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# Integral theorems of vector calculus

### From Intelligent Perception

The Fundamental Theorem of Calculus (FTC): $$\displaystyle\int_a^b f(x) dx = F(b) - F(a),$$ where $F' = f$.

The FTC is an instance of the general Stokes theorem for dimension $1$. What about dimension $2$?

The left side of the equation is an integral over a region $G = [ a, b ]$ of dimension $1$. The right-hand side can be seen as an "integral" over the "region" $\{a,b \}$ of dimension $0$, which constitutes the boundary ${\partial}G$ of the region $G = [ a, b ]$.

Suppose $G$ is $2$-dimensional now. Then, in a similar way, an area integral can be expressed as a line integral, somehow:

$$\displaystyle\int\displaystyle\int_G ... dA = \displaystyle\int_{\partial G} ... dx + ... dy .$$

## Contents

## 1 Integral over a rectangle

Before we approach the theorem over a general region, we will try to understand the missing parts in the formula above for the simplest case - $G$ is a rectangle: $$G = R = \{ ( x, y ) : a {\leq} x {\leq} b, c {\leq} y {\leq} d \}.$$

**Theorem.** Given a vector field $F = ( p, q )$ with continuous derivative. Then
$$\displaystyle\int\displaystyle\int_R \left( \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y} \right) dA = \displaystyle\int_{\partial R} p dx + q dy,$$
with ${\partial}R$ oriented counterclockwise (i.e., positively oriented).

**Proof.** The boundary ${\partial}R$ is made of $C_1, C_2, C_3$, and $C_4$, oriented accordingly.

Then, $$\begin{array}{} \displaystyle\int\displaystyle\int_R \frac{\partial q}{\partial x} dA &= \displaystyle\int_c^d \left( \displaystyle\int_a^b \frac{\partial q}{\partial x} dx \right) dy {\rm \hspace{3pt} (use \hspace{3pt} FTC \hspace{3pt} here)} \\ &= \displaystyle\int_c^d \left( q( b, y ) - q( a, y ) \right) dy {\rm \hspace{3pt} (by \hspace{3pt} using \hspace{3pt} the \hspace{3pt} FTC) \hspace{3pt} (interpret \hspace{3pt} as \hspace{3pt} a \hspace{3pt} line \hspace{3pt} integral)} \\ &= \displaystyle\int_c^d q( b, y ) dy - \displaystyle\int_c^d q( a, y ) dy \\ &= \displaystyle\int_{C_2} q( b, y ) dy - \displaystyle\int_{-C_4} q( a, y ) dy \\ &= \displaystyle\int_{C_2} q( b, y ) dy + \displaystyle\int_{C_4} q( a, y ) dy, \end{array}$$ where we have used the definition $$C_2: \{ x = b, c {\leq} y {\leq} d \}.$$

Now we want to "complete" the last expression to obtain an integral along the border ${\partial}R$. To do so, we exploit a simple orthogonality property that lets us add in some null terms: $$\begin{array}{} \displaystyle\int\displaystyle\int_R \frac{\partial q}{\partial x} dA &= \displaystyle\int_{C_2} q( b, y ) dy + \displaystyle\int_{C_4} q( a, y ) dy \\ &= \displaystyle\int_{C_1} q( x, y ) dy + \displaystyle\int_{C_2} q( b, y ) dy + \displaystyle\int_{C_3} q( x, y ) dy + \displaystyle\int_{C_4} q( a, y ) dy \\ &= \displaystyle\int_{C_1} q( x, y ) dy + \displaystyle\int_{C_2} q( x, y ) dy + \displaystyle\int_{C_3} q( x, y ) dy + \displaystyle\int_{C_4} q( x, y ) dy \\ &= \displaystyle\int_{\partial R} q( x, y ) dy, \end{array}$$ where in the last row we have used the fact that $$q( x, y ) = q( b, y ) {\rm \hspace{3pt} on \hspace{3pt}} C_2;$$ $$q( x, y ) = q( a, y ) {\rm \hspace{3pt} on \hspace{3pt}} C_4.$$

Similarly one obtains $$\displaystyle\int\displaystyle\int_R \left( - \frac{\partial p}{\partial y} \right) dA = \displaystyle\int_{\partial R} p( x, y ) dx. $$ $\blacksquare$

## 2 Integral over a curved rectangle

Next we consider a "curved rectangle" as the domain: $$R = \{ ( x, y ) : a {\leq} x {\leq} b, h_1(x) {\leq} y {\leq} h_2(x) \},$$ with $h_1(x)$ and $h_2(x)$ piecewise smooth. The statement of the theorem is exactly the same.

**Theorem.** Suppose vector field $F = ( p, q )$ has continuous derivative. Then
$$\displaystyle\int\displaystyle\int_R \left( \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y} \right) dA = \displaystyle\int_{\partial R} p dx + q dy,$$
with ${\partial}R$ oriented counterclockwise.

