=
\begin{cases}
x e^{-x-y} & x\geq 0,y\geq 0 \\
0 & \text{otherwise}.
\end{cases}
\] Find the probability that \(X+Y\leq 1.\)
Solution. The probability that \(X+Y\leq 1\) is given as
\[\begin{align*}
P(X+Y\leq 1)
& =\int _0^1\int _0^{1-x} x e^{-x-y} dy\, dx \\
& =\int_0^1 \left(-x\int _0^{1-x}e^{-x-y} (-1)dy\right) \, dx \\
& =\int_0^1 \left(-x \left.e^{-x-y}\right|^{1-x}_0 \right) \, dx \\
& =\int_0^1 \left[\left(-x e^{-x-(1-x)}\right)-\left(-x e^{-x-0}\right)\right] \, dx \\
& =\int_0^1 \left(-\frac{1}{e}+e^{-x}\right) x \, dx \\
& =-\frac{1}{e}\int_0^1 x \, dx+\int _0^1x e^{-x} dx
\end{align*}\] Let \(u=x\) and \(dv=e^{-x}dx,\) then \(du=dx\) and \(v=-e^{-x}\); and so using integration by parts we find \[\begin{align*}
P(X+Y\leq 1)
&=-\frac{1}{e}\int_0^1 x \, dx-\left.x e^{-x}\right|^1_0 +\int_0^1 e^{-x} \, dx \\
& =-\frac{1}{e}\left.\frac{x^2}{2}\right|^1_0 -x \left.e^{-x}\right|^1_0 -\left.e^{-x}\right|^1_0 \\
& =-\frac{1}{e}\left(\frac{1}{2}\right)- \left(\frac{1}{e}\right)-\left(\frac{1}{e}\right)+1
=1-\frac{5}{2 e}
\approx 0.0803014.
\end{align*}\]
Example 5.44 Suppose the joint probability density function for the random variable \(X\) and \(Y\) is modeled by \[ f(x,y)= \begin{cases} 2e^{-2x-y} & x\geq 0,y\geq 0 \\ 0 & \text{otherwise}. \end{cases} \] Find the probability that \(X+Y\leq 1.\)
Solution. We find \(P(X+Y\leq 1)\) to be \[\begin{align*} & 2\int _0^1\int _0^{1-x} e^{-2x-y} \, dy \, dx =2\int_0^1 \left(-e^{-x-1}+e^{-2 x}\right) \, dx \\ & =\frac{1}{ e^2}-\frac{2-e}{e} \approx 0.399576. \end{align*}\]
Example 5.45 Suppose \(X\) measures the time (in minutes) that a person stands in line at a certain bank and \(Y,\) the duration (in minutes) of a routine transaction at the teller’s window. You arrive at the bank to deposit a check. The joint probability density function for \(X\) and \(Y\) is modeled by
\[
f(x,y)=
\begin{cases}
\frac{1}{8} e^{-x/2-y/4} & x\geq 0,y\geq 0 \\
0 & \text{otherwise}
\end{cases}
\]
Solution. Find the probability that you will complete your business at the bank within 8 minutes.
