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Chapter 13 Line Integrals, Flux, Divergence, Gauss' and Green's Theorem

“The important thing is not to stop questioning.” - Albert Einstein

The phrases scalar field and vector field are new to us, but the concept is not. A scalar field is simply a function whose range consists of real numbers (a real-valued function) and a vector field is a function whose range consists of vectors (a vector-valued function).

Example 13.1.

\(f(x,y) = x^2 -2xy\) is a scalar field. \(f(x,\ y)=(2x,\ 3y)\) is a vector field on \({\re}^2\) and \(f(x,\ y,\ z)=(2x-y,\ y\cos(z^2),\ x-yz)\) is a vector field on \({\re}^3\text{.}\) For the point in the ocean at location \((x,y,z)\text{,}\) the vector field \(\oa f(x,y,z) = \big(p(x,y,z),q(x,y,z),r(x,y,z)\big)\) might represent the direction of the current at this point where \(p,q,r: \re^n \to \re.\)

Definition 13.2.

Let \(n\) and \(m\) be integers greater than or equal to 2. A scalar field on \({\re}^n\) is a function \(f:{\re^n}\to{\re}\text{.}\) A vector field on \({\re}^n\) is a function \(f:{\re}^n\to{\re}^n.\)

We now modify our definition for \(\nabla\text{.}\) If \(f: \re^2 \to \re\text{,}\) then we previously defined \(\dsp \nabla f = (f_x, f_y) = (\frac{ \partial f}{\partial x}, \frac{ \partial f}{\partial y}).\) We now redefine \(\nabla\) as an operator that acts on the set of differentiable functions and write

\begin{equation*} \nabla = \left(\frac{ \partial }{\partial x}, \frac{ \partial }{\partial y}\right). \end{equation*}

This encompasses our previous notation as we may now write

\begin{equation*} \dsp \nabla f = \big(\frac{ \partial }{\partial x}, \frac{ \partial }{\partial y}\big)(f)= \big(\frac{ \partial f}{\partial x}, \frac{ \partial f}{\partial y}\big)= (f_x, f_y). \end{equation*}

We previously defined the dot product only between vectors (and points) in \(\re^n\) and now redefine that notation as well. If \(\oa f: \re^2 \to \re^2\) and \(\oa f(x,y) = \left(p(x,y),q(x,y)\right)\text{,}\) then we will write

\begin{equation*} \nabla \cdot \oa f = \left(\frac{ \partial }{\partial x}, \frac{ \partial }{\partial y}\right) \cdot \big(p(x,y),q(x,y)\big) = \frac{ \partial p }{\partial x} + \frac{ \partial q }{\partial y} = p_x(x,y) + q_y(x,y). \end{equation*}

The next definition formalizes what we just wrote, stating it for \(\re^3\text{.}\)

Definition 13.3.

If \(\oa f(x,y,z) = \Big(p(x,y,z) ,q(x,y,z) ,r(x,y,z) \Big): {\re}^3\to {\re}^3\) is a differentiable vector field, then the divergence of \(\oa{f}\) is the scalar field from \({\re}^3\to {\re}\) defined by: \(\nabla \cdot \oa{f}=p_x + q_y + r_z\text{.}\)

The divergence of a function at a point is the tendency for fluid to move away from or toward that point, where a positive divergence indicates fluid is moving away from the point and a negative divergence indicates that fluid is moving toward the point.

Example 13.4.

Sketch the vector fields, \(F(x,y) = (x,y)\) and \(F(x,y) = (-x,y)\) and compute the divergence of each.

The next definition extends our definition of the cross product, \(\times.\)

Definition 13.5.

