Integration is a mathematical process that allows us to find the area under a curve, or to solve other problems involving functions. In this book, we will explore the applications of integration in detail, and learn how to apply it in various situations.

We will start with the basics of finding volumes, including the washer method and the shell method. From there, we will move on to more advanced applications. With practice, you will be able to apply these ideas and use integration to solve problems yet to be discovered.

Integrals are one of the most important concepts in mathematics, with applications in calculus, physics, engineering, and more. Simply put, an integral is a way to calculate the area under a curve. This can be used to determine the velocity of an object, the volume of a three-dimensional object, or even the length of a curve. In essence, integrals allow us to determine things that would otherwise be impossible to calculate.

Of course, integrals are just one part of calculus; there are also derivatives, limits, and so on. But without integrals, many of the things we take for granted would be impossible. So next time you see a complicated equation involving an integral, remember that it just represents someone’s attempt to calculate something important.

The applications of integration are vast and include finding the volume or mass of an object, calculating work done or energy transferred, and solving many types of differential equations.

In this book, we will explore the applications of integration and how to apply it in various situations. So if you’re ready to learn about the applications of integration, let’s get started!

Calculus is all about finding the area between curves. And while that may sound like a rather esoteric focus, the applications of integration are actually quite practical.

For example, consider a firm that wants to determine how much material to order for manufacturing a product. The company knows the length of the product and the amount of material required per unit length. To find out how much material to order, they need to integrate the function that represents the amount of material required over the length of the product.

In other words, they need to find the area between the curve representing the amount of material required and an axis. Similarly, if a company wants to calculate how much money it will save by reducing its energy consumption, it needs to find the area between the two functions representing its current and proposed energy consumption rates.

As these examples illustrate, integration is a powerful tool with numerous applications in the business world. So next time you see a calculus problem involving the area between curves, don’t be discouraged - remember that there’s a practical application for all that math!

Have you ever wondered how engineers calculate the volume of irregularly shaped objects? The answer lies in a branch of mathematics called integral calculus. One of the applications of integration is calculating the volume of a solid object by revolving a two-dimensional shape around an axis. This technique is known as the method of volumes of solids of revolution. To use this method, you first need to identify a suitable axis of revolution. This is typically a line that passes through the center of the object.

Once you have chosen an axis, you need to determine the equation that describes the shape. For example, if you were calculating the volume of a cylindrical container, you would use the equation for a circle. You then need to integrate this equation to find the area enclosed by the shape. Finally, you need to multiply this area by the length of the axis of revolution to find the volume. Although it may seem like a lot of work, the method of volumes of solids of revolution can be used to calculate the volumes of all sorts of objects, from coffee cups to oil tanks.

The Disk Method is a simple way to find volumes, but it has its limitations. First, it only works for solids that are rotationally symmetric about one axis. Second, it only works for solids with flat or circular bases. However, despite these limitations, the disk method is still a useful tool for finding volumes and has applications in many different fields.

The Washer Method is a volume of solids of revolution technique that can be applied to a wide variety of applications of integration. In essence, the washer method calculates the volume of an object that has been revolved around an axis by cleverly dividing the object into infinitely thin slices, and then calculating the volume of each slice.

The key to understanding the washer method is to remember that each slice must have a constant cross-sectional area; otherwise, the volume calculation will be inaccurate. Once this concept is understood, the applications of the washer method are nearly endless. From calculating the volumes of simple objects like cylinders and spheres to more complex objects like cones and toruses, the Washer Method is a powerful tool.

The Shell Method is also one of the applications of integration that can be used to calculate the volumes of certain solids. This method is most often used when calculating the volumes of solids of revolution, which are created when a curve is rotated about an axis. The shell method involves slicing the solid into thin shells and then calculating the volume of each individual shell. This method can be used to calculate the volumes of solids with a variety of shapes, including cylinders, spheres, and cones. While the shell method can be somewhat tricky to master, it is a powerful tool for calculating the volumes of solids of revolution.

As every math student knows, integration is a powerful tool with a wide range of applications. One of the most common applications of integration is finding the arc length of a curve. By using a small angle approximation, you can break up the arc into a series of straight line segments. To do this, you need to first break up the arc into small segments. Then, you take the length of each segment and multiply it by the corresponding width. Finally, you add up all of the lengths to get the total length of the arc.

It might sound complicated, but it’s actually quite simple. And once you know how to do it, you can apply this technique to all sorts of applications. So why not give it a try? Who knows, you might just find yourself getting addicted to integration!

One of the most practical applications is in the calculation of surface area. When an object is rotated about an axis, the resulting shape is known as a surface of revolution. By using integration, it’s possible to calculate the surface area of this type of object without having to take measurements of the individual components. This can be a particularly useful tool in engineering applications, where accurate measurements are crucial.

When calculating the surface area of an object, we are essentially finding the amount of material required to cover the object. This process can be applied to both regular and irregular objects. For example, when we calculate the surface area of a cylinder, we are finding the amount of material needed to cover the curved sides. In general, however, calculating the surface area of an arbitrary object can be a very difficult task.

However, by breaking the object down into smaller pieces, we can use integration to more easily calculate its surface area. This technique is particularly helpful for surfaces of revolution, which can be difficult to visualize. By using integration, we can more easily calculate the amount of material needed to cover these complex shapes.

These are just a few of the applications of integration. As you can see, this powerful tool can be used in a wide variety of situations. Whether you’re calculating the volume of an object or finding the arc length of a curve, integration is a powerful tool that every math student should master. So why not give it a try? You might just be surprised at how useful it can be!

In this book, I will take the applications of integration one by one, explaining in detail how to use them with the help of examples. I will also provide tips on when and how to use each application. By the end of this book, you will have a solid understanding of the applications of integration and how to apply it in various situations. So let’s get started!