The probability that you will complete your business at the bank within 8 minutes is \[\begin{align*}
P(X+Y\leq 8)
&
=\frac{1}{8}\int _0^8\int _0^{8-x}e^{-x/2-y/4} \, dy \, dx
\\
& =(-4)\frac{1}{8}\int _0^8\int _0^{8-x}e^{-x/2-y/4}\left(-\frac{1}{4}\right) \, dy \, dx
\\
& =\left(-\frac{1}{2}\right)\int _0^8\left.\left(e^{-x/2-y/4}\right)\right|^{8-x}_0 dx
\\
& =\left(-\frac{1}{2}\right)\int_0^8 \left(e^{-x/2-(8-x)/4}-e^{-x/2}\right) \, dx
\\
& =\left(-\frac{1}{2}\right)\int_0^8 \left(e^{\left(-\frac{x}{4}-2\right)}-e^{-x/2}\right) \, dx
\\
& =\left(-\frac{1}{2}\right)\left((-4)\int_0^8 e^{\left(-\frac{x}{4}-2\right)}\left(\frac{-1}{4}\right) \, dx-(-2)\int_0^8 e^{\frac{-x}{2}}\left(\frac{-1}{2}\right) \, dx\right)
\\
& =\left(-\frac{1}{2}\right)\left(\left.(-4)e^{\left(-\frac{x}{4}-2\right)}\right|^8_0 +\left.2e^{\frac{-x}{2}}\right|^8_0\right)
\\
& =2e^{\left(-\frac{x}{4}-2\right)}\left|^8_0-e^{\frac{-x}{2}}\right|^8_0
\\
& =2e^{\left(-\frac{8}{4}-2\right)}-2e^{\left(-\frac{0}{4}-2\right)}-\left(e^{\frac{-8}{2}}\right)+\left(e^{\frac{-0}{2}}\right)
\\
& =e^{-4}-2e^{-2}+1
=\left(\frac{1}{e^2}-1\right)^2
\approx 0.747645.
\end{align*}\]
5.11 Exercises
Exercise 5.35 Find the centroid for a lamina with \(\delta =4\) over the region bounded by the curve \(y=\sqrt{x}\) and the line \(x=4\) in the first octant.
Exercise 5.36 Find the centroid for a lamina with \(\delta =2\) over the region between the line \(y=2x\) and the parabola \(y=x^2.\)
Exercise 5.37 Use double integration to find the center of mass of a lamina covering the region \(x^2+y^2\leq 9,\) \(y\geq 0\) with density function \(\delta (x,y)=x^2+y^2.\)
Exercise 5.38 Use double integration to find the center of mass of a lamina covering the region bounded by \(y=0,\) \(y=x^2,\) and \(x=6\) with density function \(\delta (x,y)=3x.\)
Exercise 5.39 Use double integration to find the center of mass of a lamina covering the region bounded by \(y=\ln (x),\) \(y=0,\) and \(x=2\) with density function \(\delta (x,y)=1/x.\)
Exercise 5.40 A lamina has the shape of a semicircular region \(x^2+y^2\leq a^2,\) \(y\geq 0.\) Find the center of mass of the lamina if the density at each point is directly proportional to the square of the distance from the point to the origin.
- Find the center of mass of the cardioid \(r=1+\sin \theta\) if the density at each point \((r,\theta )\) is \(\delta (r,\theta )=r.\)
Exercise 5.41 Find the centroid of the loop of the lemniscate \(r^2=2 \sin 2 \theta\) that lies in the first quadrant.
Exercise 5.42 Find the centroid of the part of the large loop of the limacon \(r=1+2 \cos \theta\) that does not include the small loop.
Exercise 5.43 Find the center of mass of the lamina that covers the triangular region with vertices \((0,0),\) \((a,0),\) \((a,b),\) if \(a\) and \(b\) are both positive and the density at \(P(x,y)\) is directly proportional to the distance of \(P\) from the \(y\)-axis.
Exercise 5.44 For each of the following joint probability density functions with the random variables \(X\) and \(Y\) find the indicated probability.
- \(f(x,y)=\begin{cases} 2e^{-2x}e^{-y} & x\geq 0, y\geq 0 \\ 0 & \text{otherwise} \end{cases}\); \(X+Y\leq 1.\)
- \(f(x,y)=\begin{cases} x e^{-x}e^{-y} & x\geq 0, y\geq 0 \\ 0 & \text{otherwise} \end{cases}\); \(X+Y\leq 1.\)
- \(f(x,y)=\begin{cases} \frac{1}{6} e^{-x/2}e^{-y/3} & x\geq 0, y\geq 0 \\ 0 & \text{otherwise} \end{cases}\); \(X+Y\leq 3.\)
- \(f(x,y)=\begin{cases}\frac{1}{300} e^{-x/30}e^{-y/10} & x\geq 0, y\geq 0 \\ 0 & \text{otherwise} \end{cases}\); \(X+Y\leq 3.\)
5.12 Surface Area of a Differentiable Function
We apply double integrals to the problem of computing the surface area over a region. We demonstrate a formula that is analogous to the formula for finding the arc length of a one variable function and detail how to evaluate a double integral to compute the surface area of the graph of a differentiable function of two variables.
