Schedule
Today
- 31
- 32
- 34
- 35
- 36
- 37
- 38
- 39
Next time
- Whatever we don't finish today.
- We'll finish the unit up on Monday.
- Quiz will be open all weekend. Take it this weekend the first time.
- Exam on Tuesday/Wednesday - In I-Learn
Rapid Recall
1. Consider the sphere $x^2+y^2+z^2=9$. Give a cylindrical equation for the sphere. Then give a spherical equation for the sphere.
Solution
In cylindrical, an equation is $r^2+z^2=9$. In spherical, an equation is $\rho = 3$.
Remember $x^2+y^2=r^2$ (as $r$ is the distance to the $z$-axis), and $x^2+y^2+z^2=\rho^2$ (as $\rho$ is the distance to the origin).
2. The curves $y=8-x^2$ and $y=x+2$ intersect at $x=2$ and $x=-3$. The area of the region in space bounded by these two curves is $\ds \int_{-3}^{2}\int_{x+2}^{8-x^2}dydx$. Set up an integral to compute the average temperature of a metal plate in the $xy$-plane that lies in this region, provided the temperature at points on the plate is given by $f(x,y)=x+y^2$.
Solution
The AVERAGE temperature is $$\bar f = \frac{\int_{-3}^{2}\int_{x+2}^{8-x^2}(x+y^2)dydx}{ \int_{-3}^{2}\int_{x+2}^{8-x^2}dydx}.$$
3. Set up an integral to compute the average charge density on a wire that lies along the helix $\vec r(t) = (3\cos t,3\sin t, 4t)$ for $0\leq t\leq 4\pi$, provided the charge at each point on the wire is given by $\sigma(x,y,z) = x^2+y^2+z$.
Solution
The AVERAGE charge density is $$\bar \sigma = \frac{ \int_{0}^{4\pi} [(3\cos t)^2+(3\sin t)^2 +4t]\sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt}{ \int_{0}^{4\pi} \sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt}.$$
4. Set up an integral to compute the average pressure in a solid region in space inside the sphere $x^2+y^2+z^2=9$, provided the pressure at each point in the sphere is given by $P(x,y,z) = 10+x$. In case you need them, remember that in spherical coordinates, we have $x=\rho\sin\phi\cos\theta$ and the Jacobian is $|\rho^2\sin\phi|$.
Solution
The AVERAGE pressure is $$\bar P = \frac{\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3}(10+\rho\sin\phi\cos\theta)\rho^2\sin\phi d\rho d\phi d\theta}{ \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta}.$$
Additional problems
Remember, the spherical change-of-coordinates is given by $$(x,y,z) = (\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi).$$
1. Give an equation of the cone $x^2+y^2=z^2$ in both cylindrical and spherical coordinates.
Solution
Recall that $x^2+y^2=r^2$. This means the equation $x^2+y^2=z^2$ becomes $r^2=z^2$ or just $r=z$ in cylindrical coordinates. To get spherical coordinates, we know that $r=\rho\sin\phi$, and $z=\rho\cos\phi$, which mean $z=r$ becomes $\rho \cos\phi = \rho \sin \phi$ or just $\cos\phi = \sin\phi$. However, this simplifies even further to just $\phi =\pi/4$. The cone is describe by the three equations $$x^2+y^2=z^2,\quad r=z\quad, \phi=\pi/4.$$ The spherical equation is the simplest, as it is a variable equals a constant.
2. Set up an integral to find the volume of the region in space above the $xy$-plane that is bounded above by the sphere $x^2+y^2+z^2=9$ (so $\rho = 3$) and below by the cone $z^2=x^2+y^2$. Remember the Jacobian for spherical coordinates is $|\rho^2\sin\phi|$.
Solution
$$\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta$$
3. Give an equation of the plane $z=8$ in spherical coordinates.
Solution
$\rho\cos\phi = 8$ or $\rho = 8\sec\phi$.
4. 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 $z^2=x^2+y^2$.
Solution
$$\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8\sec\phi}\rho^2\sin\phi d\rho d\phi d\theta$$
Extra, if you finish the above.
- (Cylindrical Coordinates - Disc and Shell method) This sequence of problems has you work with one solid in several different ways. It also has you develop the disc and shell method as by-products of cylindrical coordinates, where the only difference is the order of integration.
- Draw 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.
- Set up a triple integral in Cartesian coordinates to compute the volume of this solid. Use software to check your answer.
- Set up a triple integral in cylindrical coordinates to compute the volume of this solid using the order $d\theta drdz$. Check with software.
- Compute the two inside integrals and simplify to show that $V = \int_{0}^{9} \pi (\sqrt{9-z})^2 dz$. Recall the disc-method $$V = \int dV =\int_a^b \underbrace{\pi (\text{rad of disc at $z$})^2}_{\text{area of disc at height $z$}} \underbrace{dz}_{\text{little height}}.$$
- Set up a triple integral in cylindrical coordinates to compute the volume of this solid using the order $d\theta dzdr$. Check with software.
- 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 $$V = \int dV = \int_a^b \underbrace{(2\pi r)(\text{height of shell at $r$})}_{\text{shell surface area = (circumference)(height)}} \underbrace{dr}_{\text{shell thickness}}.$$
- Use any remaining time to pick some problems from the list below, and tackle them.
Textbook practice
If you want more integrals to work with. These all come from Thomas's calculus, the 14th edition. Feel free to tackle these in your group today in class, if you have extra time.
- Double integrals - Swapping order - 15.2: 33-54
- Polar integrals - 15.4: 1-8, 9-22 (swap to polar and then use software to check)
- Triple integrals - 15.5: 21-36
- Cylindrical and Spherical- 15.7: 37-42 (cyl), 55-60 (sph), 65-84 (BEST ones, you have to pick the system, draw the region, set things up - use the Mathematica notebook Integration.nb to check if your bounds are right).
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