Lecture 14: Stokes' Theorem

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14.1 Introduction

From previous topic, we have learnt that double integrals over a region in the plane can be transformed into line integrals over the boundary curve of the region and conversely. We shall now see that more generally, surface integrals over a surface $S$ with boundary curve $C$ can be transformed into line integrals over $C$ and conversely.

Stokes' theorem is the "curl analogue" of the divergence theorem and relates the integral of curl of a vector field over an open surface $S$ to the line integral of the vector field around the perimeter $C$ bounding the surface.

If $\mathbf{F}$ is a vector field existing over surface $S$ and around its boundary, closed curve $c$, then

$$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r}$$

This means that we can express a surface integral in terms of a line integral round the boundary curve.

Example 14.1

A hemisphere $S$ is defined by $x^2 + y^2 + z^2 = 4$ ($z \geq 0$). A vector field $\mathbf{F} = 2y\mathbf{i} - x\mathbf{j} + xz\mathbf{k}$ exists over the surface and around its boundary $c$.

Verify Stokes' theorem, that

$$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r}$$

$$\begin{aligned} S &: x^2 + y^2 + z^2 - 4 = 0 \\ \mathbf{F} &= 2y\mathbf{i} - x\mathbf{j} + xz\mathbf{k} \\ c &\text{ is the circle } x^2 + y^2 = 4 \end{aligned}$$

Solution

(a)

$$\begin{aligned} \oint_c \mathbf{F} \cdot d\mathbf{r} &= \int_c (2y\mathbf{i} - x\mathbf{j} + xz\mathbf{k}) \cdot (\mathbf{i}\,dx + \mathbf{j}\,dy + \mathbf{k}\,dz) \\ &= \int_c (2y\,dx - x\,dy + xz\,dz) \end{aligned}$$

Converting to polar coordinates

$x = 2\cos\theta; \qquad y = 2\sin\theta; \qquad z = 0$

$dx = -2\sin\theta\,d\theta; \qquad dy = 2\cos\theta\,d\theta; \qquad \text{limits } \theta = 0 \text{ to } 2\pi$

Making the substitution and completing the integral

$$\oint_c \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left(4\sin\theta[-2\sin\theta\,d\theta] - 2\cos\theta\,2\cos\theta\,d\theta\right)$$$$\begin{aligned} &= -4\int_0^{2\pi} \left(2\sin^2\theta + \cos^2\theta\right) d\theta \\ &= -4\int_0^{2\pi} \left(1 + \sin^2\theta\right) d\theta = -2\int_0^{2\pi} \left(3 - \cos 2\theta\right) d\theta \qquad (1) \\ &= -2\left[3\theta - \frac{\sin 2\theta}{2}\right]_0^{2\pi} = -12\pi \end{aligned}$$

(b) Now we determine $\displaystyle \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$

$$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \int_S \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS$$$$\operatorname{curl} \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ 2y & -x & xz \end{vmatrix} = \mathbf{i}(0-0) - \mathbf{j}(z-0) + \mathbf{k}(-1-2) = -z\mathbf{j} - 3\mathbf{k}$$$$\mathbf{n} = \frac{\nabla S}{|\nabla S|} = \frac{2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k}}{\sqrt{4x^2 + 4y^2 + 4z^2}} = \frac{x\mathbf{i} + y\mathbf{j} + z\mathbf{k}}{2}$$

Now

$$\begin{aligned} \text{Then } \int_S \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS &= \int_S (-z\mathbf{j} - 3\mathbf{k}) \cdot \left(\frac{x\mathbf{i} + y\mathbf{j} + z\mathbf{k}}{2}\right) dS \\ &= \frac{1}{2}\int_S (-yz - 3z) \; dS \end{aligned}$$