::: {#thm- } [Surface Area] Assume that the function \(f(x,y)\) has continuous partial derivatives \(f_x\) and \(f_y\) in a region \(R\) of the \(x y\)-plane. Then the portion of the surface \(z=f(x,y)\) that lies over \(R\) has surface area \[ S=\iint _R \sqrt{ \left[ f_x(x,y)\right]^2 + \left[ f_y(x,y) \right]^2 +1 }\, dA. \] :::
Proof. Consider a surface defined by \(z=f(x,y)\) over a region \(R\) of the \(x y-\text{plane}.\) Enclose the region \(R\) in a rectangle partitioned by a grid with lines parallel to the coordinate axes. This creates a number of cells, and we let \(R_1,\) \(R_2,\) ., \(R_n\) denote those that lie entirely within \(R.\) For \(m=1,2,3,\ldots ,n,\) let \(P_m\left(x_m^*,y_m^*\right)\) be any corner of the rectangle \(R_m,\) and let \(T_m\) be the tangent plane above \(P_m\) on the surface of \(z=f(x,y).\) Let \(\text{$\triangle $S}_m\) denote the area of the patch of the surface that lies directly above \(R_m.\) The rectangle \(R_m\) projects onto a parallelogram \(A B D C\) in the tangent plane \(T_m,\) and if \(R_m\) is small we would expect the area of this parallelogram to approximate closely the element of the area \(\text{$\triangle $S}_m.\) If \(\triangle x_m\) and \(\triangle y_m\) are the lengths of the sides of the rectangle \(R_m,\) the approximating parallelogram will have sides determined by the vectors \[ \vec{A B}=\triangle x_m \vec{i}+\left[f_x\left(x_m^*,y_m^*\right)\triangle x_m\right]\vec{k} \] and \[ \vec{A C}=\triangle y_m \vec{j}+\left[f_y\left(x_m^*,y_m^*\right)\triangle y_m\right] \vec{k}. \] If \(K_m\) is the area of the approximating parallelogram, we have \[ K_m=\|\vec{A B}\times \vec{A C}\|. \] To determine \(K_m,\) we first find the cross product: \[\begin{align*} \vec{A B} \times \vec{A C} & =\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \triangle x_m & 0 & f_x\left(x_m^*,y_m^*\right)\triangle x_m \\ 0 & \triangle y_m & f_y\left(x_m^*,y_m^*\right)\triangle y_m \end{array} \right| \\ & =\triangle x_m\triangle y_m\left(-f_x\left(x_m^*,y_m^*\right)\, \vec{i}-f_y\left(x_m^*,y_m^*\right)\, \vec{j}+\vec{k}\right). \end{align*}\] Then we calculate the norm, \[ K_m=\|\vec{A B} \times \vec{A C}\| =\triangle x_m\triangle y_m \sqrt{ \left[f_x\left(x_m^*,y_m^*\right)\right]{}^2+\left[f_y\left(x_m^*,y_m^*\right)\right]{}^2+1}. \] Finally, summing over the entire partition, we see that the surface area over \(R\) may be approximated by the sum \[ \triangle S_n=\sum _{m=1}^n \sqrt{\left[f_x\left(x_m^*,y_m^*\right)\right]{}^2+\left[f_y\left(x_m^*,y_m^*\right)\right]{}^2+1}\left(\triangle A_m\right) \] where \(\triangle A_m=\triangle x_m\triangle y_m.\) This is a Riemann sum, and by taking an appropriate limit (as the partition becomes more refined), we find that the surface area \(S,\) satisfies \[\begin{align*} S& =\lim _{n\to \infty }\sum _{m=1}^n \triangle A_m\sqrt{\left[f_x\left(x_m^*,y_m^*\right)\right]^2+\left[f_y\left(x_m^*,y_m^*\right)\right]^2+1} \\ & =\iint_R\sqrt{\left[f_x(x,y)\right]^2+\left[f_y(x,y)\right]^2+1} \, d A. \end{align*}\]
Example 5.46 Find the surface area of the part of the surface \(z=x^2+2y\) that lies above the triangular region \(T\) in the \(x y-\text{plane}\) with vertices \((0,0),\) \((1,0),\) and \((1,1).\)
Solution. The region \(T\) is described by \(T=\{(x,y) \mid 0\leq x\leq 1,0\leq y\leq x\}.\) We have \[\begin{align*} S& =\iint_R\sqrt{(2x)^2+(2)^2+1} \, d A =\int _0^1\int _0^x\sqrt{4x^2+5}\, dy\, dx \\ & =\int _0^1x\sqrt{4x^2+5}\, dx =\frac{1}{12}\left(27-5\sqrt{5}\right). \end{align*}\]
Example 5.47 Find the surface area of a sphere of radius \(a\).
