Day 44 - Prep

Task 44.1

What do the curl and divergence of a vector field compute? How do we physically interpret them?

1. Head to the following webpages and read their descriptions.
   • https://mathinsight.org/curl_idea
   • https://mathinsight.org/curl_subtleties
   • https://mathinsight.org/divergence_idea
   • https://mathinsight.org/divergence_subtleties
2. After reading the 4 pages above, head to Wikipedia's article on magnetic fields, see https://en.wikipedia.org/wiki/Magnetic_field. Skim through the article (you do not need to read and comprehend all of it).
   • Are there any new symbols you don't recognize? What are they?
   • Search for the word "divergence" in the text, and read the paragraph(s) where you find it.
   • Search for the word "curl" in the text, and read the paragraph(s) where you find it.
   • Come with questions about anything you would like to discuss.

Task 44.2

As we proceed through the remainder of this unit, we'll find that three types of integrals appear quite often.

• Work integrals: $\ds\int_{C} Mdx+Ndy+Pdz$
• Surface integrals: $\ds \iint_S f(x,y,z) dS$
• Flux integrals: $\ds \iint_S \vec F \cdot \hat n dS$

For each type of integral above, use Mathematica to create a code chunk that you can use to compute each type of integral. This will allow you free yourself from performing tedious computations, letting you focus on bigger picture ideas.

• For the work integral, you can pick the vector field $\vec F(x,y,z)$, parametrization $\vec r(t)$, and bounds for $t$. Feel free to use an example from 42.3.
• For the surface integral, you can pick the surface $S$, function $f(x,y,z)$, parametriztion $\vec r(u,v)$, and bounds for $u$ and $v$. Feel free to use an example from 42.2.
• For the flux integral, you can pick the surface $S$, orientation $\hat n$, parametriztion $\vec r(u,v)$, and bounds for $u$ and $v$. Feel free to use an example from 42.1 or 42.3.

The ReplaceAll[] command in Mathematica will allow you to replace any instance of $x$, $y$, or $z$ with the corresponding value from your parametrization. This allows you to enter the vector field using Cartesian coordinates. Other commands you'll find useful are D[], Cross[], Norm[], Dot[], and Integrate[].

As an example, to dot the vector field $\vec F = (3x+4yz, 2y^2, x-4z)$ with the derivative of the parametrization $\vec r(t) = (3\cos t, 3\sin t, 4 t)$, and then make sure everything is in terms of $t$, we can use the following code.

F = {3 x + 4 y z, 2 y^2, x - 4 z}
r = {3 Cos[t], 3 Sin[t], 4 t}
ReplaceAll[Dot[F, D[r, t]], {x -> r[[1]], y -> r[[2]], z -> r[[3]]}]

(*Or using short cut notation (. for Dot and /. for ReplaceAll)*)
F . D[r, t] /. {x -> r[[1]], y -> r[[2]], z -> r[[3]]}

The notation r[[1]] accesses the first element from r, which is 3 Cos[t].

Task 44.3

Green's theorem connects circulation along a simple closed curve in 2D to an integral along the region inside the curve. The circulation per area (circulation density) is given by $N_x-M_y$ for a 2D vector field. A similar fact is true in 3D, namely if $S$ is a surface with a simple closed curve $C$ that forms the boundary of $S$, then the circulation along $C$ can be computed instead by summing circulation densities (circulation per surface area) along the surface. Given a point $(x,y,z)$ and a unit vector $\hat n$, we calculate the circulation density of $\vec F$ about $\hat n$ at $(x,y,z)$ in a similar manner, as follows.

• We construct a circle $C_a$ of radius $a$ in the plane through $(x,y,z)$ with normal vector $\hat n$.
• We compute the counter-clockwise circulation about $\hat n$, obtaining $\oint_{C_a} Mdx+Ndy+Pdz$.
• We divide by the area $A_a$ inside $C_a$, to obtain a circulation per area.
• The circulation density of $\vec F$ about $\hat n$ at $(x,y,z)$ is the limit $$\lim_{a\to 0}\frac{1}{A_a}\oint_{C_a} Mdx+Ndy+Pdz = \vec \nabla \times \vec F \cdot \hat n.$$

The curl of a vector field provides information about circulation density in every direction (just dot by the direction about which you want to know how things are rotating). Note that for a 2D vector field, we can use the normal vector $\hat n = (0,0,1)$, and then we see that the circulation density for a vector field in 2D is $$(N_z-P_y, P_x-M_z, N_x-M_y) \cdot (0,0,1) = N_x-M_y.$$

We're now ready to state Stokes's theorem.

Let's verify this theorem in some examples.

