8  Line Integrals

Let \(C\) be a smooth curve, with parametric equations \(x=x(t),\) \(y=y(t),\) and \(z=z(t)\) for \(a\leq t\leq b,\) that lies within the domain of a function \(f(x,y,z).\) We say that \(C\) is orientable if it is possible to describe direction along the curve for increasing \(t.\) Partition \(C\) into \(n\) sub-arcs, the \(k \text{th}\) of which has length \(\triangle s_k.\) Let \(\left(x_k^*,y_k^*,z_k^*\right)\) be a point chosen (arbitrarily) from the \(k \text{th}\) sub-arc. Form the Riemann sum \[ \sum _{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle s_k\] and let \(\|\triangle s\|\) denote the largest sub-arc length in the partition. The following limit \[ \lim _{\|\text{$\triangle $s}\|\to 0}\sum _{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle s_k\] is called the line integral of \(f\) over \(C\) and is denoted by \[ \int _Cf(x,y,z)\, dS. \] Additionally, if \(C\) is a closed curve, then we denote the line integral by \[ \oint _C\text{ }f \, dS. \]

Theorem 8.1 Let \(f,\) \(f_1,\) and \(f_2\) be continuous scalar functions defined on a piecewise smooth orientable curve \(C.\) Then for any constants \(k_1\)and \(k_2,\)

  • \(\int _C \left(k_1 f_1+k_2f_2\right) \, dS=k_1\int _Cf_1 \, dS+k_2\int _Cf_2 \, dS\)
  • \(\int _C f \, dS=\int _{C_1}f \,\, dS+\cdots+\int _{C_n}f \,\, dS\) where \(C\) is the union of smooth orientable sub-arcs \(C=C_1\cup C_2\cup \cdots \cup C_n\) with only endpoints in common.
  • \(\int _{-C} f \, dS=-\int _C f \, dS.\)

8.1 Evaluating Line Integrals Using Parametrization

Theorem 8.2 Suppose that the function \(f\) is continuous at each point on a smooth curve \(C\), with parametric equations \(x=x(t),\) \(y=y(t),\) and \(z=z(t)\) for \(a\leq t\leq b,\) that lies within the domain of \(f.\) Then \[ \int _Cf(x,y,z)\, dS=\int _a^bf( x(t),y(t),z(t) )\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2} \, dt . \] where \(\, dS=\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2} \, dt\).

The definition of a line integral can be extended to curves that are piecewise smooth in the sense that they are the union of a finite number of smooth curves with only endpoints in common. In particular, if \(C\) is comprised of a number of smooth sub-arcs \(C_1,C_2,\ldots,C_n,\) then \[ \int _C f(x,y,z)\, dS=\int _{C_1} f(x,y,z)\, dS+\cdots +\int _{C_n} f(x,y,z)\, dS. \]

Example 8.1 Evaluate the line integral \[ \int _Cy \sin z \, dS, \] where \(C\) is the circular helix given by the equations \(x=\cos t, y=\sin t, z=t,\) and \(0\leq t\leq 2\pi\).

Solution. We determine \[\begin{align*} \int _C y \sin z \, dS & =\int _0^{2\pi }(\sin t) \sin t \sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2+\left(\frac{d z}{d t}\right)^2}dt \\ & =\int _0^{2\pi }\sin ^2 t \sqrt{\sin ^2t+\cos ^2t+1}dt \\ & =\sqrt{2}\int _0^{2\pi }\frac{1}{2}(1-\cos 2t)dt \\ & =\frac{\sqrt{2}}{2}\left[t-\frac{1}{2}\sin 2t\right]_0^{2\pi } =\sqrt{2}\pi. \end{align*}\]

Theorem 8.3 Let \(C\) be a smooth curve and let \(f(x,y,z)\) be a continuous function with domain containing the trace of \(C.\) Then the value of the integrals \[ \int _C f \, dS, \quad \int _C f dx, \quad \int _C f dy, \quad \int _C f dz\] depen, dS only on the initial point \(A,\) terminal point \(B,\) and the trace of \(C.\) That is, two different parameterizations having the same trace from \(A\) to \(B\) yield the same values for these integrals.

Example 8.2 Suppose the smooth curves \(C_1\) and \(C_2\) are given by
\[\begin{align*} & C_1: \quad x=t, \, y=t^2 \quad \text{ for } 0\leq t\leq 1, \text{ and} \\ & C_2: \quad x=\sin t, \, y=\sin ^2 t\quad \text{ for } 0\leq t\leq \frac{\pi }{2}. \end{align*}\] Evaluate
\[ \int _{C_1} x \, dS \quad \text{ and } \quad \int _{C_2} x \, dS. \]

Solution. Both \(C_1\) and \(C_2\) are smooth curves from \((0,0)\) to \((1,1)\) with the same trace which is the portion of the parabola \(y=x^2\) for \(0\leq x\leq 1.\) For \(C_1,\) we have \(x=t\) and \(\, dS=\sqrt{1+4t^2}dt,\) therefore \[ \int _{C_1} x \, dS=\int _0^1t\sqrt{1+4t^2}dt=\frac{1}{12}\left(17^{3/2}-1\right). \] For \(C_2,\) we have \(x=\sin t\) and \(\, dS=\sqrt{\cos ^2t+4\sin ^2t\text{ }\cos ^2t}dt,\) therefore \[\begin{equation*} \int _{C_2} x \, dS=\int _0^{\pi /2} \sin t\text{ }\cos t\sqrt{1+4\sin ^2t}dt=\frac{1}{12}\left(17^{3/2}-1\right). \end{equation*}\]

8.2 Line Integrals with Respect to Coordinate Variables

Other line integrals are obtained by replacing \(\triangle s_k\) by \(\triangle x_k=x_k-x_{k-1}.\) This is called the line integral of \(f\) along \(C\) with respect to \(x\): \[ \int _C f(x,y,z) \,dx=\lim _{\| \triangle x\|\to 0}\sum _{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle x_k.\] Similarly with respect to \(y\) and \(z\), we define \[ \int _C f(x,y,z) \,dy=\lim _{\| \triangle y \|\to 0}\sum _{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle y_k\] \[ \int _C f(x,y,z) \,dz=\lim _{\| \triangle z\|\to 0}\sum _{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle z_k.\] When we want to distinguish the original line integral \(\int _Cf \,\, dS\) from these, we call it the line integral with respect to arc length.