**Proof.** The boundary ${\partial}R$ is made of $C_1, C_2, C_3$, and $C_4$, oriented accordingly. Then,
$$\begin{array}{}
\displaystyle\int\displaystyle\int_R \frac{\partial q}{\partial x} dA &= \displaystyle\int_{h_1(x)}^{h_2(x)} \left( \displaystyle\int_a^b \frac{\partial q}{\partial x} dx \right) dy \\
&= \displaystyle\int_{h_1(x)}^{h_2(x)} ( q( b, y ) - q( a, y ) ) dy {\rm \hspace{3pt} (by \hspace{3pt} using \hspace{3pt} the \hspace{3pt} FTC)} \\
&= \displaystyle\int_{h_1}^{h_2} q( b, y ) dy - \displaystyle\int_{h_1}^{h_2} q( a, y ) dy \\
&= \displaystyle\int_{C_2} q( b, y ) dy - \displaystyle\int_{-C_4} q( a, y ) dy \\
&= \displaystyle\int_{C_2} q( b, y ) dy + \displaystyle\int_{C_4} q( a, y ) dy,
\end{array}$$
where we have used the definition of $C_2, C_2: \{ x = b, h_1(x) {\leq} y {\leq} h_2(x) \}$. Now again, exploiting the orthogonality property as above, we add zero terms to the last expression and we exploit the definition of the $C_i$'s, and we arrive at:
$$\displaystyle\int\displaystyle\int_R \frac{\partial q}{\partial x} dA = ... = \displaystyle\int_{\partial R} q( x, y ) dy.$$

For the other term we have: $$\begin{array}{} \displaystyle\int\displaystyle\int_R \frac{\partial p}{\partial y} dA &= \displaystyle\int_a^b \left( \displaystyle\int_{h_1(x)}^{h_2(x)} \frac{\partial p}{\partial y} dy \right) dx \\ &= \displaystyle\int_a^b \left( p( x, h_2(x) ) - p( x, h_1(x) ) \right) dx {\rm \hspace{3pt} (by \hspace{3pt} using \hspace{3pt} the \hspace{3pt} FTC)} \\ &= \displaystyle\int_a^b p( x, h_2(x) ) dx - \displaystyle\int_a^b p( x, h_1(x) ) dx \\ &= \displaystyle\int_{-C_3} p( x, y ) dx - \displaystyle\int_{C_1} p( x, y ) dx \\ &= - \displaystyle\int_{C_3} p( x, y ) dx - \displaystyle\int_{C_1} p( x, y ) dx \\ &= - \displaystyle\int_{C_3} p( x, y ) dx - \displaystyle\int_{C_1} p( x, y ) dx - \displaystyle\int_{C_4} p( x, y ) dx - \displaystyle\int_{C_2} p( x, y ) dx \\ &= - \displaystyle\int_{\partial R} p( x, y ) dx. \end{array}$$

Again, we have used that the fact that the line integral of $p( x, y )$ along $C_2$ and $C_4$ is zero. $\blacksquare$

## 3 Integral over "simple regions"

More general than the curved rectangles are "simple region", defined as $$R = \{ ( x, y ) : g_1(y) {\leq} x {\leq} g_2(y), h_1(x) {\leq} y {\leq} h_2(x) \},$$ with $g_1(x), g_2(x), h_1(x)$ and $h_2(x)$ piecewise smooth.

**Theorem (dim 2 Stokes).** Let's consider $f = ( p, q )$ with continuous derivative. Then
$$\displaystyle\int\displaystyle\int_R \left( \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y} \right) dA = \displaystyle\int_{\partial R} p dx + q dy,$$

If the region is not simple, decompose it into simple parts.

To prove Stokes theorem one then needs to apply properties of integrals such as:

- the additivity property of area integrals - to turn the integral into the sum of integrals over the parts, and
- the orientation property of line integrals - to cancel the integrals over the edges that are shared by the parts.

In the example on the right, there is one cut and two regions. Then $$\begin{array}{} \displaystyle\int\displaystyle\int_R.. &= \displaystyle\int\displaystyle\int_{R_1} + \displaystyle\int\displaystyle\int_{\partial R_2}.. {\rm \hspace{3pt} by \hspace{3pt} (1)} \\ &= \displaystyle\int_{\partial R_1}.. + \displaystyle\int_{\partial R_2}.. {\rm \hspace{3pt} by \hspace{3pt} ST} \\ &= \displaystyle\int_{\partial R}.. {\rm \hspace{3pt} by \hspace{3pt} (2)} \end{array}$$

It's more complicated for the example below, as one has to orient the holes in the opposite direction:

**Theorem.** Assume $R$ is decomposable into a finite number of simple regions. If $f = ( p, q )$ has continuous derivative then
$$\displaystyle\int\displaystyle\int_R \left( \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y} \right) dA = \displaystyle\int_{\partial R} p dx + q dy,$$
where ${\partial}R$ is oriented positively.