1. Let $\vec F(x,y,z) = (x+2y, 3x-4z, 2z)$. Consider the surface $S$ parametrized by $\vec r(u,v) = (u\cos v, u\sin v, u^2)$ for $0\leq u\leq 3$ and $0\leq v\leq 2\pi$ with orientation $\hat n$ pointing away from the $z$-axis. The boundary $\partial S$ consists of a single curve $C_1$ parametrized by $\vec r_1(t) = (3\cos t, 3\sin t, 9)$ for $0\leq t\leq 2\pi$.
   1. Draw the surface and curve. Do your best to explain what it means for $\hat n$ and the curve to be oriented compatibly.
   2. Set up and compute the surface integral $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$.
   3. Set up and compute the line integral $\ds \int_{\partial S} Mdx+Ndy+Pdz$. Do you get the same result as the surface integral?
2. Let $\vec F(x,y,z) = (x+2y, 2x+4z, 4y)$. Consider the plane in the first octant with the three vertices $(3,0,0)$, $(0,3,0)$ and $(0,0,3)$. An equation of the plane is $x+y+z=3$, which we can parameterize using $\vec r(u,v) = (u,v,3-u-v)$ for $0\leq u\leq 3$ and $0\leq v\leq 3-u$. The boundary $\partial S$ consists of 3 curves, which we can parameterize using $$\begin{align*} \vec r_1(t) &= (3-3t,3t,0) \quad \text{for} 0\leq t\leq 1,\\ \vec r_2(t) &= (0,3-3t,3t) \quad \text{for} 0\leq t\leq 1,\\ \vec r_3(t) &= (3t,0,3-3t) \quad \text{for} 0\leq t\leq 1. \end{align*}$$
   1. Draw the surface and 3 curves, placing an arrow on the curves to demonstrate the direction of motion.
   2. Compute $\ds \int_{\partial S} Mdx+Ndy+Pdz$ by computing the three line integrals and summing the result.
   3. Compute $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$, using an orientation $\hat n$ that is compatible with the orientation of the curves.

Task 44.4

Pick some problems related to the topics we are discussing from the Text Book Practice page.

Day 45 - Prep

Task 45.1

We've discussed circulation density and connected it to the curl of a vector field. Let's construct a corresponding analog for flux.

Given a point $(x,y,z)$, we calculate the flux density of $\vec F$ at $(x,y,z)$ as follows.

• We construct a sphere $S_a$ of radius $a$ centered at $(x,y,z)$.
• We compute the outward flux of $\vec F$ across the surface of the sphere, so $\iint_{S_a} \vec F\cdot\hat n dS$.
• We divide by the volume $V_a$ inside $S_a$, to obtain a flux per volume.
• The flux density of $\vec F$ at $(x,y,z)$ is the limit $$\lim_{a\to 0}\frac{1}{V_a}\iint_{S_a}\vec F\cdot\hat n dS = \vec \nabla \cdot \vec F.$$

The divergence of a vector field provides the flux density.

We are now ready to state the divergence theorem.

Let's verify this theorem in some examples.

1. Consider the vector field $\vec F = (x^2, 2 y, x + y + z)$ and sphere of radius 5. We can parameterize the surface $S$ using $\vec r(u,v) = (5 \cos(u)\sin(v),5\sin(u)\sin(v),5\cos(v))$ for $0\leq u\leq 2\pi$ and $0\leq v\leq \pi$. The solid domain $D$ satisfies $\rho\leq 5$ (using spherical coordinates - you'll need appropriate bounds for $\theta$ and $\phi$). Set up and compute both $\iiint_D \vec \nabla \cdot \vec F dV$ and $\iint_{\partial D}\vec F\cdot\hat n dS$, using the outward pointing normal vector $\hat n$. (I got $500\pi$ for both. Pay attention to the direction in which $\vec r_u\times \vec r_v$ points.)
2. Consider the vector field $\vec F = (5x,x-z,y+z+1)$ and solid domain $D$ in space that lies below the paraboloid $z=1-r^2$ (using cylindrical coordinates) and above the $xy$-plane. The boundary of $D$ consists of two surfaces. The top surface is parametrized by $\vec r(u,v) = (u \cos v,u \sin v,1-u^2)$ for $0\leq u\leq 1$ and $0\leq v\leq 2\pi$. The bottom surface is the plane $z=0$, and for this particular problem we can parameterize it using $\vec r(u,v) = (u\cos v,u\sin v,0)$ for $0\leq u\leq 1$ and $0\leq v\leq 2\pi$. Set up and compute both $\iiint_D \vec \nabla \cdot \vec F dV$ and $\iint_{\partial D}\vec F\cdot\hat n dS$, using the outward pointing normal vector $\hat n$. (I got $3\pi$ for both. The flux integral across the top surface is $4\pi$, and across the bottom surface is $-\pi$.)

