Brain Gains (Rapid Recall, Jivin' Generation)

  1. 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.
  1. 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.