## 5.1 Area Between Curves

We explain, through several examples, how to find the area between curves (as a bounded region) using integration. We demonstrate both vertical and horizontal strips and provide several exercises.

When applying the definition for the area between curves , finding the intersection points of the curves and sketching their graphs is crucial. These graphs often reveal whether we should use vertical or horizontal strips by determining which curve is the upper curve and which is the lower.

## 5.2 Area Between Curves Using Vertical Strips

Definition 5.1 Let $$f$$ and $$g$$ be continuous functions on $$[a,b],$$ and suppose that $${f(x)\geq g(x)}$$ for all $$x$$ in $$[a,b].$$ Then the area of the region between the graphs of $$f$$ and $$g$$ and the vertical lines $$x=a$$ and $$x=b$$ is $\begin{equation} \label{vs} A=\int_a^b [f(x)-g(x)] \, dx. \end{equation}$

In particular, if $$f(x)\geq g(x)=0,$$ then $$\eqref{vs}$$ yields the area of the region bounded by: the graph of $$f,$$ above the $$x$$-axis, and between the lines $$x=a$$ and $$x=b$$ as $\begin{equation} \label{funda} A=\int_a^b f(x)\, dx. \end{equation}$ The reader is expected to be familiar with $$\eqref{funda}$$ and also with the Fundamental Theorem of Calculus.

Example 5.1 Find the area of the region enclosed by the parabola $$y=2-x^2$$ and the line $$y=-x.$$

Solution. First we find the intersection points $$(-1,1)$$ and $$(2,-2).$$ Then we see that we have $$f(x)=2-x^2 \geq g(x)=-x$$ for all $$x\in [-1,2]$$ as required.

Therefore the area between the curves is \begin{align*} \int_a^b [f(x)-g(x)] \, dx & =\int_{-1}^2[(2-x^2)-(-x)]\, dx \\ & = \int_{-1}^2 (2+x-x^2)\, dx = \left[2x+\frac{x^2}{x}-\frac{x^3}{3}\right]_{-1}^2 \\ & =\left(4+\frac{4}{2}-\frac{8}{3}\right)-\left(-2+\frac{1}{2}+\frac{1}{3}\right) =\frac{9}{2} \end{align*} as desired.

Example 5.2 Find the area of the region in the first quadrant that is bounded above by $$y=\sqrt{x}$$ and below by the $$x$$-axis and the line $$y=x-2.$$

Solution. First we find the intersection point $$(4,2).$$ From a sketch of the graphs we see that we should consider the bounded region as two separate regions.

We have $$f(x)=\sqrt{x}\geq g(x)=0$$ for all $$x$$ in $$[0,2]$$ and in the other region we have $$f(x)=\sqrt{x}\geq g(x)=x-2$$ for all $$x$$ in $$[2,4].$$ The total area is \begin{align*} & \int_0^2 \sqrt{x} \, dx +\int_2^4 (\sqrt{x}-x+2) \, dx \\ & \qquad =\left[\frac{2}{3}x^{2/3}\right]_0^2+\left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}+2x\right]_2^4 \\ & \qquad =\left[\frac{2}{3}2^{2/3}\right]+\left[\frac{2}{3}4^{3/2}-\frac{4^2}{2}+2(4)\right] -\left[\frac{2}{3}2^{3/2}-\frac{2^2}{2}+2(2)\right]\\ & \qquad =\frac{10}{3} \end{align*} as desired.

Example 5.3 Find the area of the shaded region between the curves $$y=2x^2$$ and $$y=x^4-2x^2.$$

Solution. First we find the intersection points $$(0,0)$$ and $$(\pm 2, 8).$$ From a sketch of the graph we see that we can use symmetry about the $$y$$-axis. We have ${f(x)=2x^2\geq g(x)=x^4-2x^2}$ for all $$x$$ in $$[0,2].$$

The total area is \begin{align*} & 2 \int_{0}^2 [2x^2-(x^4-2x^2)]\, dx = 2 \int_{0}^2 (4 x^2 - x^4)\, dx \\ & \qquad = 2 \left(\left.\frac{4x^3}{3}-\frac{x^5}{5}\right|_0^2 \right) = 2 \left(\frac{4(2)^3}{3}-\frac{(2)^5}{5} \right) = \frac{96}{5} \end{align*} as desired.

Example 5.4 Find the area of the shaded region between the curves $${y=2x^3-x^2-5x}$$ and $$y=-x^2+3x.$$

Solution. First we find the intersection points $$(-2,-10),$$ $$(0,0),$$ and $$(2,2).$$ From a sketch of the graphs wee see that we should break the bounded region into two regions.

Notice that for all $$x$$ in $$[-2,0]$$ we have $f(x)= 2x^3-x^2-5x\geq g(x) = -x^2+3x$ and that for all $$x$$ in $$[0,2]$$ we have $$f(x)= -x^2+3x\geq g(x) = 2x^3-x^2-5x.$$ The total area is $$16$$ given by \begin{align*} & \int_{-2}^0 [(2x^3-x^2-5x)-(-x^2+3x)] \, dx \\ & \qquad + \int_0^2 [(-x^2+3x)-(2x^3-x^2-5x)]\, dx \\ & \qquad = 8+8 =16 \end{align*} as desired.

