Brain Gains (Rapid Recall, Jivin' Generation)

  • Let $\vec F(x,y) = (2x-3y,-3x+4y^2)$. Compute the work done by $\vec F$ to go once around the circle $\vec r(t) = (3\cos t, 3\sin t)$.

Solution

The vector field has a potential, namely $f(x,y) = x^2-3xy +\frac{4}{3}y^3$. The start and end points of the curve are the same (the curve is a closed curve). The work done is zero, as the potential at the start and end points are the same.

The code below graphs the curve along with the vector field, and then computes the work directly from the formula (without finding a potential). 

F[x_, y_] := {2 x - 3 y, -3 x + 4 y^2}
r[t_] := {3 Cos[t], 3 Sin[t]}
Show[
 VectorPlot[F[x, y], {x, -4, 4}, {y, -4, 4}], 
 ParametricPlot[r[t], {t, 0, 2 Pi}]
 ]
Integrate[Dot[F @@ r@t, D[r[t], t]], {t, 0, 2 Pi}]
  • Set up an iterated integral formula in spherical coordinates ($x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$) to find the $x$-coordinate of the centroid of the solid domain $D$ in the first octant that lies above the cone $x^2+y^2=z^2$ and below the sphere $x^2+y^2+z^2=49$.

Solution

The volume is given by $\ds V= \int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}\rho^2\sin\phi d\rho d\phi d\theta$, which means $$\bar x = \frac{\iiint_D x dV}{\iiint_D dV}= \frac{\int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}(\rho\sin\phi\cos\theta)\rho^2\sin\phi d\rho d\phi d\theta}{\int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}\rho^2\sin\phi d\rho d\phi d\theta}.$$ The following Mathematica code will visualize the region, and compute the centroid. 

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Integrate[rho Sin[phi] Cos[theta] rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}]/
 Integrate[rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}]
  • Let $\vec F(x,y,z) = (a x+by+cz,dx+ey+fz,gx+hy+iz)$. Compute $D\vec F(x,y,z)$, $\vec \nabla \times \vec F$, and $\vec \nabla \cdot \vec F$.

Solution

With Mathematica, we have the following solutions. 

F[x_, y_, z_] := {a x + b y + c z, d x + e y + f z, g x + h y + i z }
D[F[x, y, z], {{x, y, z}}] // MatrixForm
Curl[F[x, y, z], {x, y, z}]
Div[F[x, y, z], {x, y, z}]

Group Problems

  1. The spherical change-of-coordinates is given by $$(x,y,z) = (\underbrace{\rho\sin\phi}_{r}\cos\theta, \underbrace{\rho\sin\phi}_{r}\sin\theta, \rho\cos\phi).$$
    1. Give Cartesian coordinates $(x,y,z)$ for the spherical coordinates $(\rho,\phi,\theta)$ given by $(2,\pi/2,\pi)$, $(2,\pi,\pi/2)$, and $(2,0,3\pi)$.
    2. Explain why an equation of the sphere $x^2+y^2+z^2=9$ in spherical coordinates is $\rho = 3$.
    3. Explain why an equation of the cone $x^2+y^2=z^2$ (so $r^2=z^2$ or $r=z$) in spherical coordinates is $\phi = \pi/4$.
    4. Set up an integral to find the volume of the region in space above the $xy$-plane that is bounded above by the sphere $x^2+y^2+z^2=9$ and below by the cone $z^2=x^2+y^2$. The Jacobian for spherical coordinates is $|\rho^2\sin\phi|$.
      • You can check your solution with Mathematica by updating the code below, or using the notebook Integration.nb 
        coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{rho, 0, 3}, {phi, Pi/4, Pi/2}, {theta, 0, Pi}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
    5. Explain why an equation of the plane $z=8$ in spherical coordinates is $\rho = 8\sec \phi$.
    6. Set up an integral to find the volume of the region in space above the $xy$-plane that is bounded above by the plane $z=8$ and below by the cone $z^2=x^2+y^2$. Check your work with Mathematica.

Some Solutions

Here is the region for part d. 

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 3}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

The corresponding integral is $$V=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta.$$ 

Integrate[rho^2 Sin[phi], {theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 3}]

Here is the region for part f. 

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 8/Cos[phi]}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

The corresponding integral is $$V=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8/ \cos\phi}\rho^2\sin\phi d\rho d\phi d\theta.$$ 

Integrate[rho^2 Sin[phi], {theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 8/Cos[phi]}]
  1. Discuss Tasks 37.1, 37.2, and/or 37.3.