Presentations
5.? - 5.?Pacing Tracker
- The quizzes have included questions for 22 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 through 4. Have you started your self-directed learning project for each unit?
- The 6th project can be over any topic from the entire semester. Feel free to get started on this one as soon as you have an idea.
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- Give the $(x,y,z)$ coordinates of the point in space whose spherical coordinates are $(\rho,\phi,\theta) = (3,\pi/2,\pi)$.
Solution
We're on a sphere of radius 3. The fact that $\rho=\pi/2$ means we've dropped 90 degrees from the positive $z$-axis, so we are in the $xy$-plane. With an angle of $\theta=\pi$, that puts us on the negative $x$-axis. The solution is $$(x,y,z)=(-3,0,0).$$
- Give $(\rho,\phi,\theta)$ coordinates for the Cartesian coordinates $(0,0,4)$, $(3,0,0)$, $(0,2,0)$, and $(0,0,-1)$.
Solution
A solution appears in the table below. When $\phi=0$ or $\phi=\pi$, the value of $\theta$ is arbitrary. You can of course add any multiple of $2\pi$ to any value of $\theta$ below, and still have a valid solution. $$\begin{array}{c|c} \text{Cartesian}&\text{Spherical}\\ (x,y,z)&(\rho,\phi,\theta)\\\hline (0,0,4)& (4,0,\text{any})\\ (3,0,0)& (3,\pi/2,0)\\ (0,2,0)& (2,\pi/2,\pi/2)\\ (0,0,-1)&(1,\pi,\text{any}) \end{array}$$
- Draw the region in space described by $1\leq \rho \leq 3$, $0\leq \theta \leq \pi$, and $\pi/4\leq \phi\leq \pi/2$.
Solution
- Set up an integral formula to compute $\bar z$ for the solid above. (Recall $\frac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)} = |\rho^2\sin\phi|$ and $z=\rho\cos\phi$.)
Solution
With $dV =(\rho^2\sin\phi) d\rho d\phi d\theta$ and $z = \rho \cos \phi$, we have $$\bar z =\frac{\iiint_DzdV}{\iiint_DdV} = \frac{\int_{0}^{\pi}\int_{\pi/4}^{\pi/2}\int_{1}^{3}(\rho\cos\phi)(\rho^2\sin\phi) d\rho d\phi d\theta}{\int_{0}^{\pi}\int_{\pi/4}^{\pi/2}\int_{1}^{3}(\rho^2\sin\phi) d\rho d\phi d\theta}.$$
- A tennis ball has been cut in half. Suppose the inner radius of the tennis ball is 3 cm, and the outer radius is 3.5 cm. Give bounds of the form $?\leq \theta\leq ?$, $?\leq \phi\leq ?$, $?\leq \rho\leq ?$, to describe the bottom half ($z\leq 0$) of the tennis ball.
Solution
One option for the bounds is $0\leq \theta\leq 2\pi$, $\pi/2\leq \phi\leq \pi$, $3\leq \rho\leq 3.5$
Group problems
- The spherical change-of-coordinates is given by $$(x,y,z) = (\underbrace{\rho\sin\phi}_{r}\cos\theta, \underbrace{\rho\sin\phi}_{r}\sin\theta, \rho\cos\phi).$$
- Give Cartesian coordinates $(x,y,z)$ for the spherical coordinates $(\rho,\phi,\theta)$ given by $(2,\pi/2,\pi)$, $(2,\pi,\pi/2)$, and $(2,0,3\pi)$.
- Explain why an equation of the sphere $x^2+y^2+z^2=9$ in spherical coordinates is $\rho = 3$.
- 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 $\phi = \pi/4$.
- 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 $|\rho^2\sin\phi|$. Remember that you can check if your bounds produce the correct solid with Integration.nb.
- Explain why an equation of the plane $z=8$ in spherical coordinates is $\rho = 8\sec \phi$.
- 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 $z^2=x^2+y^2$.
- (Cylindrical Coordinates - Disc and Shell method) This sequence of problems develops both the shell and disc method as by-products of cylindrical coordinates. The only difference is the order of integration. We'll use the solid region in space that is bounded above by $z=9-x^2-y^2$ (so $z=9-r^2$) and below by the $xy$-plane. In Cartesian coordinates, the volume of this region is given by $$\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{0}^{9-x^2-y^2}dzdydx.$$ We now work with the region using cylindrical coordinates.
- Set up a triple integral in cylindrical coordinates to compute the volume of this solid using the order $d\theta dzdr$. [Check: $\int_{0}^{3}\int_{0}^{9-r^2}\int_{0}^{2\pi}rd\theta dzdr.$ ]
- Compute the two inside integrals and simplify to show that $V = \int_{0}^{3} 2\pi r (9-r^2) dr$. Recall the shell-method $$V = \int dV = \int_a^b \underbrace{(2\pi r)(\text{height of shell at $r$})}_{\text{shell surface area = (circumference)(height)}} \underbrace{dr}_{\text{shell thickness}}.$$
- Set up a triple integral in cylindrical coordinates to compute the volume of this solid using the order $d\theta drdz$. You will end up with $r=\sqrt{9-z}$ as one of the bounds. You can use Integration.nb to check.
- Compute the two inside integrals and simplify to show that $V = \int_{0}^{9} \pi (\sqrt{9-z})^2 dz$. Recall the disc-method $$V = \int dV =\int_a^b \underbrace{\pi (\text{radius of disc at height $z$})^2}_{\text{area of disc at height $z$}} \underbrace{dz}_{\text{little height}}.$$
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