Expressing this in spherical polar coordinates and integrating, because

$x = 2\sin\theta\cos\phi; \qquad y = 2\sin\theta\sin\phi; \qquad z = 2\cos\theta; \qquad dS = 4\sin\theta\,d\theta\,d\phi$

$$\begin{aligned} \therefore \int_S \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS &= \frac{1}{2}\iint_S (-2\sin\theta\sin\phi\,2\cos\theta - 6\cos\theta)\,4\sin\theta\,d\theta\,d\phi \\ &= -4\int_0^{2\pi}\int_0^{\pi/2} \left(2\sin^2\theta\cos\theta\sin\phi + 3\sin\theta\cos\theta\right) d\theta\,d\phi \\ &= -4\int_0^{2\pi} \left[\frac{2\sin^3\theta\sin\phi}{3} + \frac{3\sin^2\theta}{2}\right]_0^{\pi/2} d\phi \qquad (2) \\ &= -4\int_0^{2\pi} \left(\frac{2}{3}\sin\phi + \frac{3}{2}\right) d\phi = -12\pi \end{aligned}$$

So we have from our two results (1) and (2)

$$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r}$$

Example 14.2

Verify the Stokes' Theorem for $\mathbf{F} = (2x-y)\mathbf{i} - yz^2\mathbf{j} - y^2z\mathbf{k}$, where $S$ is the upper half of the sphere $x^2 + y^2 + z^2 = 1$ and $C$ is its boundary.

Hemispherical surface and boundary for Example 14.2

Solution

The surface and boundary involved are illustrated in the above figure. We are required to show that

$$\oint_c \mathbf{F} \cdot d\mathbf{r} = \iint_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$$

Since $C$ is a circle of unit radius in the $(x,y)$ plane, to evaluate $\oint_c \mathbf{F} \cdot d\mathbf{r}$, we take

$x = \cos\phi, \qquad y = \sin\phi,$

so that

$\mathbf{r} = \cos\phi\,\mathbf{i} + \sin\phi\,\mathbf{j}$

giving

$d\mathbf{r} = -\sin\phi\,d\phi\,\mathbf{i} + \cos\phi\,d\phi\,\mathbf{j}$

Also, on the boundary $C$, $z = 0$, so that

$\mathbf{F} = (2x-y)\mathbf{i} = (2\cos\phi - \sin\phi)\mathbf{i}$

Thus,

$$\begin{aligned} \oint_c \mathbf{F} \cdot d\mathbf{r} &= \int_0^{2\pi} (2\cos\phi - \sin\phi)\mathbf{i} \cdot (-\sin\phi\,\mathbf{i} + \cos\phi\,\mathbf{j}) d\phi \\ &= \int_0^{2\pi} (-2\sin\phi\cos\phi + \sin^2\phi) d\phi = \int_0^{2\pi} \left[-\sin 2\phi + \frac{1}{2}(1 - \cos 2\phi)\right] d\phi = \pi \end{aligned}$$$$\operatorname{curl} \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ 2x-y & -yz^2 & -y^2z \end{vmatrix} = \mathbf{k}$$

The unit outward-drawn normal at a point $(x,y,z)$ on the hemisphere is given by $(x\mathbf{i} + y\mathbf{j} + z\mathbf{k})$ since $x^2 + y^2 + z^2 = 1$. Thus

$$\begin{aligned} \iint_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} &= \iint_S \mathbf{k} \cdot (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) dS \\ &= \iint_S z \; dS = \int_0^{2\pi}\int_0^{\pi/2} \cos\theta\sin\theta \; d\theta\,d\phi = 2\pi\left[\frac{1}{2}\sin^2\theta\right]_0^{\pi/2} = \pi \end{aligned}$$

Hence $\oint_c \mathbf{F} \cdot d\mathbf{r} = \iint_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$ and Stokes' Theorem is verified.

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14.2 Direction of Unit Normal Vectors to a Surface

When we were dealing with the divergence theorem, the normal vectors were drawn in a direction outward from the enclosed region. With an open surface as we now have, there is in fact no inward or outward direction. With any general surface, a normal vector can be drawn in either of two opposite directions. To avoid confusion, a convention must therefore be agreed upon and the established rule is as follows.