If \(\oa f(x,y,z) = (p(x,y,z) ,q(x,y,z) ,r(x,y,z) ) : \re^3 \to \re^3\text{,}\) then the curl of \(\oa f\) denoted by \(\nabla \times f\) is defined by the vector valued function

\begin{equation*} \nabla \times \oa f = \left(\frac{ \partial }{\partial x}, \frac{ \partial }{\partial y}, \frac{ \partial }{\partial z}\right) \times (p,q,r) = \left( \frac{ \partial r}{\partial y} - \frac{ \partial q}{\partial z}, \frac{ \partial p}{\partial z} - \frac{ \partial r}{\partial x}, \frac{ \partial q}{\partial x} - \frac{ \partial p}{\partial y} \right). \end{equation*}

The divergence of \(\oa f,\) denoted by \(\nabla \cdot \oa f\text{,}\) is a scalar field while the curl of \(\oa f,\) denoted by \(\nabla \times \oa f,\) is a vector field. If our vector field represents a fluid flow, then the curl of a function at a point is a the direction from that point about which the fluid rotates most rapidly. The points on the ray with base at the point and pointing in the direction of the curl tend not to move, but to spin as the fluid rotates (curls) around that ray.

Problem 13.6.

Compute the divergence and curl of \(\oa f:{\re}^3\to{\re}^3\) given by

  1. \(\oa f(x,\ y,\ z)=(x^2yz,\ x^2+y+\sqrt{z},\ \pi x^2/yz)\)

  2. \(\oa f(x,\ y,\ z)=(e^x,\ \ln(xyz),\ \sin(x^2y^z))\)

Definition 13.7.

Let \(f:{\re}^3\to{\re},\ \oa g:{\re}^3\to{\re}^3,\ \oa h:{\re}^3\to{\re}^3\text{.}\) Assume \(f,\ \oa g,\ \oa h\) are differentiable. Then

  1. \((f\oa g)(x,\ y,\ z)=f(x,\ y,\ z)\oa g(x,\ y,\ z)\)

  2. \((\oa g\cdot \oa h)(x,\ y,\ z)=\oa g(x,\ y,\ z)\cdot \oa h(x,\ y,\ z)\)

  3. \((\oa g\times \oa h)(x,\ y,\ z)=\oa g(x,\ y,\ z)\times \oa h(x,\ y,\ z)\)

Problem 13.8.

Let \(f:{\re}^3\to{\re},\ \oa g:{\re}^3\to{\re}^3,\ \oa h:{\re}^3\to{\re}^3\text{.}\) Prove or give a counter example:

  1. \(\nabla\cdot(\oa{g}+\oa{h})=\nabla\cdot\oa{g}\ +\ \nabla\cdot\oa{h}\)

  2. \(\nabla\cdot(f\oa{g})=f \ \nabla \cdot \oa{g} \hskip .2in\) (Note: \(f \ \nabla\cdot\oa{g}\) means \(f\) times \(\nabla\cdot\oa{g}\text{.}\))

  3. \(\nabla\times(\oa{g}+\oa{h})=\nabla\times\oa{g}\ +\ \nabla\times\oa{h}\)

  4. \(\nabla\cdot(f \oa{g})=f \ \nabla\cdot\oa{g} \ + \ \oa g\cdot\nabla f\)

  5. \(\nabla\times(f \oa{g})=f \nabla\times\oa{g} \ + \ \oa g\times \nabla f\)

  6. \(\nabla\cdot(\nabla \times \oa{g})=0\)

Arc Length and Line Integrals over Scalar Fields

Problem 13.9.

Use Definition 8.40 to write the integrals that would represent the arc length of \(f(x)=x^2\) from \((-1,\ 1)\) to \((1,\ 1)\) and the arc length of \(\oa f(t)=(t,\ t^2)\) for \(-1 \leq t \leq 1\text{.}\) Demonstrate your powers of integration from Calculus II by computing this integral — don't use a formula — use a trig substitution and then integration by parts.

Example 13.10.

Compute the area of a wall that has as its base the curve \(c(t) = (t,t^2), \ \ 0 \leq t \leq 1\) and has a height of \(t^3\) at the point \((t,t^2).\)

Problem 13.11.

For the wall example worked in class, suppose we replace the base \(c(t) = (t, t^2)\) with the curve \(c(t) = (u(t),v(t))\) where \(u\) and \(v\) are some real-valued functions and we replace the height \(f(t)=t^3\) with some real valued function \(h\text{.}\) What would the area of the wall be now? Your answer will be an integral in terms of \(u\text{,}\) \(v\text{,}\) and \(h\text{.}\)

Problem 13.12.