The following formulas say that the line integrals with respect to \(x,\) \(y,\) or \(z\) can also be evaluated by expressing everything in terms of \(t\): \(x=x(t),\) \(y=y(t),\) \(z=z(t),\) \(dx=x'(t),\) \(dy=y'(t),\) and \(dz=z'(t)\) yielding: \[\int _C f(x,y,z)\,d x=\int _a^b f( x(t),y(t),z(t) )x'(t)\,dt\] \[\int _C f(x,y,z)\,d y=\int _a^b f( x(t),y(t),z(t) )y'(t)\,dt\] and \[\int _C f(x,y,z)\,d z=\int _a^b f( x(t),y(t),z(t) )z'(t)\,dt\] assuming \(f\) is continuous and \(C\) lies within the domain of \(f\).

Example 8.3 Evaluate the line integral \[ \int _C y \,dx+z \,dy+x \,dz, \] where \(C\) consists of the line segment \(C_1\) from \((2,0,0)\) to \((3,4,5)\) followed by the vertical line segment \(C_2\) from \((3,4,5)\) to \((3,4,0)\)

Solution. The curve \(C\) is the union of the curves \[\begin{align*} & C_1: \vec{R}_1(t)=(1-t)\langle 2,0,0\rangle+t \langle 3,4,5\rangle = \langle 2+t,4t,5t\rangle \text{ for } 0\leq t\leq 1 \\ & C_2: \vec{R}_2(t)=(1-t)\langle3,4,5\rangle+t\langle3,4,0\rangle=\langle3,4,5-5t\rangle \text{ for } 0\leq t\leq 1. \end{align*}\] Thus \[\begin{align*} \int _C y \, dx+z\, dy+x\, dz & =\int _{C_1} y \, dx+z\, dy+x\, dz+\int _{C_2} y \, dx+z \, dy+x \, dz \\ & =\int_0^1 ( (4t)+5t(4)+(2+t)5) \, dt +\int _0^1( 0+0+3(-5) )\, dt \\ & =\int_0^1 (10+29t) \, dt +\int _0^1 (-15)\, dt %\\ & =\left[10t+29\frac{t^2}{2}\right]_0^1 +\left[-15t\right]_0^1 %= 24.5 -15 =9.5 \end{align*}\]

8.3 Line Integral of Vector Field Along a Curve

Theorem 8.4 Let \[ \vec{V}(x,y,z)=u(x,y,z)\vec{i}+v(x,y,z)\vec{j}+w(x,y,z)\vec{k} \] be a continuous vector field, and let \(C\) be a piecewise smooth orientable curve with parametric representation \[ \vec{R}(t)=x(t) \vec{i}+y(t) \vec{j}+z(t) \vec{k} \] for \(a\leq t\leq b.\) Using \(d\vec{R}=dx \vec{i}+dy \vec{j}+dz \vec{k}\) we define the line integral of \(\vec{V}\) along \(C\) by \[\begin{align*} & \int _C \vec{V}\cdot d \vec{R} =\int _C ( u \, dx +v \, dy +w \, dz) =\int _C \left( \vec{V}[\vec{R}(t)]\cdot \vec{R}'(t) \right)dt \\ & \qquad =\int _a^b \left[u[x(t),y(t),z(t)]\frac{dx}{dt}+v[x(t),y(t),z(t)]\frac{dy}{dt}+w[x(t),y(t),z(t)]\frac{dz}{dt}\right]d t.\end{align*}\]

Theorem 8.5 Let \(\vec{F}\) be a continuous force field over a domain \(D.\) Then the work \(W\) performed as an object moves along a smooth curve \(C\) in \(D\) is given by the integral \[ W=\int _C \vec{F}\cdot \vec{T} \, \, dS=\int _C \vec{F}\cdot d\vec{R} \] where \(\vec{T}\) is the unit tangent at each point on \(C\) and \(\vec{R}\) is the position vector of the object moving on \(C.\)

Example 8.4 Evaluate the line integral \[ \oint \vec{F}\cdot \, d\vec{R} \] where \(\vec{F}=y^2\vec{i} +x^2 \vec{j}-(x+z)\vec{k}\) and \(C\) is the boundary of the triangle with vertices \((0,0,0),\) \((1,0,0),\) and \((1,1,0)\), transversed once clockwise, as viewed from above.

Solution. First let \(C_1\) be the line segment from \((1,0,0)\) to \((0,0,0).\) Then \(C_1\) can be represented by the vector function
\[ \vec{R}_1(t)=(1-t)\langle 1,0,0\rangle +t\langle 0,0,0\rangle =\langle 1-t,0,0\rangle \] with \(0\leq t\leq 1.\) Let \(C_2\) be the line segment from \((0,0,0)\) to \((1,1,0)\) and so can be represented by the vector function,
\[ \vec{R}_2(t)=(1-t)\langle 0,0,0\rangle +t\langle1,1,0\rangle =\langle t,t,0\rangle \]
with \(0\leq t\leq 1.\) Let \(C_3\) be the line segment from \((1,1,0)\) to \((1,0,0)\) and so can be represented by the vector function,
\[ \vec{R}_3(t)=(1-t)\langle1,1,0\rangle+t\langle1,0,0\rangle =\langle 1,1-t,0\rangle \] with \(0\leq t\leq 1\). Then
\[\begin{align*} & \vec{F}_1=\left(1-t^2\right)\vec{j}-(1-t)\vec{k} & & d\vec{R}_1=\langle -dt,0,0\rangle \\ & \vec{F}_2=t^2\vec{i}+t^2\vec{j}-t \vec{k} & & d\vec{R}_2=\langle dt,dt,0\rangle \\ & \vec{F}_3=(1-t)^2\vec{i}+\vec{j}-\vec{k} & & d\vec{R}_3=\langle 0,-dt,0\rangle\end{align*}\] We find \[\begin{align*} \int_C \vec{F}\cdot d\vec{R} & =\underset{C_1}{\int }\vec{F}_1\cdot d\vec{R}_1+\underset{C_2}{\int }\vec{F}_2\cdot d\vec{R}_2+\underset{C_3}{\int }\vec{F}_3\cdot d\vec{R}_3 \\ & =\int_0^1 0 \, dt+\int_0^1 2t^2 \, dt+\int_0^1 (-1) \, dt=-\frac{1}{3}. \end{align*}\]

8.4 Independence of Path

Definition 8.1 The line integral is called independent of path if in a region \(D\), if for any two points \(P\) and \(Q\) in \(D\) then the line integral along every piecewise smooth curve in \(D\) from \(P\) to \(Q\) has the same value.