A unit normal $\hat{\mathbf{n}}$ is drawn perpendicular to the surface $S$ at any point in the direction indicated by applying the right-handed screw sense to the direction of integration round the boundary $c$. This is identical to right-hand grip rule. Having noted that point, we can now deal with the next example.

Example 14.3

A surface consists of five sections formed by the planes $x=0$, $x=1$, $y=0$, $y=3$, $z=2$ in the first octant. If the vector field $\mathbf{F} = y\mathbf{i} + z^2\mathbf{j} + xy\mathbf{k}$ exists over the surface and around its boundary, verify Stokes' theorem.

If we progress round the boundary along $c_1$, $c_2$, $c_3$, $c_4$ in an anti-clockwise manner, the normal to the surfaces will be as shown.

We have to verify that

$$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r}$$

Solution

(a) We will start off by finding $\displaystyle \oint_c \mathbf{F} \cdot d\mathbf{r}$

(1) Along $c_1$: $y=0$; $z=0$; $dy=0$; $dz=0$

$\therefore \int_{c_1} \mathbf{F} \cdot d\mathbf{r} = \int (0 + 0 + 0) = 0$

(2) Along $c_2$: $x=1$; $z=0$; $dx=0$; $dz=0$

$\therefore \int_{c_2} \mathbf{F} \cdot d\mathbf{r} = \int (0 + 0 + 0) = 0$

(3) Along $c_3$: $y=3$; $z=0$; $dy=0$; $dz=0$

$\therefore \int_{c_3} \mathbf{F} \cdot d\mathbf{r} = \int_1^0 (3\,dx + 0 + 0) = [3x]_1^0 = -3$

(4) Along $c_4$: $x=0$; $z=0$; $dx=0$; $dz=0$

$\therefore \int_{c_4} \mathbf{F} \cdot d\mathbf{r} = \int (0 + 0 + 0) = 0$

$\therefore \oint_c \mathbf{F} \cdot d\mathbf{r} = 0 + 0 - 3 + 0 = -3$

$\oint_c \mathbf{F} \cdot d\mathbf{r} = -3$

(b) Now we have to find $\displaystyle \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$

First we need an expression for curl $\mathbf{F}$.

$\mathbf{F} = y\mathbf{i} + z^2\mathbf{j} + xy\mathbf{k}$

$$\operatorname{curl} \mathbf{F} = \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ y & z^2 & xy \end{vmatrix} = \mathbf{i}(x-2z) - \mathbf{j}(y-0) + \mathbf{k}(0-1) = (x-2z)\mathbf{i} - y\mathbf{j} - \mathbf{k}$$

Then for each section, we obtain $\displaystyle \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \int_S \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS$

(1) $S_1$ (top) $\hat{\mathbf{n}} = \mathbf{k}$

$$\int_{S_1} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS = \int_{S_1} [(x-2z)\mathbf{i} - y\mathbf{j} - \mathbf{k}] \cdot (\mathbf{k}) \; dS$$$$\int_{S_1} (-1) \, dS = -(\text{area of } S_1) = -3$$

Then, likewise

(2) $S_2$ (right-hand end): $\hat{\mathbf{n}} = \mathbf{j}$

$$\begin{aligned} \therefore \int_{S_2} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS &= \int_{S_2} [(x-2z)\mathbf{i} - y\mathbf{j} - \mathbf{k}] \cdot (\mathbf{j}) \; dS \\ &= \int_{S_2} (-y) \; dS \end{aligned}$$

But $y=3$ for this section

$\therefore \int_{S_2} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS = \int_{S_2} (-3) \, dS = (-3)(2) = -6$

(3) $S_3$ (left-hand end): $\hat{\mathbf{n}} = -\mathbf{j}$

$$\begin{aligned} \int_{S_3} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS &= \int_{S_3} [(x-2z)\mathbf{i} - y\mathbf{j} - \mathbf{k}] \cdot (-\mathbf{j}) \; dS \\ &= \int_{S_3} (y) \; dS \end{aligned}$$

But $y=0$ over $S_3$

$\therefore \int_{S_3} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS = 0$