Two scalar line integrals:

  1. A wall over \(\oa{c}(t)=(3\ \sin(t),\ 3\ \cos(t))\) from \(t=0\) to \(\dsp{t=\frac{\pi}{2}}\) has height \(h(x,\ y)=x^2y\text{.}\) Graph the wall and determine its area.

  2. A wall over \(\oa{c}(t)=\dsp{(\sqrt{9-t^2},\ t)}\) from \(t=0\) to \(t=3\) has height \(h(x,\ y)=x^2y\text{.}\) Determine its area.

What you have just been computing are called line integrals over scalar fields.

Definition 13.13.

The line integral of a scalar field \(g:\ {\re}^2 \to{\re}\) over the curve \(\oa{c}(t)=(x(t),\ y(t))\) is defined by \(L=\dsp{\int_c{g(\oa{c}(t)) |\oa{c}'(t) |\ dt}}.\)

The line integral over a scalar field is computed with respect to arc length and we use the notation,

\begin{equation*} \int_c{ g \ ds} \; \; \mbox{ to mean } \; \; \int_c{g(\oa{c}(t))\cdot |\oa{c}'(t) |\ dt}. \end{equation*}

We may use line integrals to compute the mass of a curved piece of wire in a similar manner. Suppose we have a piece of wire in the plane that is bent into the shape of a curve, \(\oa c: \re \to \re^2.\) If \(\delta: \re^2 \to \re\) and \(\delta(x,y)\) represents the density of the wire at \((x,y),\) then the line integral of the density with respect to the arc length is the mass of the wire.

Problem 13.14.

Mass is the integral of density. Find the mass of a wire in the plane with density at \((x,\ y)\) of \(\delta(x,\ y)=2xy\) and shape \(\oa{c}(t)=(3\ \cos(t),\ 4\ \sin(t))\) from \(t=0\) to \(t=2\pi\text{.}\) Could such a wire exist?

Problem 13.15.

Find the mass of the wire in three-space with density at \((x,\ y,\ z)\) of \(\delta(x,\ y,\ z)=3z\) and in the shape \(\oa{c}(t)=(2\ \cos(t),\ 2\ \sin(t),\ 5t),\ t\in[0,4\pi]\text{.}\) Find its center of mass as well. Center of mass was defined at the end of the last chapter.

Definition 13.16.

A vector field is said to be conservative if it is the gradient of some function. Conservative fields are also called gradient fields. If \(f\) is a conservative field, then any function \(g\) satisfying \(\nabla g = f\) is called a potential for \(f.\)

Example 13.17.

\(\dsp{\oa f(x,\ y)=(-2xy+2e^y,\ -x^2+2xe^y)}\) is conservative since \(\oa f=\nabla g\) where \(g\) is given by \(\dsp{g(x,\ y)=-x^2y+2xe^y}\text{.}\)

Problem 13.18.

Is \(\oa f(x,\ y)=(xy,\ x-y)\) conservative? If \(\oa f\) is conservative, then there must be a function \(g\) so that \(\nabla g=\oa f.\) Is there a function \(g\) so that \(g_x(x,y) =xy\) and \(g_y(x,y)=x-y\text{?}\) Why or why not?

Problem 13.19.

Is \(\oa f(x,\ y)=(x^2+y^2,\ 2xy)\) conservative?

Problem 13.20.

Show that if \(\oa f(x,\ y)=(p(x,\ y),\ q(x,\ y))\) and \(p_y(x,\ y)=q_x(x,\ y)\) then \(\oa f\) is a gradient field.

