Day 46 - Prep

Continue working on the tasks from Day 45 and Day 44.

Day 47 - Prep

Task 47.1

From Ben: This task is now ready to work on.

Consider the vector field $\vec F = \frac{ (x,y,z) }{ (x^2+y^2+z^2) ^{3/2}}$. This vector field is directly proportional to gravitational and electric fields. In this task, we will show that flux of this field across any closed surface that contains the origin is $4\pi$, while the flux across any surface that does not contain the origin is zero.

Show that the divergence of $\vec F$ is 0, provided $(x,y,z)\neq(0,0,0)$. You can to this by hand, or with software, or find the solution somewhere online (one of the links in 44.1 has the work).
Consider the surface $S$ which is a sphere of radius $a$ (so $x^2+y^2+z^2=a^2$). Compute the outward flux of $\vec F$ across $S$, and show that you get $$\Phi = \iint_S \vec F\cdot \hat n dS = 4\pi.$$ In particular, notice that the answer does not depend on the radius of the sphere. There are multiple ways to do this. One option is to obtain a parametrization of the surface (look back at previous problems to get a parametrization) and then set up and compute the integral directly. Another is to reason out geometrically what $\hat n$ must equal because we're on a sphere of radius $a$, and then replace $x^2+y^2+z^2$ with $a^2$ in many places.
Let $S$ be any closed surface which does not contain $(0,0,0)$ on the surface, or inside the surface. Explain why $\iint_S \vec F\cdot \hat n dS =0$
Let $S$ be any closed surface which does contain $(0,0,0)$ inside the surface. Explain why $\iint_S \vec F\cdot \hat n dS =4\pi$. (Start by picking a small radius $a$ so that the sphere $x^2+y^2+z^2=a^2$ lies entirely inside $S$.)

Task 47.2

From Ben: This task is now ready to work on.

Let $D$ be the solid region in space that lies above the cone $z^2=x^2+y^2$ and below the paraboloid $z=6-x^2-y^2$. In cylindrical coordinates, the domain $D$ lies above the cone $z=r$ and below the paraboloid $z=6-r^2$. Let $\vec F = (x^2, 4y+z, z-x+3)$. Verify the divergence theorem for this solid region. A parametrization for the cone is $\vec r(u,v) = (u\cos v, u\sin v, u)$ for $0\leq u\leq 2$ and $0\leq v\leq 2\pi$. A parametrization for the parabaloid is $\vec r(u,v) = (u\cos v, u\sin v, 6-u^2)$ for $0\leq u\leq 2$ and $0\leq v\leq 2\pi$.

Compute $\iint_S \vec F\cdot \hat n dS$ across the cone, making sure you have an outward pointing normal vector.
Compute $\iint_S \vec F\cdot \hat n dS$ across the paraboloid, making sure you have an outward pointing normal vector.
Compute $\iiint_D \vec \nabla \cdot \vec F dV$, and show how to combine the results of the previous computations to get the same value.

In my work above, I ended up seeing the values $-20\pi/3$, $60\pi$, and $160\pi/3$.

Task 47.3

Pick a problem from OpenStax from 6.7 or 6.8, where it asks you to verify either Stokes' or the Divergence theorem, and complete it.

Task 47.4

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

Day 47 - In class

Brain Gains (Rapid Recall, Jivin' Generation)

Let $\vec F = (x+y,2x-3y+z, 4z)$.
1. Compute the flux of $\vec F$ outwards across the surface of the cube $ [0,2]\times [0,2]\times [0,2] $.
2. Compute the flux of $\vec F$ across the top surface of the cube, using the upward normal vector. A parametrization of the top surface is $\vec r(u,v) = (u, v, 2)$.
3. Compute the total outward flux of $\vec F$ across the other 5 surfaces of the cube. [How can you combine the previous results?]

Solutions

The divergence theorem gives the total outward flux, which in this case is $\Phi = \iiint_D 1-3+4 dV = 2V = 16$.

The flux across the top surface (normal vector is $(0,0,1)$ by inspection) is $\iint_S(u+v,2u-3v+2,8)\cdot(0,0,1)dS = \iint_S 8 dS$, which is 8 times the surface area (2 by 2 square). This gives the total flux upwards across the top as 32.

We need the flux across the other 5 surfaces. The divergence theorem states that the total outward flux across all 6 surfaces is 16. Because the top surface has a flux of 32, this means that the total outward flux across the other surface must be -16, so that the total sum is 16.

We can visualize all the above using Mathematica.