(4) $S_4$ (front): $\hat{\mathbf{n}} = \mathbf{i}$

$$\begin{aligned} \therefore \int_{S_4} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS &= \int_{S_4} [(x-2z)\mathbf{i} - y\mathbf{j} - \mathbf{k}] \cdot (\mathbf{i}) \; dS \\ &= \int_{S_4} (x-2z) \; dS \end{aligned}$$

but $x=1$ over $S_4$

$$\begin{aligned} \therefore \int_{S_4} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS &= \int_0^3\int_0^2 (1-2z)\,dz\,dy = \int_0^3 \left[z - z^2\right]_0^2 dy \\ &= \int_0^3 (-2) \; dy = [-2y]_0^3 = -6 \end{aligned}$$

(5) $S_5$ (back): $\hat{\mathbf{n}} = -\mathbf{i}$ with $x=0$ over $S_5$

Similar working to the above gives $\displaystyle \int_{S_5} \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS = 12$

Finally, collecting the five results together gives

$\therefore \int_S \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS = -3 - 6 + 0 - 6 + 12 = -3$

So, referring back to our result for section (a) we see that

$$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r}$$

Example 14.4

A surface $S$ consists of that part of the cylinder $x^2 + y^2 = 9$ between $z=0$ and $z=4$ for $y \geq 0$ and the two semicircles of radius 3 in the planes $z=0$ and $z=4$. If $\mathbf{F} = z\mathbf{i} + xy\mathbf{j} + xz\mathbf{k}$, evaluate $\displaystyle \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$ over the surface.

The surface $S$ consists of three sections

(a) The curved surface of the cylinder

(b) The top and bottom semicircles

We could therefore evaluate

$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$

Over each of these separately.

Solution

However, we know by Stokes' theorem that

$\mathbf{F} = z\mathbf{i} + xy\mathbf{j} + xz\mathbf{k}$

$$\begin{aligned} \oint_c \mathbf{F} \cdot d\mathbf{r} &= \oint_c (z\mathbf{i} + xy\mathbf{j} + xz\mathbf{k}) \cdot (\mathbf{i}\,dx + \mathbf{j}\,dy + \mathbf{k}\,dz) \\ &= \oint_c (z\,dx + xy\,dy + xz\,dz) \end{aligned}$$

Now we can work through this easily enough, taking $c_1$, $c_2$, $c_3$, $c_4$ in turn, and summing the results which gives

$$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r} = \oint_c (z\,dx + xy\,dy + xz\,dz)$$

(1) $C_1$: $y=0$; $z=0$; $dy=0$; $dz=0$

$\int_{c_1} \mathbf{F} \cdot d\mathbf{r} = \int_{c_1} (0 + 0 + 0) = 0$

(2) $C_2$: $x=-3$; $y=0$; $dx=0$; $dy=0$

$\int_{c_2} \mathbf{F} \cdot d\mathbf{r} = \int_{c_2} (0 + 0 - 3z\,dz) = \left[\frac{-3z^2}{2}\right]_0^4 = -24$

(3) $C_3$: $y=0$; $z=4$; $dy=0$; $dz=0$

$\int_{c_3} \mathbf{F} \cdot d\mathbf{r} = \int_{c_3} (4\,dx + 0 + 0) = \int_{-3}^{3} 4\,dx = 24$

(4) $C_4$: $x=3$; $y=0$; $dx=0$; $dy=0$

$\int_{c_4} \mathbf{F} \cdot d\mathbf{r} = \int_{c_4} (0 + 0 + 3z\,dz) = \left[\frac{3z^2}{2}\right]_4^0 = -24$

There is an alternative way of solving this example. We can consider a fictitious surface enclosed by the rectangular curve $C_1$-$C_2$-$C_3$-$C_4$ (the vertical rectangular surface formed by the closed curve). This fictitious surface shares the same closed curve as the hollow half-cylinder surface. Therefore, finding $\displaystyle \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$ of this fictitious surface bounded by $y = 0$, $-3 \leq x \leq 3$, $0 \leq z \leq 4$, which is more straight-forward than finding the original surface integral, will also give the same answer:

$$\operatorname{curl} \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ z & xy & xz \end{vmatrix} = (1-z)\mathbf{j} + y\mathbf{k}$$