This is part of a larger theory that we don't have time to cover, but is worth at least discussing. Problem 13.20 and its converse are valid in three dimensions as well. Suppose \(f: \re^3 \to \re^3\) is a vector field. If \(f\) is conservative, then \(\nabla \times f = 0.\) Conversely, if \(f\) has continuous first partial derivatives and \(\nabla \times f = 0\) then \(f\) is conservative. If \(f:\re^2 \to \re^2\) and \(f=(p,q)\text{,}\) then Problem 13.20 showed that if \(p_y=q_x\) then \(f\) was conservative. If \(p_y=q_x\text{,}\) then \(q_x-p_y=0\) which is the same statement as \(\nabla \times f \cdot \oa{k} = 0.\) In summary, understanding the notation \(\nabla \times f\) is enough to determine when a vector field is conservative.

Problem 13.21.

Is \(\dsp{\oa f(x,\ y)=\left(\frac{y}{x^2+y^2},\ \frac{-x}{x^2+y^2}\right)}\) a gradient field?

The next two examples remind us that work may be computed by integrating force.

Example 13.22.

From physics we know that \(Work=Force \cdot Distance\text{.}\) So lifting a \(5\ lb\) book \(2\ feet\) requires \(10\ ft \cdot lbs\) of work. Suppose we have a spring that has force \(f\) proportional to the square of the distance it is stretched from equilibrium so that \(f(x)=kx^2\) (where \(k>0\) is the spring's coefficient). If \(0=x_0 \leq x_1 \leq \dots \leq x_N=4\text{,}\) then the work done to stretch it \(4 \ units\) would be

\begin{align*} W \amp = \lim_{N\to\infty}\sum_{i=1}^Nk\hat x_i^2\ (x_{i+1}-x_i) \mbox{ where } \hat x_i \in [x_i, x_{i+1}]\\ \amp =\int_0^4kx^2\ dx \mbox{ by the definition of the integral} . \end{align*}

Thus in one dimension, work is the integral of force: \(\dsp{W=\int{f(x)\ dx}}\text{.}\)

Example 13.23.

Suppose we have a cone full of water with radius \(10\ ft\text{,}\) height \(15\ ft\text{,}\) its point on the ground, and standing so that it holds the water. How much work is done in pumping all the water out of the cone? Tools you will need are: Force = Mass \(\cdot\) Acceleration (due to gravity) and Mass of one slice = volume of slice \(\cdot\) Density of \(H_2O\text{.}\) Add (integrate) the amount or work done in moving each horizontal slice of water out of the tank. The density of water is \(62.5\ lbs/ft^3\) and acceleration due to gravity is \(32\ ft/sec^2.\)

We now move to the study of vector (force) fields because we would like to be able to compute the total work done by an object as it moves through some field that acts on the object by either aiding or hindering its motion. As examples, think of a metal ball passing through a magnetic field, an electron passing through an electric field, a submarine passing through a fluid field, or a person passing through a gravitational field. Suppose we have an object passing along a curve through a vector field. Let \(\oa f: \re^2 \to \re^2\) be our (differentiable) vector field and \(c: \re \to \re^2\) be our (differentiable) planar curve or path of the object. Therefore \(\oa f\) is of the form, \(\oa f(x,\ y)=\left(p(x,\ y),\ q(x,\ y)\right)\) for some \(p,q: \re^2 \to \re\) and \(c\) is of the form, \(c(t)=(a(t),\ b(t))\) for some \(a, b : \re \to \re.\) Thus we may compute the work done as the particle moves through the vector field by integrating over the field evaluated at the particle, dotted with the speed of the particle. This gives the component of the force that is acting in the direction of travel of the particle.

\begin{align*} W \amp = \int \oa f \big(\oa c(t)\big) \cdot \oa c'(t)\ dt\\ \amp =\int \big( \ p(a(t) \ ,\ b(t)),\ q(a(t),\ b(t)) \ \big) \ \cdot \ \big( a'(t) \ ,\ b'(t) \big) \ dt\\ \amp = \int p\big( \ a(t),b(t) \ \big) \ a'(t) + q\big( \ a(t),b(t) \ \big) \ b'(t) \ dt \end{align*}

There are two additional ways in which vector line integrals are often written. First, the independent variable \(t\) is often omitted:

\begin{equation*} W = \int_{\oa c} \oa f(\oa c(t)) \cdot \oa c'(t) \ dt = \int_{\oa c} \oa f(\oa c) \cdot d\oa c. \end{equation*}