Solution. Let \(z=f(x,y)=\sqrt{a^2-x^2-y^2}.\) Then \[ f_x=\frac{-x}{\sqrt{a^2-x^2-y^2}}, \qquad f_y=\frac{-y}{\sqrt{a^2-x^2-y^2}},\] and \[ \sqrt{f_x^2+f_y^2+1}=\sqrt{\frac{x^2+y^2+\left(a^2-x^2-y^2\right)}{a^2-x^2-y^2}}=\frac{a}{\sqrt{a^2-x^2-y^2}}. \] In polar form we have \[\begin{align*} & 8\int _0^{\pi /2}\int _0^a\frac{a}{\sqrt{a^2-r^2}}r \, drd\theta \\ & =-4a\int _0^{\pi /2}\int _0^a\frac{(-2r)}{\sqrt{a^2-r^2}} \, drd\theta =-8a\int_0^{\pi /2} (-a) \, d\theta =4\pi a^2. \end{align*}\]
Example 5.48 Find the surface area of a cylinder of radius \(a\) and height \(h\).
Solution. Let \(z=f(x,y)=\sqrt{a^2-x^2}.\) Then \[ f_x=\frac{-x}{\sqrt{a^2-x^2}}, f_y=0, \] and \[ \sqrt{f_x^2+f_y^2+1}=\sqrt{\frac{x^2+a^2-x^2}{a^2-x^2}}=\frac{a}{\sqrt{a^2-x^2}}.\] We find the surface area as \[\begin{align*} 4\int _0^a\int _0^h\frac{a}{\sqrt{a^2-x^2}} \, dy\, dx & =4a\int _0^a\frac{h a}{\sqrt{a^2-x^2}} dx \\ & =\left. 4 a h \sin ^{-1}\left(\frac{x}{a}\right) \right|_0^a =2\pi a h. \end{align*}\]
Example 5.49 Find the surface area the part of the paraboloid \(z=x^2+y^2\) that lies under the plane \(z=9.\)
Solution. The plane intersects the paraboloid in the circle \(x^2+y^2=9\) when \(z=9.\) Therefore, the given surface lies above the disk \(D\) with center at the origin and radius 3. Converting to polar coordinates we have \[\begin{align*} S& = \iint_R \sqrt{(2x)^2+(2y)^2+1} \, d A =\int _0^{2\pi }\int _0^3\sqrt{4r^2+1}r \, dr \, d\theta \\ & =\int _0^{2\pi }d \theta \int _0^3\frac{1}{8}\sqrt{4r^2+1}(8r) \, dr =\frac{\pi }{6}\left(37\sqrt{37}-1\right). \end{align*}\]
Example 5.50 Find the surface area the portion of the sphere \(x^2+y^2+z^2=4\) that lies inside the cylinder \(x^2+y^2=2y.\)
Solution. Let \(z=f(x,y)=\sqrt{4-x^2-y^2}.\) Then \[f_x=\frac{-x}{\sqrt{4-x^2-y^2}}, \qquad f_y=\frac{-y}{\sqrt{4-x^2-y^2}}, \] and \[ \sqrt{f_x^2+f_y^2+1}=\sqrt{\frac{x^2+y^2+\left(4-x^2-y^2\right)}{4-x^2-y^2}}=\frac{2}{\sqrt{4-x^2-y^2}}. \] The projected region in polar form is \(r=2 \sin \theta .\) Since half the surface is above the \(x y-\text{plane}\) and half the surface is below we have \[\begin{align*} S& =2\int _0^{\pi }\int _0^{2 \sin \theta }\frac{2r}{\sqrt{4-r^2}}drd\theta =-8\int_0^{\pi } \left(\sqrt{4-4\sin ^2\theta }-2\right) \, d\theta \\ & =-8\int_0^{\pi /2} (2 \cos \theta -2) \, d\theta -8\int_{\pi /2}^{\pi } (-2 \cos \theta -2) \, d\theta =8(\pi -2). \end{align*}\]
5.13 Surface Area Defined Parametrically
We show a way to find the surface area of the graph of a surface given a parameterization of the surface. We also illustrate how to find the surface area with two examples and then show how to find the surface area of a torus with given inner and outer radii.