For this vertical rectangular surface, unit normal vector, $\hat{\mathbf{n}} = \mathbf{j}$.

$$\begin{aligned} \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} &= \int_S \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS \\ &= \int_{-3}^3\int_0^4 (1-z)\,dz\,dx \\ &= \int_{-3}^3 \left[z - \frac{z^2}{2}\right]_0^4 dx \\ &= \int_{-3}^3 -4 \; dx = -24 \end{aligned}$$

This alternative solution demonstrates an interesting property related to Stokes' theorem: if two or more surfaces share the same closed curve, their surface integrals (of the Stokes' theorem) give the same value.

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14.3 Green's Theorem (Circulation Form)

Green's theorem enables an integral over a plane area to be expressed in terms of a line integral round its boundary curve. Let $P$ and $Q$ be two functions of $x$ and $y$ that are, along with their first partial derivatives, finite and continuous inside and on the boundary $c$ of a region $R$ in the $x$-$y$ plane.

If the first partial derivatives are continuous within the region and on the boundary, then Green's theorem states that

$$\iint_R \left(\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}\right) dx\,dy = -\oint_c (P\,dx + Q\,dy)$$

That is, a double integral over the plane region $R$ can be transformed into a line integral over the boundary $c$ of the region – and the action is reversible.

If $P$ and $Q$ are two single-valued functions of $x$ and $y$, continuous over a plane surface $S$, and $c$ is its boundary curve, then

$$\oint_c (P\,dx + Q\,dy) = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy$$

where the line integral is taken round $c$ in an anticlockwise manner. In vector terms, this becomes:

$S$ is a two-dimensional space enclosed by a simple closed curve $c$.

RHS: $dS = dx\,dy$

$d\mathbf{S} = \hat{\mathbf{n}} \; dS = \mathbf{k} \; dx\,dy$

If $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ where $P = P(x,y)$ and $Q = Q(x,y)$, then

$$\operatorname{curl} \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ P & Q & 0 \end{vmatrix} = \mathbf{i}\left(0 - \frac{\partial Q}{\partial z}\right) - \mathbf{j}\left(0 - \frac{\partial P}{\partial z}\right) + \mathbf{k}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$

But in the $x$-$y$ plane, $\dfrac{\partial Q}{\partial z} = \dfrac{\partial P}{\partial z} = 0$. $\therefore \operatorname{curl} \mathbf{F} = \mathbf{k}\left(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\right)$

So $\displaystyle \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \int_S \operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}} \; dS$ and in the $x$-$y$ plane, $\hat{\mathbf{n}} = \mathbf{k}$

$$\therefore \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \int_S \mathbf{k}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \cdot (\mathbf{k}) \, dS = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy$$ $$\therefore \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy \qquad (1)$$

Now by Stokes' theorem

$\int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r}$

LHS: in this case

$$\begin{aligned} \oint_c \mathbf{F} \cdot d\mathbf{r} &= \oint_c (P\mathbf{i} + Q\mathbf{j}) \cdot (\mathbf{i}\,dx + \mathbf{j}\,dy + \mathbf{k}\,dz) \\ &= \oint_c (P\,dx + Q\,dy) \end{aligned}$$

$\therefore \oint_c \mathbf{F} \cdot d\mathbf{r} = \oint_c P\,dx + Q\,dy \qquad (2)$

Therefore from (1) and (2)

Stokes' theorem $\displaystyle \int_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_c \mathbf{F} \cdot d\mathbf{r}$ in two dimensions becomes Green's theorem

$$\iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy = \oint_c P\,dx + Q\,dy$$

Example 14.5

Verify Green's theorem for the integral $\displaystyle \oint_c \left[(x^2 + y^2)dx + (x + 2y)dy\right]$ taken round the boundary curve $c$ defined by

$$\begin{aligned} y &= 0 & 0 &\leq x \leq 2 \\ x^2 + y^2 &= 4 & 0 &\leq x \leq 2 \\ x &= 0 & 0 &\leq y \leq 2 \end{aligned}$$

Green's theorem: $\displaystyle \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy = \oint_c (P\,dx + Q\,dy)$

Solution

In this case $(x^2 + y^2)dx + (x + 2y)dy = P\,dx + Q\,dy$

$\therefore P = x^2 + y^2 \quad \text{and} \quad Q = x + 2y$

We now take $c_1$, $c_2$, $c_3$ in turn.