Second, if we \(f(x,\ y)=(p(x,\ y),\ q(x,\ y))\) and \(\oa c(t)=(x(t),\ y(t))\) then

\begin{align*} W \amp = \int f(\oa{c})\cdot d\oa{c}\\ \amp = \int \big(p(x,y),q(x,y)\big) \cdot \big(x'(t),y'(t)\big)\ dt\\ \amp = \int p\big(x(t),y(t)\big) \ x'(t)+q\big(x(t),y(t)\big) \ y'(t)\ dt\\ \amp = \int p \ dx + q \ dy \end{align*}

We now have many different types of integrals to compute. Here is a table of notations to help you determine which integral we are computing. Some of these notations we have already seen, some are coming soon.

  • \(dx\) or \(dt\) means the usual, for example:

    \begin{equation*} \dsp \int_1^2 x^2 \ dx = \int_1^2 t^2 \ dt \end{equation*}

  • \(ds\) means we integrate with respect to arc length, for example:

    \begin{equation*} \dsp \int_{\oa c} f \ ds = \int_{\oa c} f(\oa c(t)) |\oa c'(t) | \ dt \end{equation*}

  • \(d \oa c\) means a line integral over a vector field along some curve, for example:

    \begin{equation*} \dsp \int_{\oa c} \oa f \cdot d \oa c = \int_{\oa c} \oa f(\oa c(t)) \cdot \frac{\oa c'(t)}{|\oa c'(t)|} \ ds = \int_{\oa c} \oa f(\oa c(t)) \cdot \frac{\oa c'(t)}{|\oa c'(t)|} |\oa c'(t)| \ dt = \int_{\oa c} \oa f(\oa c(t)) \cdot \oa c'(t) \ dt \end{equation*}

  • \(dA\) means we integrate over a two dimensional domain, for example the double integral:

    \begin{equation*} \dsp \int_D f \ dA = \iint f(x,y) \ dx \ dy \end{equation*}

  • \(dV\) means we integrate over a three dimensional domain, for example the triple integral:

    \begin{equation*} \dsp \int_D f \ dV = \iiint f(x,y,z) \ dx \ dy \ dz \end{equation*}
Problem 13.24.

Suppose we have a particle traveling through the force field, \(\oa f(x,\ y)=(xy,\ 2x-y)\text{.}\)

  1. Compute the work done as the particle travels through the force field \(f\) along the curve \(\oa c(t)=(t,\ t^2)\) from \(t=0\) to \(t=1\text{.}\)

  2. Compute the work done as the particle travels through the force field \(f\) along the curve \(\oa c(t)=(2t,\ 4t^2)\) from \(t=0\) to \(t=\frac{1}{2}\text{.}\)

What you have just computed is called a line integral over a vector field.

Definition 13.25.

The line integral of the vector field \(\oa f: \re^2 \to \re^2\) given by \(\oa f(x,y) = (p(x,y), q(x,y))\) over the curve \(\oa c : \re \to \re^2\) given by \(\oa c(t) = (x(t),y(t))\) is defined by

\begin{equation*} L = \int \oa f(\oa{c})\cdot d\oa{c} = \int \Big(p\big(x(t),y(t)\big),q\big(x(t),y(t)\big)\Big) \cdot \big(x'(t),y'(t)\big)\ dt \end{equation*}
Problem 13.26.

Compute the line integral of \(\oa f(x,\ y)=(4x+y,\ x+2y)\) over \(\oa c\) where \(\oa c\) is the curve bordering the rectangle with corners at \((0,\ 0),\ (1,\ 0),\) \((1,\ 2),\ (0,\ 2).\) Start your path at the origin and work counter-clockwise around the rectangle. You will need to compute and sum four integrals along four parameterized lines.

You have been using the Fundamental Theorem of Calculus for three semesters now. This powerful result in one dimension is all that is needed to prove the version we need for three dimensions.