::: {#thm- } Independence of Path If \(\vec{V}\) is a continuous vector field on the open connected set \(D,\) then the following three conditions are either all true or all false:

  • \(\vec{V}\) is conservative on \(D\)
  • \(\int _C \vec{V}\cdot d\vec{R}=0\) for every piecewise smooth closed curve \(C\) in \(D\).
  • \(\int _C \vec{V}\cdot d\vec{R}\) is independent of path within \(D\).

:::

Example 8.5 Let \(F=\langle y,-x\rangle\) and let \(C_1\) and \(C_2\) be the following two paths joining \((0,0)\) to \((1,1)\); \(C_1\): \(y=x\) for \(0\leq x\leq 1\) and \(C_2\): \(y=x^2\) for \(0\leq x\leq 1\). Show that \[ \int _{C_1}\vec{F}\cdot d \vec{R}\neq \int _{C_2}\vec{F}\cdot d \vec{R}. \] Explain what this means? On \(C_1:\) \(x=t,\) \(y=t\) with \(0\leq t\leq 1,\) \[ \int_{C_1} y \, dx-x \, dy=\int_0^1 (t-t) \, dt=0. \] On \(C_2:\) \(x=t,\) \(y=t^2,\) with \(0\leq t\leq 1,\) \[ \int_{C_2}y \,dx-x\,dy=\int_0^1 \left(t^2-t(2t)\right) \, dt=-\frac{1}{3}. \]

Solution. Since these two line integrals do not have the same value we see that not all line integrals are independent of path.

8.5 Fundamental Theorem of Line Integrals

::: {#thm- } Fundamental Theorem of Line Integrals Let \(C\) be a piecewise smooth curve that is parametrized by the vector function \(\vec{R}(t)\) for \({a\leq t\leq b}\) and let \(\vec{V}\) be a vector field that is continuous on \(C\). If \(f\) is a scalar function such that \(\vec{V}=\nabla f,\) then \[ \int _C \vec{V}\cdot d\vec{R}=f(Q)-f(P)\] where \(Q=R(b)\) and \(P=R(a)\) are the endpoints of \(C\). :::

Proof. Suppose \(\vec{R}(t)=x(t) \vec{i}+y(t) \vec{j}+z(t) \vec{k}\) and let \(G\) be the composite function \(G(t)=f(x(t),y(t),z(t))\). We have \[\begin{align*} \int _C \vec{F}\cdot d\vec{R} & =\int _C\nabla f\cdot d\vec{R} =\int _C \left[\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\right] \\ & =\int_a^b \left[\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}\right] \, dt \\ & =\int_a^b \frac{d G}{d t} \, dt =G(b)-G(a) =f(Q)-f(P). \end{align*}\]

Example 8.6 Show that the vector field \[ \vec{F}(x,y,z)=\left( \frac{y}{1+x^2}+\tan ^{-1}z\right)\vec{i}+\left(\tan ^{-1}x\right) \vec{j}+\left(\frac{x}{1+z^2}\right)\vec{k} \]
is conservative and find a scalar potential \(f\) for \(F\). Then evaluate the line integral \[ \int _C \vec{F}\cdot d \vec{R}. \] where \(C\) is any piecewise smooth path connecting \(A(1,0,-1)\) to \(B(0,-1,1)\).

Solution. Since \(\text{curl} \vec{V}=\vec{0},\) \(\vec{V}\) is conservative. Now we set out to find \(f\). Since
\[ \frac{\partial f}{\partial x}=\frac{y}{1+x^2}+\tan ^{-1}z \] we set \[ f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c(y,z). \] Since
\[ \frac{\partial f}{\partial y}=\tan ^{-1}z=\frac{\partial }{\partial y}\left(y \tan ^{-1}x+x \tan ^{-1}z+c\right)=\tan ^{-1}x+\frac{\partial c}{\partial y}, \]
so \(\frac{\partial c}{\partial y}=0\) and \(c=c_1(z)\) and so we set
\(f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c_1(z)\). Since \[ \frac{\partial f}{\partial z}=\frac{x}{1+z^2}=\frac{\partial }{\partial z}\left[y \tan ^{-1}x+x \tan ^{-1}z+c_1(z)\right]=\frac{x}{1+z^2}+c'_1(z),\] so \(c_1'(z)=0,\) \(c_1=0\) and so we set
\(f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z\).
\[\begin{equation*} \int _C\vec{F}\cdot d \vec{R}=f(1,0,-1)-f(0,-1,1) =\frac{3\pi }{4}+\frac{\pi }{4} =\pi . \end{equation*}\]

Example 8.7 Show that the vector field \[ \vec{V}(x,y,z)=\left(x y^2+y z\right)\vec{i}+\left(x^2y+x z+3y^2z\right)\vec{j}+\left(x y+y^3\right)\vec{k} \] is conservative and find a scalar potential \(f\) for \(\vec{V}\). Then evaluate the line integral \[ \int _C \vec{F}\cdot d \vec{R}. \] where \(C\) is any piecewise smooth path joining \(A(8,6,1)\) to \(B(1,3,5)\).

Solution. First we determine \(\text{curl} \vec{V}\). We find \[\begin{align*} \text{curl} \vec{V} & =\left| \begin{array}{cc} \begin{array}{c} \vec{i} \\ \begin{array}{c} \partial / \partial x \\ x y^2+y z \end{array} \end{array} & \begin{array}{cc} \vec{j} & \vec{k} \\ \begin{array}{c} \partial / \partial y \\ x^2y+x z+3y^2z \end{array} & \begin{array}{c} \partial / \partial z \\ x y+y^3 \end{array} \end{array} \end{array} \right| \\ & =\left(x+3y^2-x-3y^2\right)\vec{i}-(y-y)\vec{j}+(2x y+z-2x y-z)\vec{k} =\vec{0}\end{align*}\] and so the vector field is conservative and we can find the scalar potential function. Now we set out to find \(f\) with \(\nabla f=\vec{V}\). Since \(f_x=x y^2+y z\) we know that
\[ f(x,y,z)=\frac{x^2}{2}y^2+x y z+C_1(y,z). \] Then we find \[ f_y=x^2y+x z+\frac{ \partial C_1}{\partial y}; \] and so comparing this with the given \(x^2y+x z+3y^2z\) we determine that \[ \frac{ \partial C_1}{\partial y}=3y^2z. \] So \(C_1=y^3z+C_2(z)\). So far we have \[ f=\frac{x^2}{2}y^2+x y z+y^3z+C_2(z). \] Also since \(f_z=x y+y^3+\frac{dC_2}{d z}\) and comparing this to the given \(x y+y^3\) we determine \(\frac{dC_2}{dz}=0\). So \(C_2\) is a constant with respect to \(x,\) \(y,\) and \(z\). Therefore a scalar potential function is \[ f(x,y,z)=\frac{x^2}{2}y^2+x y z+y^3z\] (taking the constant to be zero). Finally \[\begin{equation} \int _C \vec{V}\cdot d\vec{R}=f(1,3,5)-f(8,6,1) %=\frac{2523}{2}. \end{equation}\]