(1) $c_1$: $y=0$; $dy=0$

$\therefore \int_{c_1} (P\,dx + Q\,dy) = \int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$

(2) $c_2$:

$$\begin{aligned} x^2 + y^2 &= 4 \quad \therefore y^2 = 4 - x^2 \quad \therefore y = (4 - x^2)^{1/2} \\ x + 2y &= x + 2(4 - x^2)^{1/2} \\ dy &= \frac{1}{2}(4 - x^2)^{-1/2}(-2x)dx = \frac{-x}{\sqrt{4 - x^2}} \, dx \end{aligned}$$$$\begin{aligned} \therefore \int_{c_2} (P\,dx + Q\,dy) &= \int_{c_2} \left[4 + \left(x + 2\sqrt{4-x^2}\right)\left(\frac{-x}{\sqrt{4-x^2}}\right)\right] dx \\ &= \int_{c_2} \left[4 - 2x - \frac{x^2}{\sqrt{4-x^2}}\right] dx \end{aligned}$$

Putting $x = 2\sin\theta$, $\sqrt{4-x^2} = 2\cos\theta$, $dx = 2\cos\theta\,d\theta$

Limits: $x=2$, $\theta = \dfrac{\pi}{2}$; $\qquad x=0$, $\theta = 0$

$$\begin{aligned} \therefore \int_{c_2} (P\,dx + Q\,dy) &= \int_{\pi/2}^{0} \left[4 - 4\sin\theta - \frac{4\sin^2\theta}{2\cos\theta}\right] 2\cos\theta \, d\theta \\ &= 4\left[2\sin\theta - \sin^2\theta - \frac{1}{2}\left(\theta - \frac{\sin 2\theta}{2}\right)\right]_{\pi/2}^{0} \\ &= 4\left[-\left(2 - 1 - \frac{\pi}{4}\right)\right] = \pi - 4 \end{aligned}$$

Finally

(3) $c_3$: $x=0$; $dx=0$

$\therefore \int_{c_3} (P\,dx + Q\,dy) = \int_2^0 2y \, dy = \left[y^2\right]_2^0 = -4$

Therefore, collecting our three partial results

$$\oint_c (P\,dx + Q\,dy) = \frac{8}{3} + \pi - 4 - 4 = \pi - \frac{16}{3}$$

That is one part done. Now we have to evaluate $\displaystyle \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy$

$P = x^2 + y^2 \qquad \therefore \frac{\partial P}{\partial y} = 2y$

$Q = x + 2y \qquad \therefore \frac{\partial Q}{\partial x} = 1$

$\iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy = \iint_S (1 - 2y) \, dy\,dx$

It will be more convenient to work in polar coordinates, so we make the substitutions

$x = r\cos\theta; \qquad y = r\sin\theta; \qquad dS = dx\,dy = r\,dr\,d\theta$

$$\begin{aligned} \therefore \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy &= \int_0^{\pi/2}\int_0^2 (1 - 2r\sin\theta) \, r \, dr \, d\theta \\ &= \int_0^{\pi/2} \left[\frac{r^2}{2} - \frac{2r^3}{3}\sin\theta\right]_0^2 d\theta \\ &= \int_0^{\pi/2} \left[2 - \frac{16}{3}\sin\theta\right] d\theta = \left[2\theta + \frac{16}{3}\cos\theta\right]_0^{\pi/2} = \pi - \frac{16}{3} \end{aligned}$$

So we have established once again that

$$\oint_c (P\,dx + Q\,dy) = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy$$