The next two theorems are important results associated with line integrals. We don't assert all the necessary hypothesis — curves are smooth, functions are assumed to be integrable, etc.

Theorem 13.31 states that if we reverse the path along which we compute a line integral, then we change the sign of the result. Thinking of the line integral as work, this makes sense, because the work done by traveling one direction along a path should be the opposite of the work done traveling the other way. When we have a curve \(\oa{c}\) and we write \(\oa{-c}\) we mean the same set of points in the plane, but we simply reverse the direction. If we have the curve \(\oa{c}(t) = (x(t),y(t))\) from \(t=a\) to \(t=b\text{,}\) then \(\oa{-c}\) is simply the same curve where \(t\) goes from \(t=b\) to \(t=a\text{.}\)

Problem 13.32.

Show that \(\dsp{\int_{\oa{c}}{\oa f(\oa{c})\cdot d{\oa{c}}}=-\int_{\oa{-c}}\oa f(\oa{c})\cdot{d{\oa{c}}}}\) where \(\oa f(x,\ y)=(xy,\ y-x),\ \oa c(t)=(t^2,\ t),\ \ t\in[0,\ 2]\text{.}\) Since \(c\) is the path from \((0,\ 0)\) to \((4,\ 2)\text{,}\) then \(-c\) is the same path but from \((4,\ 2)\) to \((0,\ 0)\text{.}\)

Theorem 13.33 states that line integrals over conservative fields are independent of path. Suppose you and I start at the same point at the bottom of a mountain and we walk to the top following different paths. Did we do the same amount of work? Yes. Because gravity is a conservative (gradient) field it does not matter what path we follow as long as we both start at the same place and end at the same place. By the same logic, if we start at a point on the mountain, walk around, and return to the same spot, then the total work is zero.

Problem 13.34.

Let \(f\) be the vector field \(\oa f(x,\ y)=(3x^2y+x,\ x^3)\) and \(c\) be the planar curve \(\oa c(t)=(\cos(t),\sin(t))\) from \(t=0\) to \(t=\pi\text{.}\)

  1. Write out and simplify, but don't compute, the line integral of \(f\) over \(\oa c\text{.}\)

  2. Compute this line integral by choosing the simpler path \(\oa p(t)= (-t,0)\) from \(t=-1\) to \(t=1\) and applying Theorem 13.33.

  3. Recompute this line integral by finding a function \(g\) so that \(\nabla g = (3x^2y+x,\ x^3)\) and applying Theorem 13.28.

Problem 13.35.

Let \(\oa f(x,\ y)=(x-y,\ x+y).\)

  1. Compute \(\dsp \int_{\oa c_1} \oa f(\oa{c})\ d\oa c\) where \(\oa c_1 (t)=(t,\ t^2),\ \ t\in[0,\ 1].\)

  2. Compute \(\dsp \int_{\oa c_2} \oa f(\oa{c})\ d\oa c\) where \(\dsp \oa c_2 (t)=\big(\sin(t),\ \sin^2(t)\big)\ \ t\in[2\pi,\ \frac{5\pi}{2}]\text{.}\)

  3. Explain your answer.

When we compute line integrals, we often integrate along a curve that is the boundary of some region. Here are a few of the buzz words about curves that we will use.

Definition 13.36.

Let \(\oa{c}:\ [a,b] \to{\re}^2\text{.}\) We say that \(\oa c\) is a simple closed curve if \(\oa c\) starts and ends at the same point (i.e. \(\oa{c}(a)=\oa{c}(b)\)) and never crosses itself.

When we are integrating around a simple closed curve \(\oa{c}\) with respect to the arc length of the curve, we will use the notation \(\dsp \oint_{\oa c} \cdots \ ds.\) The circle indicates that the curve is closed, starting and ending at the same point in the plane.

Definition 13.37.

Let \(\oa{c}:\ [a,b] \to{\re}^2\) be a simple closed curve on \([a,\ b]\text{.}\) We say that \(\oa c\) is positively oriented if as \(t\) increases from \(a\) to \(b\) and we traverse the curve \(\oa c\) then the enclosed region is on our left.