Example 8.8 Show that the vector field is conservative and find a scalar potential \(f\) for \(F\). Then evaluate the line integral \[ \int _C \vec{F}\cdot d \vec{R} = \vec{F}(x,y)=\frac{(y+1)\vec{i}-x \vec{j}}{(y+1)^2} \]
where \(C\) is any smooth path connecting \(A(0,0)\) to \(B(1,1)\).

Solution. Let \(\vec{F}(x,y)=u(x,y)\vec{i}+v(x,y)\vec{j}\), then since \[ \frac{\partial u}{\partial y}=\frac{-1}{(y+1)^2}=\frac{\partial v}{\partial x}, \] we know that \(F\) is conservative. By definition, we know \[ f_x(x,y)=\frac{1}{y+1} \qquad \text{and}\qquad f_y(x,y)=\frac{-x}{(y+1)^2}. \] Integrating with respect to \(x,\) \(f(x,y)=\frac{x}{y+1}+c(y)\). Since,
\[ f_y(x,y)=-\frac{x}{(y+1)^2}+c'(y)=-\frac{x}{(y+1)^2}, \] so \(c'(y)=0\).
For \(c(y)=0;\) and then \(f(x,y)=\frac{x}{y+1}\) is a scalar potential function for \(\vec{F}\). By the fundamental theorem of line integrals, \[\begin{equation*} \int_C \vec{F}\cdot d\vec{R}=f(1,1)-f(0,0)=\frac{1}{2}-0=\frac{1}{2}. \end{equation*}\]

8.6 Work

Example 8.9 Find the work done on an object moves in the force field \[ \vec{F}(x,y,z)=x \vec{i}+y \vec{j}+(x z-y)\vec{k} \] along the curve \(C\) defined parametrically by \(\vec{R}(t)=t^2\vec{i}+2t \vec{j}+4t^3 \vec{k}\) for \(0\leq t\leq 1\).

Solution. We determine \[\begin{align*} & d\vec{R}=\left(2t \vec{i}+2\vec{j}+12t^2\vec{k}\right)dt, \qquad \vec{F}(t)=t^2\vec{i}+2t\vec{j}+\left(4t^5-2t\right)\vec{k}, \end{align*}\] (from \(x(t)=t^2, y(t)=2t,\) and \(z=4t^3\)) and \[ \vec{F}\cdot d\vec{R}=\left(2t^3+4t+48t^7-24t^3\right)dt. \] Thus, \[\begin{equation*} W=\int _C \vec{F}\cdot d\vec{R}=\int_0^1 \left(2t^3+4t+48t^7-24t^3\right) \, dt=\frac{5}{2}. \end{equation*}\]

Example 8.10 Find the work done when an object moves along a closed path in a connected domain where the force field is conservative.

Solution. In such a force field \(\vec{F},\) where \(f\) is a scalar potential of \(\vec{F}\) and because the path of motion is closed, it begins and ends at the same point \(P\). Thus, the work is given by \[ W=\oint _C \vec{F}\cdot d\vec{R}=f(P)-f(P)=0. \]

Example 8.11 Let
\[ \vec{F}(x,y)=\frac{-y \vec{i}+x \vec{j}}{x^2+y^2}. \] Evaluate the line integral \[ \int _{C_1}\vec{F}\cdot d \vec{R} \] where \(C_1\) is the upper semicircle \(y=\sqrt{1-x^2}\) transversed counterclockwise. What is the value of \[ \int _{C_2}\vec{F}\cdot d \vec{R} \] if \(C_2\) is the lower semicircle \(y=-\sqrt{1-x^2}\) also transversed counterclockwise?

Solution. On the upper semi-circle \(C_1\), we let \(x=\cos (t)\) and \(y=\sin (t)\) for \(0\leq t\leq \pi ,\) and then
\[ \int_{C_1} \frac{-y \, dx+x \, dy}{x^2+y^2}=\int_0^{\pi } \frac{-\sin (t)(-\sin (t))+\cos (t)(\cos (t))}{\sin ^2(t)+\cos ^2(t)} \, dt=\int _0^{\pi }dt=\pi . \] For \(\pi \leq t\leq 2\pi ,\) we have the lower semi-circle \(C_2\):
\[\begin{equation*} \int_{C_2}\frac{-y\, dx+x\, dy}{x^2+y^2}=\int _{\pi }^{2\pi }dt=\pi . \end{equation*}\]

8.7 Finding Area with Line Integral

Theorem 8.6 If \(R\) is a region bounded by a piecewise smooth simple closed curve \(C\) oriented counterclockwise, then the area of \(R\) is given by \[ A=\oint _C x dy=-\oint _C y dx=\frac{1}{2}\oint _C x dy-ydx. \]

8.8 Exercises

Exercise 8.1 Evaluate the line integral \[ \int _C\frac{1}{3+y}\, dS \] where \(C\) is the curve with parametric equations \(x=2t^{3/2}\) and \(y=3t\) with \(0\leq t\leq 1.\)

Exercise 8.2 Evaluate the line integral \[ \int _C\left(x^2+y^2\right)\, dS \] where \(C\) is the curve with parametric equations \(x=e^{-t}\cos t\) and \(y=e^{-t}\sin t\) with \(0\leq t\leq \frac{\pi }{2}.\)

Exercise 8.3 Evaluate the line integral \[ \int _C xdy-ydx \] where \(C\) is the curve defined by \(2x-4y=1\) with \(4\leq x\leq 8.\)

Exercise 8.4 Evaluate the line integral \[ \int _C -xdy+\left(y^2-x^2\right)dx \] where \(C\) is the quarter-circle \(x^2+y^2=4\) from \((0,2)\) to \((2,0).\)

Exercise 8.5 Evaluate the line integral \[ \int _C (x+y)^2dx-(x-y)^2 dy \] where \(C\) is the curve defined by \(y=|2x|\) from \((-1,2)\) to \((1,2).\)

Exercise 8.6 Evaluate the line integral \[ \oint \vec{F} \cdot d\vec{R} \] where \(\vec{F}=y^2 \vec{i} +x^2 \vec{j}-(x+z)\vec{k}\) and \(C\) is the boundary of the triangle with vertices \((0,0,0),\) \((1,0,0),\) and \((1,1,0)\), transversed once clockwise, as viewed from above.