## 5.3 Area Between Curves Using Horizontal Strips

If a region’s bounding curves are described by functions of $$y,$$ the approximating rectangles are horizontal instead of vertical and the basis formula has $$y$$ instead of $$x.$$

Definition 5.2 Let $$f$$ and $$g$$ be continuous functions on $$[c,d],$$ and suppose that $${f(y)\geq g(y)}$$ for all $$y$$ in $$[c,d].$$ Then the area of the region between the graphs of $$f$$ and $$g$$ and the horizontal lines $$y=c$$ and $$y=d$$ is $\begin{equation} A=\int_c^d [f(y)-g(y)] \, dy. \end{equation}$

In this equation $$f$$ always denotes the right-hand curve and $$g$$ the left-hand curve, so that $$f(y)-g(y)$$ is nonnegative.

Example 5.5 Find the area of the region in the first quadrant that is bounded above by $$y=\sqrt{x}$$ and below by the $$x$$-axis and the line $$y=x-2.$$

Solution. In this example we consider horizontal strips. Here we have $f(y)=y+2\geq g(y)=y^2$ for all $$y$$ in $$[0,2]$$ as is seen in the sketch of the region.

The total area is \begin{align*} & \int_0^2 [y+2-y^2] \, dy = \left[2y+\frac{y^2}{2}-\frac{y^3}{3}\right]_0^2 = 4+\frac{4}{2}-\frac{8}{3} = \frac{10}{3} \end{align*} as desired.

## 5.4 Exercises

Exercise 5.1 Find the areas of the regions enclosed by the lines and curves. - $$x=2y^2,$$ $$x=0,$$ and $$y=3$$ - $$y^2-4x=4,$$ $$4x-y=16$$ - $$y=\frac{x}{x^2+1},$$ $$y=-\frac{1}{2} x^2,$$ $$x=1$$ - $$x=y^3-y^2$$ and $$x=2y$$ - $$y=x\sqrt{4-x^2},$$ $$y=0$$ - $$y=\sqrt{x+3},$$ $$y=(x+3)/2$$ - $$x+y^2=3$$ and $$4x+y^2=0$$ - $$\sqrt{x}+\sqrt{y}=1,$$ $$x+y=1$$ - $$y=-x^3+x,$$ $$y=x^4-1$$ - $$y=\sin x,$$ $$y=x,$$ $$x=\pi/2,$$ $$x=\pi$$ - $$y=\cos x,$$ $$y=2-\cos x,$$ $$0\leq x \leq 2\pi$$ - $$x=1-y^2,$$ $$x=y^2-1$$

Exercise 5.2 Find the are of the region in the first quadrant bounded on the left by the $$y$$-axis and on the right by the curves $$y=\sin x$$ and $$y=\cos x.$$

Exercise 5.3 Find the area of the region enclosed by the curve $$y^2=x^2(1-x^2).$$

Exercise 5.4 Find the area, integrating with respect to $$x$$ and then $$y,$$ of the region between the curve $$y=3-x^2$$ and the line $$y=-1.$$

Exercise 5.5 Use integration to find the area of the triangle with the given vertices $$(-2,4),$$ $$(0,-2),$$ and $$(6,2).$$

Exercise 5.6 Find the area of the region in the first quadrant bounded on the left by the $$y$$-axis, below by the curve $$x = 2\sqrt{y},$$ above left by the curve $$x = (y-1)^2,$$ and above right by the line $$x = 3- y.$$

Exercise 5.7 Find the area of the region bounded by the graph of the curve $$y^2=x^3-x^2$$ and the line $$x=2.$$

Exercise 5.8 Find the value of $$c$$ such that the parabola $$y =c x^2$$ divides the region by the parabola $$y=\frac{1}{9} x^2,$$ and the lines $$y=2,$$ and $$x=0$$ into two subregions of equal area.

Exercise 5.9 Find the area of the region bounded by the parabola $$y=x^2,$$ the tangent line to this parabola at $$(1,1)$$ and the $$x$$-axis.

Exercise 5.10 Find the values of $$c$$ such that at the area of the region bounded by the parabolas $y=x^2 - c^2$ and $$y= c^2 - x^2$$ is 576.

Exercise 5.11 Find $$a$$ such that the line $$x=a$$ divides the region bounded by the graphs of the equations into two regions of equal area given $$y=x,$$ $$y=4,$$ and $$x=0.$$

## 5.5 Volumes of Solids of Revolution

Finding the volume of the solid generated by rotating a bounded planar region about an axis of rotation is discussed. We cover the disk method, the washer method, and method of cylindrical shells. We provide several examples of solids generated by revolving around both vertical and horizontal lines.

The solid generated by rotating a plane region about an axis in its plane is called a solid of revolution .

Definition 5.3 The volume of a solid of known integrable cross-sectional area $$A(x)$$ from $$x=a$$ to $$x=b$$ is the integral of $$A$$ from $$a$$ to $$b$$, $\begin{equation} \label{defvol} V=\int_a^b A(x) \, dx \end{equation}$

Throughout our discussion we assume that $$A(x)$$ is integrable and in particular, we assume that the reader is familiar with the Fundamental Theorem of Calculus.

## 5.6 The Disk Method

::: {#thm- } Let $$A(x)$$ be the cross-sectional area of a disk of radius $$R(x)$$, the distance of the planar regions’s boundary from the axis of revolution. The volume of the solid of revolution about the $$x$$-axis is $\begin{equation} V = \int_a^b A(x) \, dx = \int_a^b \pi [ R(x) ]^2 \, dx. \end{equation}$ :::

Example 5.6 Find the volume of the solid that results when the region enclosed by $$y=9-x^2$$ and $$y=0$$ is revolved about the $$x$$-axis.

Solution. First we sketch the region bounded by the given curves. We use $$R(x)=9-x^2$$ and find the volume \begin{align*} V & = \pi \int_{-3}^3 (9-x^2)^2 \, dx \\ & = \pi \int_{-3}^3 (81-18x^2+x^4) \, dx = \frac{1296\pi}{5} \end{align*} as desired.