As we will show in a forthcoming lecture, Gauss' Divergence Theorem is equivalent to Green's Theorem. They are simply the same theorem stated in two different ways. Gauss' Divergence Theorem says that if we have a certain simple closed curve representing the boundary of a region over which we have a vector field (a flow), then the flow across boundary (the flux) must equal the integral of the divergence of the fluid (or the electricity or whatever) over the region enclosed by the boundary.

For both Green's and Gauss' Theorems, when integrating along the boundary of the region the simple closed curve must be positively oriented. That is, you must integrate in a counter-clockwise direction so that the region is on your left as you travel along the parametric curve.

Example 13.39.

Verify Gauss' Divergence theorem for \(\oa f(x,y) = (-2x,-2y)\) over the triangular domain with vertices (0,0), (0,1) and (1,1).

Problem 13.40.

Verify Gauss' Divergence Theorem by computing both integrals (the flux integral and the divergence integral) with \(\oa f(x,\ y)=-4(x,\ y)\) and \(\oa c(t)=(\cos(t),\ \sin(t))\) assuming \(t \in [0, 2\pi].\)

Problem 13.41.

Verify Gauss' Divergence Theorem for \(\oa f(x,y) = (xy, 2x-y)\) over the region \(D\) in the first quadrant bounded by \(y=0\text{,}\) \(x=0\) and the line \(y=1-x\text{.}\)

Problem 13.42.

Verify Gauss' Divergence Theorem for the flow \(f(x,y) = (0,y)\) over the rectangular region, \(\{(x,y) : 0 \le x \le 2, 0 \le y \le 3 \}.\)

Example 13.44.

Verify Green's theorem for \(\oa f(x,y) = (4y, 3x+y)\) over the region bounded by the x-axis and \(y=\cos(x)\text{.}\) Compute both sides of Green's theorem for \(\oa f(x,y) = (x-y,x+y)\) over the circle \(\oa c(t) = (2\sin(t),2\cos(t))\) and explain why they are not equal.

Problem 13.45.

Verify Green's Theorem for the flow \(\oa f(x,y) = (-y^2, xy)\) over the region bounded by \(x=0, y=0, x=3, \mbox{ and } y=3.\)

Problem 13.46.

Verify Green's Theorem for the vector field \(\oa f(x,y) = (-y,x)\) over the region bounded by the curve \(\dsp \frac{x^2}{4} + \frac{y^2}{9} = 1.\)

Problem 13.47.

Verify Green's Theorem for the vector field \(\oa f(x,y) = (xy,3x)\) over the region formed by the three points, \((-1,0), (1,0),\) and \((0,4).\)

Problem 13.49.

Find the area of the ellipse \(\dsp \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) via the formula \(\dsp{A=\frac{1}{2}\int{x\ dy-y\ dx}}\) and the parametrization \(\oa c(t)=(a\ \cos(t),\ b\ \sin(t)),\ t\in[0,\ 2\pi]\text{.}\)

Gauss' Divergence Theorem is valid in higher dimensions as well, although it is often the case that integrating over certain parts of the boundary is challenging.

Problem 13.51.

Verify Gauss' Divergence Theorem for the vector field \(\oa f(x,y,z) = (x^2y, 2xz, yz^3)\) over the three dimensional box, \(0 \le x \le 1, 0 \le y \le 2, \mbox{ and } 0 \le z \le 3.\)

Problem 13.52.

Let \(S\) be the cylinder of radius 2 with base centered at the origin and height 3. Let \(\oa f(x,y,z) = (x^3 + \tan(yz), y^3 - e^{xz}, 3z + x^3)\text{.}\) Use the divergence theorem to compute the flux across the side of the cylinder.

Problem 13.53.

Let \(S\) be the solid bounded by \(2x+2y+z=6\) in the first octant. Let \(\oa f(x,y,z) = (x, y^2, z)\text{.}\) Sketch the solid and verify Gauss' Divergence Theorem.

Congratulations, for some of you this note constitutes having independently worked through three semesters of Calculus, which is quite an accomplishment.

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