F = {x + y, 2 x - 3 y + z^2, 4 z};
r = {u , v, 2};
uB = {u, 0, 2};
vB = {v, 0, 2};
p1 = ParametricPlot3D[Evaluate[r, uB, vB]];
surface = ParametricRegion[r, Evaluate[{uB, vB}]];
SliceVectorPlot3D[F, surface, {x, 0, 2}, {y, 0, 2}, {z, 0, 2}, 
 VectorScaling -> Automatic, VectorSizes -> Automatic, 
 VectorMarkers -> Placed["Arrow3D", "Start"]]
SliceVectorPlot3D[F, "CenterCutBox", {x, 0, 2}, {y, 0, 2}, {z, 0, 2}, 
 VectorScaling -> Automatic, VectorSizes -> Automatic, 
 VectorMarkers -> Placed["Arrow3D", "Start"]]
SliceVectorPlot3D[F, "BackPlanes", {x, 0, 2}, {y, 0, 2}, {z, 0, 2}, 
 VectorScaling -> Automatic, VectorSizes -> Automatic, 
 VectorMarkers -> Placed["Arrow3D", "Start"]]

For reference (if you want to practice computing flux), the outward flux across each surface is given below, which we can visualize using the code above.

The surfuce $z=0$ (parametrization $\vec r(u,v) = (u,v,0)$ has flux 0.
The surfuce $z=2$ (parametrization $\vec r(u,v) = (u,v,2)$ has flux 32.
The surfuce $y=0$ (parametrization $\vec r(u,v) = (u,0,v)$ has flux -40/3.
The surfuce $y=2$ (parametrization $\vec r(u,v) = (u,2,v)$ has flux -32/3.
The surfuce $x=0$ (parametrization $\vec r(u,v) = (0,u,v)$ has flux -4.
The surfuce $x=2$ (parametrization $\vec r(u,v) = (2,u,v)$ has flux 12.

Let $\vec F = (2y, x^2+1)$. Let $C$ be the piecewise curve which forms the boundary of the right side of the disc $x^2+y^2\leq 9$, which involves two curves namely the half circle parametrized by $\vec r(t) = (3 \cos t, 3\sin t)$ for $-\pi/2\leq t\leq \pi/2$, and the straight line connecting $(0,3)$ to $(0,-3)$ (I'll let you parametrize it). Verify Green's theorem for this vector field and curve $C$. This will require you to set up and compute all integrals involved in the theorem.

Solution

We have $N_x-M_y = 2x-2$, as such the double integral from Green's theorem is (you can pick whichever coordinate system you want to use) $$\ds \iint_R 2x-2 = \int_{0}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}(2x-2)drd\theta = \int_{-\pi/2}^{\pi/2}\int_{0}^{3}(2 r\cos\theta -2)rdrd\theta = 36-9\pi.$$

We now compute $\int_C Mdx+Ndy$ for each of the two curves. For the semicircle, we have $$\ds \int_C Mdx+Ndy = \int_{-\pi/2}^{\pi/2}[(2(3\sin t))(-3\sin t)+((3\cos t)^2+1)(3\cos t)]dt = 42-9\pi.$$ For the second curve, we have $\vec r(t) = (0,3)+(0,-6)t$ for $0\leq t\leq 1$, which gives $$\ds \int_C Mdx+Ndy = \int_{0}^{1}[(2(3-6t))(0)+((0)^2+1)(-6)]dt = -6.$$ The sum of these two integrals is $36-9\pi$, precisely the double integral above.

The following Mathematica code computes all the integrals above. I use rr for the polar coordinate $r$, so that I can use r for parametrizations and plotting.

Integrate[(2 x - 2), {x, 0, 3}, {y, -Sqrt[9 - x^2], Sqrt[9 - x^2]}]
Integrate[(2 rr Cos[theta] - 2) rr, {theta, -Pi/2,  Pi/2}, {rr, 0, 3}]
Integrate[(2 (3 Sin[t])) (-3 Sin[t]) + ((3 Cos[t])^2 + 1) (3 Cos[t]), {t, -Pi/2, Pi/2}]
Integrate[(2 (3 - 6 t)) (0) + (0^2 + 1) (-6), {t, 0, 1}]

Group Problems

We'll have people share the work they've prepared.

Day 48 - Prep

Learning Target Checkoff

A learning target quiz will appear in I-Learn. Complete and submit the quiz before the due date.

Today

Sep24, Oct, Nov, Dec

20221205