Example 5.7 Find the volume of the solid generated when the region bounded by the graph of $$f(x)=\sqrt{x}+1$$ and the line $$y=2$$ on the interval $$[0,1]$$, is revolved about the line $$y=2$$.

Solution. First we sketch the bounded region. The radius is given by $\begin{equation} R(x)= 2-(\sqrt{x}+1) = 1-\sqrt{x} \end{equation}$

We find the volume using the Disk Method \begin{align*} V = \pi \int_0^1(1-\sqrt{x})^2 \, dx = \pi \int_0^1 (1-2\sqrt{x}+x) \, dx = \frac{\pi}{6} \end{align*} as desired.

::: {#thm- } Let $$A(y)$$ be the cross-sectional area of a disk of radius $$R(y)$$, the distance of the planar regions’s boundary from the axis of revolution. The volume of the solid of revolution about the $$y$$-axis is $\begin{equation} V = \int_a^b A(y) \, dy = \int_a^b \pi [ R(y) ]^2 \, dy. \end{equation}$ :::

Example 5.8 Find the volume of the solid that results when the region enclosed by the curves $$y=0$$, $$y=\ln x$$, $$y=2$$, and $$x=0$$ is revolved about the $$y$$-axis.

Solution. First we sketch the bounded region. The radius is given by $$R(y)=e^y$$.

We find the volume using the Disk Method $\begin{equation} V= \pi \int_0^2 e^{2y} \, dy = \frac{\pi(e^4-1)}{2} \end{equation}$ as desired.

Example 5.9 Find the volume of the solid that results when the region enclosed by $$y=\sqrt{x}$$, $$y=0$$, and $$x=9$$ is revolved about the line $$x=9$$.

Solution. First we sketch the bounded region. The radius is given by $${R(y)=9-y^2}$$.

We find the volume using the Disk Method \begin{align*} V & = \pi \int_0^3 (9-y^2)^2 \, dy \\ & = \pi \int_0^3 (81-18y^2+y^4)\, dy = \frac{648\pi}{5} \end{align*} as desired.

## 5.7 The Washer Method

::: {#thm- } Let $$A(x)$$ be the cross-sectional area of a disk of outer radius $$R(x)$$ and inner radius r(x).
The volume of the solid of revolution about the $$x$$-axis is $\begin{equation} V = \int_a^b A(x) \, dx = \int_a^b \pi \left( [ R(x) ]^2 - [ r(x) ]^2 \right) \, dx. \end{equation}$ :::

Example 5.10 Use the washer method to find the volume of the solid generated when the region bounded by $$y=x$$ and $$y=2\sqrt{x}$$ is revolved about the $$x$$-axis.

Solution. First we sketch the bounded region. The outer radius is $$R(x)=2\sqrt{x}$$ and the inner radius is $$r(x)=x$$.

We find the volume using the Washer Method \begin{align*} V & = \pi \int_0^4 ( (2\sqrt{x})^2-x^2)\, dx \\ & = \pi \left.(2x^2-x^3/3)\right\vert_0^4 = \pi(32-64/3) = \frac{32\pi}{3}. \end{align*} as desired.

Example 5.11 Find the volume for the region bounded by the graphs of $$y=2\sin x$$ and the $$x$$-axis on $$[0,\pi]$$ is revolved around the line $$y=-2$$.

Solution. First we sketch the bounded region.

The washer has outer radius $$R(x)=2+2\sin x$$ and inner radius $$2$$, so using the Washer Method the volume is \begin{align*} V & = \pi \int_0^\pi ((2+2\sin x)^2 -2^2) \, dx = \pi \int_0^8 (8\sin x+4\sin^2 x)\, dx \\ & = \pi \int_0^8 (8\sin x+2(1-\cos 2x))\, dx = \pi \left.(-8\cos x+2x-sin 2x)\right\vert_0^\pi \\ & = \pi(8+2\pi+8) = 2\pi (\pi +8) \end{align*} as desired.

::: {#thm- } Let $$A(y)$$ be the cross-sectional area of a disk of outer radius $$R(y)$$ and inner radius r(y).
The volume of the solid of revolution about the $$y$$-axis is $\begin{equation} V = \int_c^d A(y) \, dy = \int_c^d \pi \left( [ R(y) ]^2 - [ r(y) ]^2 \right) \, dy. \end{equation}$ :::

Example 5.12 Find the volume of the solid that results when the region enclosed by the curves $$x=1-y^2$$, $$x=2+y^2$$, $$y=1$$, $$y=-1$$ is revolved about the $$y$$-axis.

Solution. First we sketch the bounded region.

The washer has outer radius $$R(y)=2+y^2$$ and inner radius $$r(y)=1-y^2$$, so using the Washer Method the volume is \begin{align*} V & = \pi \int_{-1}^1 [(2+y^2)^2-(1-y^2)^2] \, dy \\ & = \pi \int_{-1}^1 (3+6y^2) \, dy = 10\pi \end{align*} as desired.

Example 5.13 Find the volume of the solid generated by the region bounded by $$y=\ln x$$ and the $$y$$-axis on the interval $$0\leq y \leq 1$$ and is revolved about the line $$x=-1$$.

Solution. First we sketch the bounded region.

The washer has outer radius $$R(y)=1+e^y$$ and inner radius $$r(y)=1$$, so using the Washer Method the volume is \begin{align*} V & = \pi \int_0^1 ((e^y+1)^2-1^2) \, dy \\ & = \pi \int_0^1 (2 e^y +e^{2y}) \, dy = \pi \left.\left( 2e^y +\frac{1}{2} e^{2y}\right)\right\vert_0^1 \\ & = \pi \left(2e+\frac{1}{2} e^2 -\frac{5}{2}\right) = \frac{\pi}{2} (e^2+4e-5). \end{align*} as desired.

