Brain Gains (Rapid Recall, Jivin' Generation)

  • For the vector field $\vec F(x,y,z)=(x+yz,xz+z,xy+y)$, start by computing the second derivative (to check if there is a potential), and then find a potential.

Solution

This is part of Task 34.2. Let's finish discussing the solution to the entire problem.

  • Find the area of the parallelogram whose edges are the vector $(1,2,3)$ and $(-2,0,1)$. Then find the volume of the parallelepiped whose edges are the vectors $(1,2,3)$, $(-2,0,1)$, and $(-1,1,2)$.

Solution

The area is $|(1,2,3)\times(-2,0,1)|=\sqrt{69}$, and the volume is $|[(1,2,3)\times (-2,0,1)]\cdot (-1,1,2)|$. 

Dot[Cross[{1, 2, 3}, {-2, 0, 1}], {-1, 1, 2}]

Let's look at the solution to Task 34.3

  • For the change-of-coordinates $u=x+y$ and $v=x-y$, compute the Jacobians $\ds \frac{\partial (u,v)}{\partial (x,y)}$ and $\ds \frac{\partial (x,y)}{\partial (u,v)}$.

Solution

This is part of Task 35.1.

  • Compute the Jacobian $\dfrac{\partial(x,y,z)}{\partial(r,\theta,z)}$ for cylindrical coordinates $x=r\cos\theta$, $y=r\sin\theta$, $z=z$.

Solution

This is part of Task 35.2.

  • Let $\vec F(x,y,z) = (2x+yz,2z+xz,2y+xy)$ and $C$ be the straight segment from $(2,-5,0)$ to $(1,2,3)$. Compute the work done by $\vec F$ along $C$ by first finding a potential for $\vec F$.

Solution

This is part of Task 35.3.

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 36.1, 36.2, and/or 36.3.