**Proof.** Idea of: decompose $R$ into $R_1, \ldots , R_n$ simple regions, and then use the Stoke Theorem on each $R_i$:
$$\displaystyle\int\displaystyle\int_{R_i} ... = \displaystyle\int_{\partial R_i} ... , i = 1, ... n,$$
and then add them.

What about this?

To figure out which way to orient which part, use the rule: as you move, the region has to be to your left.

For the left-hand side we have $$\displaystyle\int\displaystyle\int_R = \displaystyle\int\displaystyle\int_{R_1} + \displaystyle\int\displaystyle\int_{R_2}$$ while for the right-hand side we have $$\displaystyle\int_{\partial R} = \displaystyle\int_{C_1} + \displaystyle\int_{C_2},$$ so that we have:

## 4 Applications

Suppose we want to evaluate the area of the region $R$ without double integration. (Imagine you have to find the area of a fortress and you can't get inside, or a pond.) Then $$Area = \displaystyle\int\displaystyle\int_R 1 dA.$$ Here we want to use Stokes Theorem!

The above expression can be set equal to $$\displaystyle\int\displaystyle\int_R \left( \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y} \right) dA,$$ where the appropriate $p$ and $q$ are to be found. One immediately finds that $$q = x,$$ $$p = 0,$$ is a suitable choice. Then, by applying the Stokes Theorem, we get $$Area = \displaystyle\int_{\partial R} x dx = \displaystyle\int_a^b ( {\alpha}, 0 ) ( {\alpha}', {\beta}' ) dt = \displaystyle\int_a^b {\alpha}{\alpha}' dt.$$

Walking around the wall means using a specific parametrization, with time as a parameter: $${\alpha} = {\alpha}(t), {\beta} = {\beta}(t).$$

The above approach can be used for analysis of digital images.

Problem: Suppose we have a binary image, i.e., a table of $0$s and $1$'s. How do we compute the area of the region $R$ of $1$'s (pixels are treated as squares)?

To apply the Stokes theorem, we treat the image as a function of two variables. This function is constant on each pixel, so it's integrable. Then $$\displaystyle\int\displaystyle\int_R 1 dA = \displaystyle\int_{\partial R} x dx,$$ and the integrals turn into sums:

where $x_k$ is the $x$-coordinate of the $k$-th point.

The result is a simple algorithm.

**Example.** Consider
$$F = ( p, q ) = ( xy, -xy )$$
defined on the disk
$$B = \{ (x,y): x^2 + y^2 {\leq} 9 \}.$$
The left-hand side of the Stokes Theorem yields
$$\displaystyle\int\displaystyle\int_B ( -y - x ) dA = \displaystyle\int_{-3}^3 \displaystyle\int_{\sqrt{9-x^2}} ^{\sqrt{9-x^2}} ( -y - x ) dy dx.$$

With $\frac{\partial p}{\partial y} = x, \frac{\partial q}{\partial x} = -y$, this equals

$$\begin{array}{} \displaystyle\int_{-3}^3 \left. \left( -\frac{y^2}{2} - xy \right) \right|_{\sqrt{9-x^2}}^{\sqrt{9-x^2}} dx &= \displaystyle\int_{-3}^3 \left( -( 9 - x^2 - x ( 9 - x^2 )^{\frac{1}{2}} \right) - \left( -( 9 - x^2 - x ( 9 + x^2 )^{\frac{1}{2}} \right) dx \\ &= \displaystyle\int_{-3}^3 -2x ( 9 + x^2 )^{\frac{1}{2}} dx, \end{array}$$ $u = 9 - x^2$ ...

With $${\partial}B = \{ (x,y): x^2 + y^2 = 9 \},$$ $$p(t) = ( 3 \cos t, 3 \sin t ),$$ the right-hand side yields $$\begin{array}{} \displaystyle\int_{\partial B} xy dx - xy dy &= \displaystyle\int_0^{2 \pi} \left( 3 \cos t 3 \sin t ( - 3 \sin t ) - 3 \cos t 3 \sin t 3 \cos t \right) dt \\ &= -27 \displaystyle\int_0^{2 \pi} ( \cos t \sin ^2 t - \cos ^2 t \sin t ) dt. \end{array}$$

Now $u = \sin t, v = \cos t$. Finish as an **exercise**...

## 5 Divergence Theorem

Suppose we have a flow on the plane and we want to compute how much of the liquid leaves (or enters) a particular region.