## 5.8 The Shell Method

::: {#thm- } Let $$f$$ be a continuous nonnegative function of $$x$$ on $$[a,b]$$, where $$0\leq a\leq b$$, and let $$R$$ be the region under the graph of $$f$$ on the interval $$[a,b]$$. The volume $$V$$ of the solid of revolution generated by revolving $$R$$ about the $$y$$-axis is $\begin{equation} \label{shellmethod} V = \lim_{n\to \infty} \sum_{k=1}^n 2\pi c_k f(c_k) \Delta x = \int_a^b 2\pi x f(x) \, dx. \end{equation}$ :::

Example 5.14 Find the volume of the solid obtained by rotating about the $$y$$-axis the region bounded by $$y=2x^2-x^3$$ and $$y=0$$.

Solution. First we sketch the bounded region.

The circumference is $$2\pi x$$ and height is $$f(x)=2x^2-x^3$$ so by the Shell Method the volume is \begin{align*} V & = \int_0^2 (2\pi x)(2x^2-x^3) \, dx = 2\pi \int_0^2 (2x^3-x^4) \, dx = \frac{16}{5}\pi \end{align*} as desired.

Example 5.15 Find the volume of the solid formed by revolving the region bounded by the graphs of $$y=x^3+x+1$$, $$y=1$$, and $$x=1$$ about the line $$x=2$$.

Solution. First we sketch the bounded region.

Notice the variable of integration should be $$x$$ and so we find the volume using the Shell Method, \begin{align*} V & = 2\pi \int_0^1 (2-x)(x^3+x+1-1)\, dx \\ & = 2\pi \int_0^1 (-x^4+2x^3-x^2+2x)\, dx = \frac{29\pi}{15} \end{align*} as desired.

::: {#thm- } Let $$f$$ be a continuous nonnegative function of $$y$$ on $$[c,d]$$, where $$0\leq c\leq d$$, and let $$R$$ be the region bounded by the graph of $$f$$ on the interval $$[c,d]$$ and the $$y$$-axis. The volume $$V$$ of the solid of revolution generated by revolving $$R$$ about the $$x$$-axis is $\begin{equation} V = \lim_{n\to \infty} \sum_{k=1}^n 2\pi a_k f(a_k) \Delta y = \int_c^d 2\pi y f(y) \, dy. \end{equation}$ :::

Example 5.16 The region bounded by the curve $$y=\sqrt{x}$$, the $$x$$-axis, and the line $$x=4$$ is revolved about the $$x$$ axis to generate a solid. Find the volume of the solid.

Solution. First we sketch the bounded region.

We find the volume using the Shell Method \begin{align*} V & = \int_0^2 2\pi y (4-y^2)\, dy = \int_0^2 2\pi (4y-y^3)\, dy = 8 \pi \end{align*} as desired.

Example 5.17 The region bounded by the curve $$y=x^2$$, the $$x$$-axis, and the line $$x=1$$ is revolved about the line $$y=2$$ to generate a solid. Find the volume of the solid.

Solution. First we sketch the bounded region.

\begin{align*} V & = 2\pi \int_0^1 (2-y)(1-\sqrt{y}) \, dy \\ & = 2\pi \int_0^1 (2-y-2y^{1/2}+y^{3/2})\, dy = \frac{17\pi}{15} \end{align*} as desired.

## 5.9 Exercises

Exercise 5.12 Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the the given axis. - $$y=x^2+1$$, $$y=x+3$$, about the $$x$$-axis - $$y=4-x^2$$, $$y=2-x$$, about the $$x$$-axis - $$x=1/y$$, $$x=0$$, $$y=1$$, $$y=2$$, about the $$y$$-axis - $$x=\sqrt{4-y^2}$$, $$x=0$$, $$y=0$$, about the $$y$$-axis

Exercise 5.13 Use the disk method to verify that the volume of a sphere is $$\frac{4}{3} \pi r^2$$, where $$r$$ is the radius.

Exercise 5.14 Use the disk method to verify that the volume of a cone is $$\frac{1}{3} \pi r^2 h$$, where $$r$$ is the radius of the base and $$h$$ is the height.

Exercise 5.15 Find the volume of the solid generated by revolving the region bounded by $$y=\sqrt{x}$$ and the lines $$y=2$$ and $$x=0$$ about

• the $$x$$-axis
• the line $$y=2$$
• the $$y$$-axis
• the line $$x=4$$

Exercise 5.16 Find the volume of the solid region. The solid lies between planes perpendicular to the $$x$$-axis at $$x=0$$ and $$x=4$$. The cross-sections perpendicular to the axis on the interval $$0\leq x \leq 4$$ are squares whose diagonals run form the parabola $$y-\sqrt{x}$$ to the parabola $$y=\sqrt{x}$$.

Exercise 5.17 Find the volume of the solid generated by revolving the region bounded by $$y=4+2x-x^2$$ and the line $$y=4-x$$.