Exercise 8.7 Let \(\vec{F}=y \, \vec{i}-x \, \vec{j}\) and let \(C_1\) and \(C_2\) be the following two paths joining \((0,0)\) to \((1,1).\) \[ C_1: y=x \text{ for } 0\leq x\leq 1 \quad \text{and} \quad C_2: y=x^2 \text{ for } 0\leq x\leq 1. \] Show that \[ \int _{C_1}\vec{F}\cdot d \vec{R}\neq \int _{C_2}F\cdot d \vec{R}. \] Explain what this means?

Exercise 8.8 Evaluate the closed line integral \[ \oint _C \vec{F}\cdot \, dS \] where \(\vec{F}=-x \, \vec{i}+2\,\vec{j}\) and \(C\) is the boundary of the trapezoid with vertices \((0,0),\) \((1,0),\) \((2,1)\) and \((0,1)\) transversed once clockwise, as viewed from above.

Exercise 8.9 Find the work done by the force field \[ \vec{F}(x,y,z)=\left(y^2-z^2\right) \vec{i}+2 y z \vec{j}-x^2 \vec{k} \] on any object moving along the curve \(C\) where \(C\) is the path given by \(x(t)=t,\) \(y(t)=t^2,\) and \(z(t)=t^3,\) for \(0\leq t\leq 1.\)

Exercise 8.10 Find the work done by the force field \[ \vec{F}(x,y,z)=2 x y \, \vec{i}+\left(x^2+2\right)\, \vec{j}+y \,\vec{k} \] on any object moving along the curve \(C\) where \(C\) is the line segment from \((1,0,2)\) to \((3,4,1).\)

Exercise 8.11 Show that the vector field \[ \vec{F}(x,y)=(x+2y)\, \vec{i}+(2x+y)\, \vec{j} \] is conservative and find a scalar potential function \(f\) for \(F.\) Then evaluate the line integral \(\int _C \vec{F}\cdot d\vec{R}\) where \(C\) is any smooth path connecting \(A(0,0)\)to \(B(1,1).\)

Exercise 8.12 Show that the vector field \[ \vec{F}(x,y)=\left(y-x^2\right) \, \vec{i}+\left(x+y^2\right) \, \vec{j} \] is conservative and find a scalar potential function \(f\) for \(\vec{F}.\) Then evaluate the line integral \[ \int _C \vec{F}\cdot dR \] where \(C\) is any smooth path connecting \(A(0,0)\)to \(B(1,1).\)

Exercise 8.13 Show that the vector field \[ \vec{F}(x,y)=e^{-y}\, \vec{i}-x e^{-y}\, \vec{j} \] is conservative and find a scalar potential function \(f\) for \(\vec{F}.\) Then evaluate the line integral \[ \int _C\vec{F}\cdot d\vec{R} \] where \(C\) is any smooth path connecting \(A(0,0)\)to \(B(1,1).\)

Exercise 8.14 Show that the vector field \[ \vec{F}(x,y,z)=e^{x y}y z \, \vec{i}+e^{x y}x z \, \vec{j}+e^{x y} \, \vec{k} \] is conservative and find a scalar potential function \(f\) for \(\vec{F}.\)

Exercise 8.15 Show that the vector field \(\vec{F}\) with component functions \(f(x,y,z)=\left(2 x z^3-e^{-x y}y \sin z\right)\), \(g(x,y,z)=-x e^{-x y}\sin z\), and \(h(x,y,z)=\left(3x^2z^2+e^{-x y}\cos z\right)\) is conservative and find a scalar potential function \(f\) for \(\vec{F}.\) Then evaluate the line integral \[ \int _C\vec{F}\cdot d\vec{R} \] where \(C\) is any smooth path connecting \(A(1,0,-1)\)to \(B(0,-1,1).\)

Exercise 8.16 Show that the vector field \[ \vec{F}(x,y)=\frac{(y+1)\vec{i}-x \vec{j}}{(y+1)^2} \] is conservative and find a scalar potential \(f\) for \(\vec{F}.\) Then evaluate the line integral \[ \int _C\vec{F}\cdot d \vec{R} \] where \(C\) is any smooth path connecting \(A(0,0)\) to \(B(1,1).\)

