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 $$x^2+y^2=2$$.