::: {#thm- } [Surface Area (Parametrically)] Let \(S\) be a surface defined parametrically by \[ \vec{R}(u,v)=x(u,v)\vec{i}+y(u,v)\vec{j}+z(u,v)\vec{k} \] on the region \(D\) in the \(u v\)-plane, and assume that \(S\) is smooth in the sense that \(\vec{R}_u\) and \(\vec{R}_v\) are continuous with \(\vec{R}_u\times \vec{R}_v\neq \vec{0}\) on \(D.\) Then the surface area \(S\) is given by \[ S=\underset{D}{\iint }\left\|\vec{R}_u\times \vec{R}_v\right\| d u d v. \] The quantity \(\left\|\vec{R}_u\times \vec{R}_v\right\|\) is called the fundamental cross product . :::
Proof. Suppose a surface \(S\) is defined parametrically by the vector function \[ \vec{R}(u,v)=x(u,v)\vec{i}+y(u,v)\vec{j}+z(u,v)\vec{k} \] for parameters \(u\) and \(v.\) Let \(D\) be a region in the \(x y\)-plane on which \(x,\) \(y,\) and \(z,\) as well as their partial derivatives with respect to \(u\) and \(v\) are continuous. The partial derivatives of \(R(u,v)\) are given by \[ \vec{R}_u=\frac{\partial x}{\partial u}\vec{i}+\frac{\partial y}{\partial u}\vec{j}+\frac{\partial z}{\partial u}\vec{k} \qquad \text{ and } \qquad \vec{R}_v=\frac{\partial x}{\partial v}\vec{i}+\frac{\partial y}{\partial v}\vec{j}+\frac{\partial z}{\partial v}\vec{k}. \] Suppose the region \(D\) is subdivided into cells. Consider a typical rectangle in this partition, dimension \(\triangle x\) and \(\triangle y\), where \(\triangle x\) and \(\triangle y\) are small. If we project this rectangle onto the surface \(\vec{R}(u,v),\) we obtain a curvilinear parallelogram with adjacent sides \(\vec{R}_u(u,v)\triangle u\) and \(\vec{R}_v(u,v)\triangle v.\) The area of this rectangle is approximated by \[ \triangle S=\left\|\vec{R}_u(u,v)\triangle u\times \vec{R}_v(u,v)\triangle v\right\| =\left\|\vec{R}_u(u,v)\times \vec{R}_v(u,v)\right\|\triangle u \triangle v. \] By taking an appropriate limit, we find the surface area to be a double integral.
Example 5.51 Find the area of the surface given parametrically by the equation \[ \vec{R}(u,v)= u v \vec{i}+(u-v) \vec{j}+(u+v) \vec{k} \] for \(u^2+v^2\leq 1\).