8.9 Green’s Theorem

Suppose \(C\) is a piecewise smooth closed curve oriented counterclockwise in the Cartesian plane whose bounded region \(R\) is simply connected . Suppose further that we have a continuously differentiable vector field of the form \(\vec{F}=M(x,y)\vec{i}\).
Let \([a,b]\) be the projection of \(R\) onto the \(x\)-axis, let \(y_1(x)\) represent the lower part of the curve \(C\), and let \(y_2(x)\) represent the upper part of \(C\). We can write \[\begin{align} \label{gr1} \oint_C \vec{F}\cdot \, d\vec{r} & = \int_a^b M(x,y_1(x)) \, dx + \int_b^a M(x,y_2(x)) \, dx \notag \\ & = \int_a^b \left[M(x,y_1(x)) - M(x,y_2(x)) \right] \, dx \notag \\ & = - \int_a^b M(x,y)\vert_{y_1(x)}^{y_2(x)} \, dx \notag \\ & = - \int_a^b \int_{y_1(x)}^{y_2(x)} \frac{\partial M}{\partial y} \, dy \, dx \notag \\ & =-\iint_R \frac{\partial M}{\partial y} \, dy \, dx \end{align}\] Now let us suppose that we have a continuously differentiable vector field of the form \(\vec{F}=N(x,y)\vec{j}\).
Let \([a,b]\) be the projection of \(R\) onto the \(y\)-axis, let \(x_1(y)\) represent the lower part of the curve \(C\), and let \(x_2(y)\) represent the upper part of \(C\). We can write \[\begin{align} \label{gr2} \oint_C \vec{F}\cdot \, d\vec{r} & = \int_a^b N(x_1(x),y) \, dy + \int_b^a N(x_2(x),y) \, dy\notag \\ & = \int_a^b \left[N(x_1(y),y) - N(x_2(y),y) \right] \, dy \notag \\ & = - \int_a^b N(x,y)\vert_{x_1(y)}^{x_2(y)} \, dy \notag \\ & = \int_a^b \int_{x_2(y)}^{x_1(y)} \frac{\partial N}{\partial x} \, dx \, dy \notag \\ & =\iint_R \frac{\partial N}{\partial x} \, dx \, dy \end{align}\] Finally, suppose that \(\vec{F}\) is a continuously differentiable vector field in the plane, say \[ \vec{F}(x,y)=M(x,y)\vec{i}+N(x,y)\vec{j}. \] Using \(\ref{gr1}\) and \(\ref{gr2}\) we find \[\begin{align*} \oint_C \vec{F}\cdot \, d\vec{r} & = \oint_C M(x,y)\, dx +N(x,y)\, dy \\ & = \oint_C M(x,y) \, dx +\oint_C N(x,y)\, dy \\ & = -\iint_R \frac{\partial M}{\partial y} \, dy \, dx + \iint_R \frac{\partial N}{\partial x} \, dx \, dy \\ & = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A \end{align*}\] as needed to prove the following theorem.

Theorem 8.7 (Green’s Theorem) Let \(R\) be a simply connected region with a piecewise smooth boundary curve \(C\) oriented counterclockwise and let \[ \vec{F}= M \vec{i}+ N \vec{j}+0 \vec{k} \] be a continuously differentiable vector field on \(R,\) then \[\begin{equation} \label{gr} \oint _C M dx+Ndy = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A. \end{equation}\]

Example 8.12 Use Green’s theorem to evaluate the line integral \[ \oint _C4 y \, dx-3x \, dy \] around the curve \(C\) defined by the ellipse \(2x^2+y^2=4\) oriented counterclockwise.

Solution. Let \(\vec{F}(x,y)=4y \vec{i}+(-3x)\vec{j}\) and let \(R\) be the region enclosed by \(C\). Notice that \(\vec{F}\) and \(R\) satisfy the hypothesis of Green’s theorem, so that \[\begin{equation*} \oint _C 4 y \, dx-3x \, dy = \iint_R (-3-4)d A =-7(2)\left(\sqrt{2}\right)\pi. \end{equation*}\]

Example 8.13 Use Green’s theorem to evaluate the line integral \[ \oint _C4x y \, dx\] around the curve \(C\) defined by the unit circle oriented clockwise.

Solution. Let \(\vec{F}(x,y)=4x y \vec{i}+0 \vec{j}\) and let \(R\) be the region enclosed by \(C\). Notice that \(\vec{F}\) and \(R\) satisfy the hypothesis of Green’s theorem, so that \[\begin{equation*} \oint _C4x y \, dx % & =\oint _C M \, dx+N \, dy %=-\iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right) \,d A %=-\iint_R (-4x) \, d A \\ =4\iint_R x \, d A =4\int _0^{2\pi }\int _0^1 r^2 \cos \theta drd\theta =0 \end{equation*}\]

Example 8.14 Use Green’s theorem to evaluate the line integral \[ \oint _Cy^2 \, dx+x\, dy \] around the curve \(C\) defined by the square with vertices \((0,0),\) \((2,0),\) \((2,2),\) \((0,2)\) oriented counterclockwise.

Solution. Let \(\vec{F}(x,y)=y^2 \vec{i}+x \vec{j}\), then \(M(x,y)=y^2\) which is continuously differentiable over the square as well as \(N(x,y)=x.\) Therefore \(\vec{F}\) and \(R\) satisfy the hypothesis of Green’s theorem and \[\begin{equation*} \oint _C y^2 \, dx+x \, dy =\int _0^2\int _0^2(1-2y) \, dy \, dx =\int_0^2 (-2) \, dx =-4. \end{equation*}\]

Example 8.15 Find the work done on an object that moves in the force field \[ \vec{F}(x,y)=y^2 \vec{i}+x^2 \vec{j} \] once counterclockwise around the circular path \(x^2+y^2=2\).

Solution. Let \(\vec{F}(x,y)=y^2 \vec{i}+x^2 \vec{j}\), then \(M(x,y)=y^2\) which is continuously differentiable over the circle as well as \(N(x,y)=x^2.\) Let \(R\) be the region bounded by the curve \(x^2+y^2=2.\) Then \(\vec{F}\) and \(R\) satisfy the hypothesis of Green’s theorem and \[\begin{align*} W& =\oint _C \vec{F}\cdot d\vec{R} =\oint _C \left(y^2 \vec{i}+x^2 \vec{j}\right)\cdot ( dx \vec{i}+dy \vec{j}) =\oint _C \left(y^2dx +x^2dy\right) \\ & =\oint _C M \, dx+N\,dy = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)\, d A =2 \iint_R (x-y) \, dA \\ & =2\int _0^{2\pi }\int _0^{\sqrt{2}}r^2(\cos \theta -\sin \theta )\, dr \, d\theta =\frac{4\sqrt{2}}{3}\int_0^{2\pi } (\cos \theta -\sin \theta ) \, \, d\theta =0. \end{align*}\]

Example 8.16 Find the work done on an object that moves in the force field \[ \vec{F}(x,y)=\left(x+2y^2\right) \vec{j} \] once counterclockwise around the circular path \((x-2)^2+y^2=1\).

Solution. Let \(\vec{F}(x,y)=0 \vec{i}+\left(x+2y^2\right) \vec{j}\). Then \(M(x,y)=0\) which is continuously differentiable over the circle as well as \(N(x,y)=x+2y^2.\) Let \(R\) be the region bounded by the curve \((x-2)^2+y^2=1.\) Then \(\vec{F}\) and \(R\) satisfy the hypothesis of Green’s theorem and \[\begin{align*} W& =\oint _C \vec{F}\cdot d\vec{R} =\oint _C \left(x+2y^2\right)\vec{j}\cdot ( dx \vec{i}+dy \vec{j}) =\oint _C \left(x+2y^2\right) dy \\ & =\oint _C M dx +N dy = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A = \iint_R dA =\pi. \end{align*}\]

Example 8.17 Evaluate the closed line integral \[ \oint _C \frac{-y dx+(x-1)dy}{(x-1)^2+y^2} \] where \(C\) is any Jordan curve whose interior does not contain the point \((1,0)\) transversed counterclockwise.