Suppose $F = ( p, q )$ is the velocity field for the flow. The idea is that we first break the region into small pieces and answer the question for each and then put the outcomes together. As the in-flow/out-flow of adjacent pieces cancel each other, the total of these numbers is the total amount of liquid leaving the region.

As we make each of these pieces smaller and smaller and it approaches a point, the out-flow will approach the *divergence* of $F$ defined as:
$${\rm div \hspace{3pt}} F = \frac{\partial p}{\partial x} + \frac{\partial q}{\partial y},$$
pointwise.

As this process continues, the sums of out-flows will approach the integral of the divergence: $$\displaystyle\int\displaystyle\int_R {\rm div \hspace{2pt}} D {\hspace{3pt}} dA.$$

On the other hand, another way to answer the question is simply watch how much liquid is crossing the boundary of the region. In other words: $$\displaystyle\int\displaystyle\int_R {\rm \hspace{3pt} div \hspace{3pt}} D dA = \displaystyle\int_{\partial R} ... ds.$$

Before we approach the problem in its generality, consider the case of dimension $1$. Instead of a flow on the plane we have a flow in a piece of a pipe, $R = [a,b]$. Once again we can look at the out-flow at each point and then add (integrate) the result or we can look at what is happening at the border, i.e., the end points: $$\displaystyle\int\displaystyle\int_{[a,b]} F' dx = F(b) - F(a),$$ i.e., we have the Fundamental Theorem of Calculus again.

Back to dimension $2$. At the crossing of the boundary, note that the tangent (to ${\partial}R$) component of the velocity of the flow does not contribute:

Specifically, if the boundary is given by as a parametric curve $$p(t) {\rm \hspace{3pt} with \hspace{3pt}} v = p'(t), $$ then the normal vector is $u$ with $$u \perp {\partial}R$$ and in order to find $u$ we rotate $v$ through $90$ degrees. So, if $$v = ( a, b ), $$ then $$u = ( -b, a ).$$

**Divergence Theorem.** Let
$$F = ( f_1, f_2 ) $$
be a function with continuous derivative, and $R$ a simple region. Then
$$\displaystyle\int\displaystyle\int_R {\rm \hspace{3pt} div \hspace{3pt}} F da = \displaystyle\int_{\partial R} F \cdot N ds,$$
where $N$ denotes a unit vector normal to ${\partial}R$ and ds denotes the arc-length.

**Proof.** On the left-hand side it is
$$\displaystyle\int\displaystyle\int_R \left( \frac{{\partial}f_1}{{\partial}x} + \frac{{\partial}f_2}{{\partial}y} \right) dA = \displaystyle\int_{\partial R} -f_2 dx + f_1 dy.$$

Recall the Stokes Theorem: $$\displaystyle\int\displaystyle\int_R \left( \frac{\partial q}{\partial x} + \frac{\partial p}{\partial y} \right) dA = \displaystyle\int_{\partial R} p dx + q dy.$$

Matching the two integrals and substituting $$f = ( p, q ) = ( -f_2, f_1 ),$$ we obtain

$$\begin{array}{} \displaystyle\int\displaystyle\int_R \left( \frac{{\partial}f_1}{{\partial}x} + \frac{{\partial}f_2}{{\partial}y} \right) dA &= \displaystyle\int_{\partial R} f_1 dy + f_2 (-dx) \\ &= \displaystyle\int_{\partial R} F( dy, -dx ), \end{array}$$ where $( dy, -dx )$ is normal to ${\partial}R$.

Parametrizing the boundary ${\partial}R$ via $$p(t) = ( {\alpha}(t), {\beta}(t) ), a {\leq} t {\leq} b,$$ the above integral is equal to $$\displaystyle\int_a^b ( f_1( {\alpha}(t) ), f_2( {\beta}(t) ) ) \cdot ( {\beta}'(t), -{\alpha}'(t) ) dt.$$ This shows what is "behind the right-hand side". It is normal to ${\partial}R$, and $p(t)$ is rotated by $90$ degrees, hence the integral is equal to $$\displaystyle\int_a^b ( f_1( {\alpha}(t) ), f_2( {\beta}(t) ) ) || p'(t) || \frac{{(\beta}'(t), -{\alpha}'(t) )}{|| p'(t) ||} dt,$$ where $$|| p'(t) || dt$$ is the arc-length.

## 6 The general Stokes Theorem

Together:

It is written in terms of differential forms and the exterior derivative: $$\int_ {\sigma} d{\omega} = \int_{{\partial}{\sigma}} {\omega}.$$ Period.

## 7 Exercises

1. Give an example of that shows that the integral theorems do not hold when the region is unbounded.

2. Verify that the integral theorems remain true even on an unbounded when the integrand is of order $O(\frac{1}{r^3})$, where $r$ is the magnitude of $(x,y,z)$. What about $O(\frac{1}{r^2})$?