• the $$x$$-axis
• the line $$y=1$$

Exercise 5.18 Using integration, find the volume of the solid generated by revolving the triangular region with vertices $$(0,0), (b,0), (0,h)$$ about

• the $$x$$-axis
• the $$y$$-axis

Exercise 5.19 Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated line. - $$y=x$$, $$y=x^2$$, the line $$y=2$$ - $$y=x^2$$, $$y=\frac{1}{2} x^2+2$$, about the line $$y=5$$ - $$x=\sqrt{9-y^2}$$, $$x=0$$, $$y=0$$, about the $$x$$-axis - $$y=\sqrt{x-1}$$, $$y=x-1$$, about the $$y$$-axis - $$x=y-y^2$$, $$x=0$$; about the $$y$$-axis - $$y=1/x$$, $$y=0$$, $$x=1$$, $$x=3$$, about the line $$y=-1$$ - $$y=x$$, $$y=0$$, $$x=2$$, $$x=4$$, about the line $$x=1$$

Exercise 5.20 Use the shell method to find the volume of the solids generated by revolving the regions bounded by the curves and lines. - $$x=1+y^2$$, $$x=0$$, $$y=1$$, $$y=2$$, about the $$x$$-axis - $$x=1+(y-2)^2$$, $$x=2$$, about the $$x$$-axis - $$y=\sqrt{9-x^2}$$, $$y=\frac{2}{3}\sqrt{9-x^2}$$, $$x\geq 0$$, about the $$y$$-axis - $$y=x^2+1$$, $$y=0$$, $$x=0$$, $$x=2$$, about the line $$x=3$$ - $$y=\frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$, $$y=0$$, $$x=0$$, $$x=1$$, about the $$y$$-axis - $$y=\frac{1}{3} x^2$$, $$y=6x-x^2$$, about the line $$x=3$$

Exercise 5.21 Let $$f$$ be the function defined by $\begin{equation} f(x) =\begin{cases} \sqrt{x} & \text{ if 0\leq x \leq 1} \\ x^2-2x+2 & \text{ if 1< x \leq 2.} \end{cases} \end{equation}$ Find the volume of the solid generated by revolving the region under the graph of $$f$$ on $$[0,2]$$ about the $$x$$-axis.

Exercise 5.22 Compute the volume of the solid generated by revolving the region bounded by $$y=x$$ and $$y=x^2$$ about each coordinate axis using

• the shell method
• the washer method

Exercise 5.23 Compute the volume of the solid generated by revolving the triangular region bounded by the lines $$2y=x+4$$, $$y=x$$, and $$x=0$$ about

• the $$x$$-axis using the washer method
• the $$y$$-axis using the shell method
• the line $$x=4$$ using the shell method
• the line $$y=8$$ using the washer method

Exercise 5.24 Each integral represents the volume of a solid. Describe the solid. - $$2\pi \int_0^2 \frac{y}{1+y^2} \, dy$$ - $$\int_0^{\pi/4} 2\pi (\pi-x) (\cos x -\sin x) \, dx$$ - $$\pi \int_{-r}^r [(R+\sqrt{r^2-x^2})^2 -(R-\sqrt{r^2-x^2})^2 ] \, dx$$

Exercise 5.25 Use the disk method or the shell method to find the volumes of the solid generated by revolving the region bounded by the graphs of $$y=10/x^2$$, $$y=0$$, $$x=1$$, $$x=5$$.

• the $$x$$-axis
• the $$y$$-axis
• the line $$y=10$$

Exercise 5.26 Use the disk method or the shell method to find the volumes of the solid generated by revolving the region bounded by the graphs of $$x^{1/2}+y^{1/2}=a^{a^1/2}$$, $$x=0$$, $$y=0$$.

• the $$x$$-axis
• the $$y$$-axis
• the line $$x=a$$

Exercise 5.27 A solid has a circular base of radius 2, and its parallel cross-sections perpendicular to its base are isosceles right triangles oriented so that the endpoints of the hypotenuse of a triangle lie on the circle. Find the volume of the solid.

Exercise 5.28 The region in the first quadrant that is bounded above by the curve $$y=1/\sqrt{x}$$ on the left by the line $$x=1/4$$, and below by the line $$y=1$$ is revolved about the $$y$$-axis to generate a solid. Find the volume of the solid by

• the washer method
• the shell method

Exercise 5.29 Use the method of cylindrical shells to find the volume of the ellipsoid obtained by revolving the elliptical region enclosed by the graph of $\begin{equation} \frac{x^2}{a^2}+ \frac{y^2}{b^2} =1, \quad x\geq 0 \end{equation}$ about the $$y$$-axis.

Exercise 5.30 Set up an integral for the volume of a solid torus (doughnut) with radii $$r$$ and $$R$$. By interpreting the integral as an area, find the volume of the torus.

Exercise 5.31 A bowl is shaped like a hemisphere with diameter 30 cm. A ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of $$h$$ centimeters. Find the volume of water in the bowl.

Exercise 5.32 A hole of radius $$r$$ is bored through the center of a sphere of radius $$R>r$$. Find the volume of the remaining portion of the sphere.

## 5.10 Arc Length and Surfaces of Revolution

We motivate an integration formula for finding the arc length of a smooth curve in a plane. Similarly, we use a formula for finding the surface area of the solid obtained by revolving a curve about an axis. These formulas are demonstrated with several examples and exercises are provided at the end to develop integration skills.

## 5.11 Arc Length

To find a nice usable formula for finding the we will work with .

Definition 5.4 Let $$y=f(x)$$ be a plane curve over an interval $$[a,b]$$. We say that $$f$$ is smooth whenever $$f'$$ is continuous on $$[a,b]$$.

To find the arc length of a smooth curve we begin by dividing the curve into small segments, or subcurves .
Then we approximate the curve segments by line segments and sum the lengths of the line segments to form a Riemann sum. We then let the number of segments increase, the corresponding Riemann sums approach a definite integral (under our assumption of smooth curve) whose value we will take to be the arc length $$L$$ of the curve. Let’s see this in more detail.