Task 45.2

Stokes's theorem can be applied when the boundary of a surface consists of disjoint curves. Remember to check if the parametrization associated with each curve is compatible with the orientation on the surface. When not compatible, a simple negative sign can quickly fix any issues.

Let $\vec F = (2x-3y,x^2+2z,4xy)$.

1. Consider the surface $S$ parametrized by $\vec r(u,v) = (u\cos v, u, u \sin v)$ for $1\leq u\leq 3$ and $0\leq v\leq 2\pi$, oriented by $\hat n$ pointing away from the $y$-axis. The boundary $\partial S$ of $S$ is the union of the two disconnected curves $C_1$ with parametrization $\vec r_1(t) = (\cos t, 1, \sin t)$ for $0\leq t\leq 2\pi$ and $C_2$ with parametrization $\vec r_2(t) = (3\cos t, 3, 3\sin t)$ for $0\leq t\leq 2\pi$. Provided the curves are oriented compatibly with $\hat n$, Stokes's theorem states that $$\ds \int_{\partial S} Mdx+Ndy+Pdz = \iint_S \vec \nabla \times \vec F \cdot \hat n dS.$$
   1. Draw the surface and curves.
   2. Compute $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$. (I got $104\pi$.)
   3. Compute $\ds \int_{C_1} Mdx+Ndy+Pdz$ and $\ds \int_{C_2} Mdx+Ndy+Pdz$, using the parametrizations provided.
   4. Explain how to combine the results of the previous computations to obtain $\ds \int_{\partial S} Mdx+Ndy+Pdz$.
2. Now instead let $S$ be half of the surface above, parametrized by $\vec r(u,v) = (u\cos v, u, u \sin v)$ for $1\leq u\leq 3$ and $0\leq v\leq \pi$. The boundary of $S$ now consists of 4 connected curves, two arcs and two straight lines. One arc has parametrization $\vec r_1(t) = (\cos t, 1, \sin t)$ for $0\leq t\leq \pi$. One of the straight lines has parametrization $\vec r_2(t) = (t, t, 0)$ for $1\leq t\leq 3$.
   1. Compute $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS$ (change the bounds of what you did in the previous part).
   2. Give parametrizations for the other two curves that make up $\partial S$.
   3. Compute $\ds \int_{C} Mdx+Ndy+Pdz$ for each of the 4 curves in $\partial S$. (You'll end up with $6+2\pi$, $54+54\pi$, $14/3$, and $86/3$, all of which could be negative instead of positive, based on your parametrization.)
   4. Explain how to combine the results of the previous computations to obtain $\ds \int_{\partial S} Mdx+Ndy+Pdz$.

Task 45.3

At this point, we've seen all the topics for the semester, and just need to make sure we practice them all. If you haven't yet, please create blocks of Mathematica code that will help you quickly compute work, surface integrals, and flux integrals, as well as quickly see $\vec r_u\times \vec r_v$ so you can make decisions about in which direction this normal vector points. I'll add more problems for later days. If you want more to work on before then, head to OpenStax and tackle any of the problems in 6.6, 6.7, or 6.8.

Task 45.4

Pick some problems related to the topics we are discussing from the Text Book Practice page.

With each questions above, the key is to first identify the right theorem.

1. Use the divergence theorem. Note that the divergence is $\vec\nabla \cdot \vec F = 2+3+1=6$. As such, the outward flux is $\iiint_D 6 dV = 6V = 6(2)(2)(2) = 48$.
2. Use Stokes' Theorem. The curl is $\vec\nabla \times \vec F = (0,8,0)$. Because the curve is a circle, let's use the flat planar surface $S$ inside this circle. The only thing left is to figure out which orientation to use for the surface. The curve is a circle centered at $(2,5,7)$ of radius 3, in a plane parallel to the $xz$-plane, which means a unit normal vector can be chosen as $(0,1,0)$ or $(0,-1,0)$. The orientation of the curve is compatible with $(0,-1,0)$, which we can use in Stokes's theorem. We obtain $$\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS = \ds \iint_S (0,8,0) \cdot (0,-1,0) dS = \ds \iint_S -8 dS = -8\pi(3)^2 = -72\pi.$$ We reduced the problem to a surface area question.
3. Using Green's theorem, we have $$\ds \iint_R N_x-M_y dA = \iint_R \frac{1}{2}-(-\frac{1}{2}) dA = A = \pi (2)^2 = 4\pi.$$ This particular integral is often used by a planimeter to compute the area inside odd shape regions.