Solution. We find \[ \vec{R}_u\times \vec{R}_v=\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ v & 1 & 1 \\ u & -1 & 1 \end{array} \right|=2\vec{i}+(u-v)\vec{j}-(u+v)\vec{k}. \] Therefore the fundamental cross product is \[ \left\|\vec{R}_u\times \vec{R}_v\right\|=\sqrt{4+(u-v)^2+(-u-v)^2}=\sqrt{4+2u^2+2v^2}. \] Using polar coordinates the surface area is \[\begin{align*} S& =\int _0^{2\pi }\int _0^1\sqrt{4+2r^2}rdrd\theta =\frac{1}{4}\int _0^{2\pi }\int _0^1\sqrt{4+2r^2}4rdrd\theta \\ & =\frac{1}{6}\int_0^{2\pi } \left(6\sqrt{6}-8\right) \, d\theta =\frac{2\pi }{3}\left(3\sqrt{6}-4\right). \end{align*}\]
Example 5.52 Find the area of the surface given parametrically by the equation \[ \vec{R}(u,v)=(u \sin v)\vec{i}+(u \cos v)\vec{j}+ v \vec{k} \] for \(0\leq u\leq a\) and \(0\leq v\leq b\).
Solution. We find \[ \vec{R}_u=(\sin v)\vec{i}+(\cos v)\vec{j}, \quad \vec{R}_v=(u \cos v)\vec{i}+(-u \sin v)\vec{j}+\vec{k},\] and \[ \vec{R}_u\times \vec{R}_v=\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \sin v & \cos v & 0 \\ u \cos v & -u \sin v & 1 \end{array} \right|=(\cos v)\vec{i}-(\sin v)\vec{j}-u \vec{k}.\] Therefore the fundamental cross product is \(\left\|\vec{R}_u\times \vec{R}_v\right\|=\sqrt{1+u^2}.\) So the surface area is \[\begin{align*} S& =\int _0^b\int _0^a\sqrt{1+u^2} \, du dv =\int_0^b \left[\frac{\ln \left|u+\sqrt{1+u^2}\right|}{2}+\frac{u\sqrt{1+u^2}}{2}|_0^a\right] \, dv \\ & =\frac{b}{2}\left[\ln \left(a+\sqrt{1+a^2}\right)+a\sqrt{1+a^2}\right]. \end{align*}\]
Example 5.53 Find the area of the torus which is given parametrically by \[ \vec{R}(u,v)=(a+b \cos v)\cos u \vec{i}+(a+b \cos v)\sin u \vec{j}+ (b \sin v) \vec{k} \] for \(0<b<a,\) \(0\leq u\leq 2\pi ,\) and \(0\leq v\leq 2\pi.\)
Solution. We find \[\begin{align*} & \vec{R}_u=-(a+b \cos v)\sin u \vec{i}+(a+b \cos v)\cos u\vec{j}, \\ & \vec{R}_v=-b \sin v \cos u \vec{i}-b \sin v \sin u\vec{j} +b \cos v \vec{k}, \text{ and} \\ & \vec{R}_u\times \vec{R}_v =\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ -(a+b \cos v) \sin u & (a+b \cos v)\cos u & 0 \\ -b \sin v \cos u & -b \sin v \sin u & b \cos v \end{array} \right| \\ & \qquad \qquad =\left(b^2\cos ^2v+a b \cos v\right)(\cos u)\vec{i} +\left(b^2\cos ^2v+a b \cos v\right)(\sin u)\vec{j} \\ & \qquad \qquad \qquad \qquad +\left(b^2\sin v \cos v+a b \sin v\right)\vec{k}.\end{align*}\] Therefore \(\left\|\vec{R}_u\times \vec{R}_v\right\|=\left|a b+b^2\cos v\right|.\) So the surface area is \[\begin{align*} S& =\int _0^{2\pi }\int _0^{2\pi }\left|a b+b^2\cos v\right| du dv =\int _0^{2\pi }2\pi \left(a b+b^2\cos v\right)dv =4\pi ^2 a b. \end{align*}\]
5.14 Exercises
Exercise 5.45 Find the surface area of the part of the paraboloid \(z=x^2+y^2\) that lies under the plane \(z=9.\)
Exercise 5.46 Find the surface area of the surface of the portion of the plane \(4x+y+z=9\) that lies in he first octant.