Consider the $$k$$-th subinterval $$[x_{k-1}, x_k]$$ and the line segment between the points $\begin{equation} (x_{k-1}, f(x_{k-1})) \qquad \text{and} \qquad (x_k, f(x_k)). \end{equation}$ The change in the $$y$$-coordinate is denoted by $\Delta y_k = f(x_k)-f(x_{k-1}).$ Using the Pythagorean theorem, the length of each line segment is then $\begin{equation} \label{pthm} \sqrt{(\Delta x)^2+(\Delta y_k)^2} \qquad \text{ for k=1,2, \ldots ,n.} \end{equation}$ Summing these lengths we obtain an approximation for the arc length $$L$$ as $\begin{equation} \label{approx} L\approx \sum_{k=1}^n \sqrt{(\Delta x)^2+(\Delta y_k)^2}. \end{equation}$

Manipulating the inside expression we obtain $\begin{equation} L \approx \sum_{k=1}^n \sqrt{(\Delta x)^2\left(1+\left(\frac{\Delta y_k}{\Delta x}\right)^2\right)} = \sum_{k=1}^n \sqrt{1+\left(\frac{\Delta y_k}{\Delta x}\right)^2} \Delta x. \end{equation}$ Using the Mean Value Theorem we obtain $\begin{equation} L \approx \sum_{k=1}^n \sqrt{1+ [f'(x_k^*)]^2} \Delta x \end{equation}$ where $$x_k^*$$ is in the $$k$$-th subinterval $$[x_{k-1}, x_k]$$.

::: {#thm- } If $$y=f(x)$$ is a smooth curve on the interval $$[a,b]$$, then the arc length $$L$$ of this curve over $$[a,b]$$ is given as $\begin{equation} \label{arclength} L=\int_a^b \sqrt{1+[f'(x)]^2}\, dx. \end{equation}$ :::

Example 5.18 Find the exact arc length of the curve $$y=\frac{x^6+8}{16x^2}$$ over the interval $$x=2$$ to $$x=3$$.

Solution. Notice that $$f'(x)= \frac{1}{4} x^3-x^{-3}$$ and so \begin{align*} L & = \int_2^3 \sqrt{1+ \left(\frac{1}{16} x^6-\frac{1}{2}+x^{-6}\right)} \, dx \\ & = \int_2^3 \sqrt{\left(\frac{1}{4}x^3+x^{-3}\right)^2} \, dx = \int_2^3 \left(\frac{1}{4}x^3 + x^{-3}\right) \, dx = \frac{595}{144}. \end{align*} as desired.

Example 5.19 Find the arc length of the graph of $$y=\ln \cos x$$ on $$[0,\frac{\pi}{4}]$$.

Solution. First we find that $$y'=-\tan x$$. Now we apply the Arc Length Formula to find \begin{align*} L & = \int_0^{\pi/4} \sqrt{1+(-\tan x)^2}\, dx = \int_0^{\pi/4} \sec x \, dx \\ & = \int_0^{\pi/4} \sec x \left(\frac{\sec x+\tan x}{\sec x+\tan x} \right) \, dx. \end{align*} Now to continue we let $$u =\sec x+\tan x$$. Then we find $$du = (\sec x \tan x+\sec^2 x)\, dx$$ and so we find the arc length as \begin{align*} L & = \int_1^{1+\sqrt{2}} \left(\frac{1}{u}\right) \, du = \ln (1+\sqrt{2}) \end{align*} as desired.

Switching the roles of $$x$$ and $$y$$, we obtain the arc length for a smooth curve having the form $$x=g(y)$$.

::: {#thm- } If $$x=f(y)$$ is a smooth curve on the interval $$[c,d]$$, then the arc length $$L$$ of this curve over $$[c,d]$$ is given as $\begin{equation} L=\int_c^d \sqrt{1+[g'(y)]^2}\, dy. \label{arclengtheq} \end{equation}$ :::

Example 5.20 Find the exact arc length of the curve $$x=\frac{1}{8}y^4+\frac{1}{4}y^{-2}$$ over the interval $$y=1$$ to $$y=4$$.

Solution. Let $$x=g(y)=\frac{1}{8}y^4+\frac{1}{4}y^{-2}$$ and notice that $$g'(y)= \frac{1}{8} y^2-2y^{-2}$$ and so \begin{align*} L & = \int_1^4 \sqrt{1+\left(\frac{1}{4}y^6 -\frac{1}{2}+\frac{1}{4}y^{-6}\right)}\, dy \\ & = \int_1^4 \left(\frac{1}{2}y^3+\frac{1}{2} y^{-3}\right) \, dy = \frac{2055}{64} \end{align*} as desired.

Example 5.21 Find the arc length of the curve given by $$y=\ln(x-\sqrt{x^2-1})$$, for $$1\leq x \leq \sqrt{2}$$ by integrating wth respect to $$y$$. :::

Solution. Notice that $$e^y=x-\sqrt{x^2-1}$$. Also notice that $\begin{equation} e^{-y} = \frac{1}{x-\sqrt{x^2-1}} = \frac{1}{x-\sqrt{x^2-1}} \left(\frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}}\right) \end{equation}$ We obtain $$e^{-y} = x+\sqrt{x^2-1}$$. Now by adding $$e^y$$ and $$e^{-y}$$, we obtain $$x=\frac{1}{2}(e^y +e^{-y})$$. Therefore, $$\frac{dx}{dy} = \frac{e^y-e^{-y}}{2}$$ and so the arc length is \begin{align*} L & = \int_{\ln(\sqrt{2}-1)}^0 \sqrt{1+\left(\frac{e^y-e^{-y}}{2}\right)^2}\, dy = \left.\left(\frac{e^y-e^{-y}}{2}\right)\right|^0_{\ln (\sqrt{2}-1)} = 1 \end{align*} as desired.