Solution. Let \[ \vec{F}(x,y)=\frac{-y }{(x-1)^2+y^2} \vec{i}+ \frac{x-1 }{(x-1)^2+y^2} \vec{j}, \] then \[ M(x,y)=\frac{-y }{(x-1)^2+y^2} \] which is continuously differentiable over the circle as well as \[ N(x,y)=\frac{x-1 }{(x-1)^2+y^2}. \] Let \(R\) be the region bounded by the given curve \(C.\) Then \(\vec{F}\) and \(R\) satisfy the hypothesis of Green’s theorem and \[\begin{align*} & \oint _C \frac{-y dx+(x-1)dy}{(x-1)^2+y^2} \\ & \qquad =\oint _C \frac{-y }{(x-1)^2+y^2}dx+\frac{x-1}{(x-1)^2+y^2}dy \\ & \qquad = \iint_R \left(\partial _x \left(\frac{x-1 }{(x-1)^2+y^2}\right)-\partial _y \left(\frac{-y }{(x-1)^2+y^2}\right)\right)dA \\ & \qquad = \iint_R 0 \, dA =0. \end{align*}\]

8.10 Doubly-Connected Regions

Definition 8.2 A Jordan curve is a closed curve \(C\) that does not intersect itself and a simply connected region \(R\) has the property that it is connected and the interior of every Jordan curve \(C\) in \(R\) also lies in \(R.\)

::: {#thm- } [Green’s Theorem for Doubly-Connected Regions] Let \(R\) be a doubly-connected region with a piecewise smooth outer boundary curve \(C_1\) oriented counterclockwise and a piecewise smooth inner boundary curve \(C_2\) oriented clockwise and let \(\vec{F}= M \vec{i}+ N \vec{j}+0 \vec{k}\) be a continuously differentiable vector field on \(R,\) then \[ \oint _{C_1} M \, dx+N \, dy+\oint _{C_2} M \, dx+N \, dy= \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right) \, d A.\] :::

Example 8.18 Evaluate the closed line integral \[ \oint _C \frac{x dx+ydy}{x^2+y^2} \] where \(C\) is any Jordan curve whose interior contains the point \((0,0)\) transversed counterclockwise

Solution. Let \(C_1\) be a circle centered at \((0,0)\) with radius \(r\) so small that all of \(C_1\) is contained within \(C.\) Let \(C_1\) be oriented clockwise and let \(R\) be the region between \(C_1\) and \(C.\) Then by Green’s theorem for doubly-connected regions, \[\begin{align*} & \oint _C \frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy+\oint _{C_1} \frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy \\ & \qquad = \iint_R \left(\frac{\partial }{\partial x}\left(\frac{y}{x^2+y^2}\right)-\frac{\partial }{\partial y}\left(\frac{x}{x^2+y^2}\right)\right)d A \\ & \qquad = \iint_R \left(\frac{-2 x y}{x^2+y^2}-\frac{-2 x y}{x^2+y^2}\right)d A =0.\end{align*}\] Thus, \[ \oint _C \frac{x dx+ydy}{x^2+y^2}=-\oint _{C_1} \frac{x dx+ydy}{x^2+y^2} \] which is something that can be easily evaluated, using say, the parametrization \(C_1: x=r \sin \theta , y=r \cos \theta; 0\leq \theta \leq 2\pi\). Therefore \[\begin{align*} \oint _C \frac{x dx+ydy}{x^2+y^2} & =-\oint _{C_1} \frac{x dx+ydy}{x^2+y^2} \\ & =\int_0^{2\pi } \frac{(r \sin \theta ) (r \cos \theta )+(r \cos \theta )(-r \sin \theta )}{\sin ^2\theta +\cos ^2\theta } \, d\theta =0. \end{align*}\]

Example 8.19 Evaluate the closed line integral \[ \oint _C\left(x^2y dx-y^2xdy\right) \] where \(C\) is the boundary of the region between the \(x\)-axis and the semicircle \(y=\sqrt{a^2-x^2},\) traversed counterclockwise (including the \(x\)-axis).

Solution. Using Green’s Theorem, we find \[\begin{align*} \oint _C\left(x^2y dx-y^2xdy\right) & =\int \int \left(-y^2\right)-\left(x^2\right)dA =-\int \int \left(x^2+y^2\right) \, dA \\ & =-\int _0^{\pi }\int _0^ar^3drd\theta =-\int_0^{\pi } \frac{a^4}{4} \, d\theta =-\frac{a^4 \pi }{4} \end{align*}\]

Example 8.20 Evaluate the closed line integral \[ \oint _C\frac{-(y+2)dx+(x-1)dy}{(x-1)^2+(y+2)^2}\] where \(C\) is any Jordan curve whose interior does not contain the point \((1,-2)\).

Solution. Green’s theorem applies because we are using any Jordan curve which does not contain the point \((1,-2).\) \[\begin{align*} & \oint _C\frac{-(y+2)dx+(x-1)dy}{(x-1)^2+(y+2)^2} \\ & \quad =\oint _C-\frac{(y+2)}{(x-1)^2+(y+2)^2}dx+\frac{(x-1)}{(x-1)^2+(y+2)^2}dy \\ & \quad =\iint_D \left(\partial _x \left(\frac{(x-1)}{(x-1)^2+(y+2)^2}\right)\right)-\left(\partial _y \left(-\frac{(y+2)}{(x-1)^2+(y+2)^2}\right)\right)dA \\ & \quad =\iint_D\left(\frac{-x^2+2 x+y^2+4 y+3}{\left(x^2-2 x+y^2+4 y+5\right)^2}\right)-\left(\frac{-x^2+2 x+y^2+4 y+3}{\left(x^2-2 x+y^2+4 y+5\right)^2}\right) \, dA =0 \end{align*}\]

Example 8.21 Evaluate the closed line integral \[ \oint _C\left[(x-3y)dx+\left(2x-y^2\right)dy\right] \] where \(A\) is the region \(D\) enclosed by a Jordan curve \(C\).

