Brain Gains (Rapid Recall, Jivin' Generation)

With appropriate assumptions place on the corresponding vector fields and objects involved, here are the three big theorems we've seen.

  • Green's Theorem: $\ds \iint_R N_x-M_y dA = \oint_{\partial R} M dx+Ndy $
  • Stokes' Theorem: $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS = \int_{\partial S} Mdx+Ndy+Pdz$.
  • Divergence Theorem: $\ds \iiint_D \vec \nabla \cdot \vec F dV = \iint_{\partial D}\vec F\cdot\hat n dS$.

One (of many) use of each theorem is that sometimes one side of the equality is extremely simple to compute. Let's see this in a few settings.

  1. Let $\vec F = (2x-y^2z, 3y-4x^3, e^{x^2y}+z)$. Compute the outward flux of $\vec F$ across the surface $S$ of the cube $-1\leq x\leq 1$, $-1\leq y\leq 1$, $-1\leq z\leq 1$, so compute $\iint_{S}\vec F\cdot\hat n dS$.
  2. For the vector field $\vec F = (x^2+4z, 3y, -4x)$, compute $\int_C M dx+Ndy+Pdz$ along the curve $C$ parametrized by $\vec r(t) = (2+3\cos t, 5, 7+3\sin t)$ for $0\leq t\leq 2\pi$.
  3. Compute the counterclockwise circulation $\oint_C -\frac{y}{2}dx+\frac{x}{2}dy$ for the curve $C$ defined by $(x-3)^2 + (y+7)^2=4$.

Solutions

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.
  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

  1. Pick your own vector field $\vec F(x,y,z) = (?, ?, ?)$, compute the derivative, the divergence, and the curl. Use Mathematica to check your work. 
    F = {x^2 y, x y z, 3 x + 4 z }
D[F, {{x, y, z}}] // MatrixForm
Div[F, {x, y, z}]
Curl[F, {x, y, z}]
  2. Consider the vector field $\ds \vec F = \frac{1}{2\pi}\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$.
    1. Let $C$ be the curve parametrized by $\vec r(t) = (3\cos t, 3\sin t)$ for $0\leq t\leq 2\pi$. Compute the work done by $\vec F$ to move an object along $C$. [Hint: The answer is NOT zero.]
    2. Let $C$ be the curve parametrized by $\vec r(t) = (7\cos t, 7\sin t)$ for $0\leq t\leq 4\pi$. Compute the work done by $\vec F$ to move an object along $C$.
    3. Let $C$ be the curve parametrized by $\vec r(t) = (a\cos t, a\sin t)$ for $0\leq t\leq 2k\pi$. Compute the work done by $\vec F$ to move an object along $C$.
    4. Show that $D\vec F(x,y)$ is symmetric.
    5. Does $\vec F$ have a potential?

Solutions

If $\vec F$ had a potential, then the work done along a closed curve would be zero. This is not true, as the first 3 parts illustrate. The vector field does NOT have a potential.

The function and derivative are not defined at $(0,0)$, which means the domain of the vector field is not simply connected. We cannot use the symmetry of the derivative to conclude that there is a potential, because the domain is not simply connected.

Here are some Mathematica computations relevant to this problem. 

ClearAll[F, r]
F[x_, y_] := 1/(2 Pi (x^2 + y^2)) {-y, x}
r[t_] := {a Cos[t], a Sin[t]}

(*Is the derivative symmetric*)
D[F[x, y], {{x, y}}] // Simplify // MatrixForm

(*Try to find a potential*)
Integrate[F[x, y][[1]], x]
Integrate[F[x, y][[2]], y]

(*Compute work done by wrapping k times around a circle of radius a*)
Dot[F @@ r@t, D[r[t], t]] // Simplify
Integrate[Dot[F @@ r@t, D[r[t], t]], {t, 0, 2 k Pi}]
  1. Consider the vector field $\vec F = (x,x-z,y+z)$, the surface $S$ parametrized by $\vec r(u,v)=(u^2, u\cos v, u\sin v)$ for $0\leq u\leq 2$ and $0\leq v\leq 2\pi$, and the curve $C$ parametrized by $\vec r(t) = (4,2\cos t, 2\sin t)$ for $0\leq t\leq 2\pi$.
    1. Draw the surface $S$ and curve $C$. How are these two objects related?
    2. Compute $\vec N = \vec r_u\times \vec r_v$ and determine if $\vec N$ points inward toward the $x$-axis, or outwards away from the $x$-axis.
    3. Set up and compute the integral $\ds \int_C Mdx+Ndy+Pdz$, computing the work done by $\vec F$ along $C$.
    4. Set up and compute $\ds \iint_S \vec \nabla \times \vec F\cdot \hat n dS$, computing the flux of the curl of $\vec F$ across $S$ in the direction $\hat n$ outwards away from the $x$-axis.
  2. For the cone $\vec r(u,v) = (u, u\cos v, u\sin v)$ for $0\leq u\leq 5$ and $0\leq v\leq 2\pi$, set up the integrals needed to compute the center-of-mass of the cone if density at points in the cone is given by $\delta = (x+z+10)$.
  3. Set up an integral to compute the flux of $\vec F = (2x+3y, -3x+4z, 4y+z^2)$ across the surface of the cone from above, using the outward (away from $x$-axis) orientation.

End