## Surfaces of Revolution

::: {#thm- } Let $$f$$ be a nonnegative smooth function on $$[a,b]$$. The surface area of the surface obtained by revolving the graph of $$f$$ about the $$x$$-axis is $\begin{equation} S=2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} \, dx. \end{equation}$

Example 5.22 Find the surface area generated by revolving the curve $$y= \sqrt{2x-x^2}$$ on $$0.5\leq x \leq 1.5$$.

Solution. We have $$\frac{dy}{dx}=\frac{1-x}{\sqrt{2x-x^2}}$$ and so using the Surface Area Theorem we have \begin{align*} S & = \int_{0.5}^{1.5} 2\pi \sqrt{2x-x^2} \sqrt{1+\frac{(1-x)^2}{2x-x^2}}\, dx \\ & = 2\pi \int_{0.5}^{1.5} \sqrt{2x-x^2} \left(\frac{\sqrt{2x-x^2+1-2x+x^2}}{\sqrt{2x-x^2}}\right) \, dx \\ & = 2\pi \int_{0.5}^{1.5} \, dx = 2\pi \end{align*} as desired.

Example 5.23 Verify that the surface area of a spher of radius $$r$$ is $$S=4\pi r^2$$ by evaluating a definite integral.

Solution. Let $$y=\sqrt{r^2-x^2}$$. Then $$y'=-\frac{x}{\sqrt{r^2-x^2}}$$ and so $\begin{equation} 1+(y')^2 = \frac{r}{r^2-x^2}. \end{equation}$ Therefore we find that $\begin{equation} S = 2\pi \int_{-r}^r y \left(\frac{r}{\sqrt{r^2-x^2}}\right)\, dx = 2\pi r \int_{-r}^r \, dx = 4\pi r^2 \end{equation}$ as desired.

::: {#thm- } Let $$g$$ be a nonnegative smooth function on $$[c,d]$$. The surface area of the surface obtained by revolving the graph of $$g$$ about the $$y$$-axis is $\begin{equation} S=2\pi \int_c^d f(y) \sqrt{1+[g'(y)]^2} \, dy. \end{equation}$ :::

Example 5.24 Find the area of the surface obtained by revolving the curve $\begin{equation} x=\frac{1}{6} y^3+\frac{1}{2y} \end{equation}$ for $$1\leq y \leq 2$$ about the $$y$$-axis.

Solution. Notice that $$x'=\frac{y^4-1}{2y^2}$$ and so $\begin{equation} 1 + (x')^2 = 1+\frac{(y^4-1)^2}{4y^4} = \frac{(y^4+1)^2}{4y^4} \end{equation}$ Using the Surface Area Theorem we find the surface area to be \begin{align*} S & = \pi \int_1^2 \left(\frac{y^3}{6}+\frac{1}{2y}\right)\left(y^2+\frac{1}{y^2}\right) \, dy \\ & = \pi \int_1^2 \left(\frac{y^5}{6}+\frac{2y}{3}+\frac{1}{2y^3}\right) \, dy = \frac{47\pi}{16} \end{align*} as desired.

## 5.12 Exercises

Exercise 5.33 Find the length of the curves. - $$y=(1/3)(x^2+2)^{3/2}$$ from $$x=0$$ to $$x=3$$ - $$y=(x^3/3)+x^2+x+1/(4x+4)$$ from $$x=0$$ to $$x=2$$ - $$y=(2/3) (x^2+1)^{3/2}$$, from $$x=1$$ to $$x=4$$ - $$y=x^2\cos x$$ from $$x=0$$ to $$x=\pi$$ - $$y=x^3-x^2$$ from $$x=-1$$ to $$x=1$$ - $$y=(1/2)(e^x + e^{-x})$$ from $$x=0$$ to $$x=\ln 2$$ - $$x=\sqrt{36-y^2}$$, $$0\leq y \leq 3$$ - $$x=(1/3)\sqrt{y}(y-3)$$, $$1\leq y \leq 4$$ - $$x=2e^{\sqrt{2}y} +(1/16) e^{-\sqrt{2}y}$$, $$0\leq y \leq (\ln 2)/\sqrt{2}$$ - $$x=y^4/4+1/(8y^2)$$, $$1\leq y \leq 2$$

Exercise 5.34 Use integration to find the lateral surface area of the cone generated by revolving the line segment $$y=x/2$$, $$0\leq x \leq 4$$ about the $$y$$-axis.

Exercise 5.35 Find the areas of the surfaces generate by revolving the curves about the indicated axis. - $$y=x^3/9$$, $$0\leq x \leq 2$$, $$x$$-axis - $$y=\sqrt{x}$$ on $$[4,9]$$, $$x$$-axis - $$x=y^3/3$$, $$0\leq y \leq 1$$, $$y$$-axis - $$y=\sqrt{4x+6}$$, $$0\leq x \leq 5$$, $$x$$-axis - $$y=(1/3)\sqrt{y(3-y)^2}$$ on $$0\leq y \leq 3$$, $$y$$-axis - $$x=2\sqrt{4-y}$$, $$0\leq y \leq 15/4$$, $$y$$-axis - $$y=x^2/4$$, $$2\leq x \leq 4$$, $$y$$-axis

Exercise 5.36 Find the area of the surface generated by revolving about the $$x$$-axis the portion of the astroid $$x^{2/3}+y^{2/3}=1$$, $$0\leq x \leq 1$$.

Exercise 5.37 Find the area of the surface obtained by revolving the graph of $$y=\sqrt{4-x^2}$$ on $$[0,1]$$ about the $$x$$-axis

Exercise 5.38 Which curve has greater length on the interval $$[-1,1]$$, $$y=1-x^2$$ or $$y=\cos(\pi x/2)$$?

Exercise 5.39 Find the surface area of an ellipsoid. Use the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to represent the curve that is rotated about an axis to generate the solid, under a suitable restriction.