Exercise 5.47 Find the surface area of the surface of the portion of the paraboloid \(z=3x^2+3y^2\) that lies inside the cylinder \(x^2+y^2=1.\)
Exercise 5.48 Find the surface area of the surface of the portion of the plane \(2x+2y-z=0\) that is above the square in the plane with vertices \((0,1,0),\) \((0,0,0),\) \((1,0,0),\) and \((1,1,0).\)
Exercise 5.49 Find the surface area of the surface of the portion of the surface \(z=x^2\) that lies over the triangular region in the plane with vertices \((0,0,0),\) \((0,1,0),\) and \((1,0,0).\)
Exercise 5.50 Find the surface area of the surface of the portion of the sphere \(x^2+y^2+z^2=25\) inside the cylinder \(x^2+y^2=9.\)
Exercise 5.51 Find the surface area of the surface of the portion of the cone \(z=2\sqrt{x^2+y^2}\) inside the cylinder \(x^2+y^2=4.\)
Exercise 5.52 Find the surface area of the surface of the portion of the sphere \(x^2+y^2+z^2=a^2\) inside the cylinder \(x^2+y^2=a x\) and above the \(x y\)-plane.
Exercise 5.53 Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by \(z=e^{-x}\sin y\) over the triangle with vertices \((0,0,0),\) \((0,1,1),\) \((0,1,0).\)
Exercise 5.54 Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by \(x=z^3-y z+y^3\) over the square \((0,0,0),\) \((0,0,2),\) \((0,2,0),\) and \((0,2,2).\)
Exercise 5.55 Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by \(z=\cos \left(x^2+y^2\right)\) over the disk \(x^2+y^2\leq \frac{\pi }{2}.\)
Exercise 5.56 Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by \(z=e^{-x}\cos y\) over the disk \(x^2+y^2\leq 2.\)
Exercise 5.57 Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by \(z=x^2+5 x y+y^2\) over the region in the \(x y\)-plane bounded by the curve \(x y=5\) and the line \(x+y=6.\)
Exercise 5.58 Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by \(z=x^2+3 x y+y^2\) over the region in the \(x y\)-plane bounded by \(0\leq x\leq 4,\) \(0\leq y\leq x.\)
Exercise 5.59 Compute the magnitude of the fundamental cross product for the surface defined parametrically by \(\vec{R}(u,v)=(2 u \sin v)\, \vec{i}+(2 u \cos v)\, \vec{j}+u^2\, \vec{k}.\)
Exercise 5.60 Compute the magnitude of the fundamental cross product for the surface defined parametrically by \(\vec{R}(u,v)=(4 \sin u \cos v )\vec{i}+(4 \sin u \sin v) \, \vec{j}+(5 \cos u) \vec{k}.\)
Exercise 5.61 Compute the magnitude of the fundamental cross product for the surface defined parametrically by \(\vec{R}(u,v)=u \, \vec{i}+ v^2 \, \vec{j}+u^3 \, \vec{k}.\)
Exercise 5.62 Compute the magnitude of the fundamental cross product for the surface defined parametrically by \(\vec{R}(u,v)=(2 u \sin v)\, \vec{i}+(2 u \cos v)\, \vec{j}+\left(u^2\sin 2 v\right)\vec{k}.\)
Exercise 5.63 Find the area of the surface given parametrically by the equation \(\vec{R}(u,v)=u v \, \vec{i}+(u-v)\, \vec{j}+(u+v) \vec{k}\) for \(u^2+v^2\leq 1.\)
Exercise 5.64 A spiral ramp has the vector parametric equation \(\vec{R}(u,v)=(u \cos v) \, \vec{i}+(u \sin v) \, \vec{j}+v \, \vec{k}\) for \(0\leq u\leq 1,\) \(0\leq v\leq \pi .\) Find the surface area of this ramp.