Solution. Green’s theorem applies and
\[\begin{align*} \oint _C(x-3y)dx+\left(2x-y^2\right)dy & =\iint_D \left(\partial _x \left(2x-y^2\right)\right)-\left(\partial _y (x-3y)\right)dA \\ & =\iint_D (2+3)dA =5A. \end{align*}\]

Example 8.22 Use a line integral to find the area enclosed by the region \(R\) defined by the circle \(x^2+y^2=4\)

Solution. We can parametrize the circle by \(x=2 \cos t\) and \(y=2 \sin t\) for \(0\leq t\leq 2\pi .\) Then the area is \[ \frac{1}{2}\oint _C x dy-ydx= \frac{1}{2}\int _0^{2\pi } [4\cos (t)\cos (t)+4 \sin (t)\sin (t)] \, dt=4\pi. \] - We can parametrize the line segments by \[\begin{align*} & C_1: x=t, y=t; 0\leq t\leq 1 \\ & C_2: x=1-t, y=1+t; 0\leq t\leq 1 \\ & C_3: x=0, y=2(1-t); 0\leq t\leq 1 \end{align*}\] Then the area is \[\begin{align*} \oint _C x dy & =\int _{C_1} x dy+\int _{C_2} x dy+\int _{C_3} x dy \\ & =\int_0^1 t \, dt+\int_0^1 (1-t) \, dt+\int _0^10(-2)dt =\frac{1}{2}+\frac{1}{2}+0 =1. \end{align*}\]

Example 8.23 Use a line integral to find the area enclosed by the region \(R\) defined by the curve \(C: x=\cos ^3t,\) \(y=\sin^3 t\) for \(0\leq t\leq 2\pi\)

Solution. The required area is \[\begin{align*} A& =\oint _C x dy =\int _0^{2\pi }\cos ^3t \left(3 \sin ^2 t \cos t\right)dt \\ & =\int _0^{2\pi }\cos ^4 t \left(1-\cos ^2 t \right) dt =\int _0^{2\pi }\left(\cos ^4 t -\cos ^6 t \right) dt %& =3\int _0^{2\pi }\left(\cos ^4 t -\cos ^6 t \right) dt %& =3\int _0^{2\pi }\left(\cos ^4(t)-\frac{5}{6}\cos ^4(t)\right) dt \\ %& =\frac{1}{2}\int _0^{2\pi }\cos ^4(t) dt \\ %=\frac{1}{2}\left(\frac{3}{4}\right)\int _0^{2\pi }\cos ^2(t) dt % =\frac{1}{2}\left(\frac{3}{4}\right)\left(\frac{1}{2}\right)\int _0^{2\pi } dt %=\frac{3 \pi }{8}.\end{align*} \end{align*}\] by using the reduction formula, \[ \int \cos ^nx \, dx=\frac{\cos ^{n-1}x \sin x}{n}+\frac{n-1}{n}\int \cos ^{n-2}xdx. \]

8.11 Exercises

Exercise 8.17 Use Green’s theorem to evaluate the closed line integral \[ \oint _C y^2 dx+x^2 dy\] where \(C\) is the boundary of the square with vertices \((0,0),\) \((1,0),\) \((1,1),\) and \((0,1)\) traversed counterclockwise.

Exercise 8.18 Use Green’s theorem to evaluate the closed line integral \[\oint _C4x y dx \] where \(C\) is the boundary of the circle \(x^2+y^2=1\) traversed clockwise.

Exercise 8.19 Use Green’s theorem to evaluate the closed line integral \[\oint _C x \sin x dx-e^{y^2} dy\] where \(C\) is the boundary of the triangle with vertices \((-1,-1),\) \((1,-1),\) and \((2,5)\) traversed counterclockwise.

Exercise 8.20 Use Green’s theorem to evaluate the closed line integral \[\oint _C \sin x \cos y \, dx+\cos x \sin y \, dy \] where \(C\) is the boundary of the square with vertices \((0,0),\) \((2,0),\) \((2,2),\) and \((0,2)\) traversed clockwise.

Exercise 8.21 Use Green’s theorem to evaluate a closed line integral that represents the area enclosed by the region defined by the curve \(x^2+y^2=4.\)

Exercise 8.22 Use Green’s theorem to evaluate a closed line integral that represents the area enclosed by the region defined by the trapezoid with vertices \((0,0),\) \((4,0),\) \((1,3),\) and \((0,3).\)

Exercise 8.23 Evaluate the closed line integral \[ \oint _C\frac{x dx+ydy}{x^2+y^2}\] where \(C\) is any piecewise smooth Jordan curve enclosing the origin, traversed counterclockwise.

Exercise 8.24 Evaluate the closed line integral \[ \oint _C x^2y \, dx-y^2x \, dy, \] where \(C\) is the boundary of the region between the \(x\)-axis and the semicircle \(y=\sqrt{a^2-x^2},\) traversed counterclockwise (including the \(x\)-axis).

Exercise 8.25 Evaluate \[\oint _C\frac{-(y+2)dx+(x-1)dy}{(x-1)^2+(y+2)^2}\] where \(C\) is any Jordan curve whose interior does not contain the point \((1,-2).\)

Exercise 8.26 If \(C\) is a Jordan curve, show that \[ \oint _C (x-3y)dx+\left(2x-y^2\right)dy =5A\] where \(A\) is the region \(D\) enclosed by \(C.\)

Exercise 8.27 Suppose \(\vec{F}=M(x,y)\, \vec{i}+N(x,y)\, \vec{j}\) is continuously differentiable in a doubly-connected region \(R\) and that \[ \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y} \] throughout \(R.\) How many distinct values of \(I\) are there for the integral \[ I=\oint _C M(x,y)\, dx+N(x,y)\, dy \] where \(C\) is a piecewise smooth Jordan curve in \(R?\)

Exercise 8.28 Evaluate the line integral \[ \oint _C x^2y \, dx-y^2x \, dy, \] where \(C\) is the boundary of the region between the \(x\)-axis and the semicircle \(y=\sqrt{a^2-x^2},\) transversed counterclockwise (including the \(x\)-axis).

Exercise 8.29 Find the work done if an object that moves in the force field \(\vec{F}(x,y)=y^2 \vec{i}+x^2 \vec{j}\) once counterclockwise around the circular path \(x^2+y^2=2\).