9:00 AM Jamboard Links
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12:45 PM Jamboard Links
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Presenters
9AM
Thanks for sharing things in Perusall. Here are the presenters for today. Remember, I start picking presenters at 8am. If you upload something after that time, I may not see it.
- 5.37 - Braydon
- 5.38 - Tanner, Ethan
- 5.39 - Kai, Jeremy
- 6.1 - Olivia
- 6.2 - Jae
- 6.3 -
- 6.5 -
- 6.6 -
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today. Remember, I start picking presenters at 12 noon. If you upload something after that time, I may not see it.
- 5.35 -
- 5.37 -
- 5.38 -
- 5.39 -
- 6.1 -
- 6.2 -
- 6.3 -
- 6.5 -
Learning Reminders
- We are in the 13th week of the semester. If you are on track for an A, then ideally you're finishing your SDL project for the 5th unit, and proposed something for the 6th unit.
- There are only 2 weeks left, which means you can submit at most 2 more SDL projects.
- The final SDL project (6th) can be over any topic from the entire semester. You can use it to expand what we do in the 6th unit, or you may choose to revisit something from a prior unit that you would like to spend more time with.
Brain Gains (Rapid Recall)
- Consider the region $R$ in the plane that lies below $y = f(x)$ and above $y=g(x)$, for $a\leq x\leq b$. Set up formulas involving iterated double integrals to compute both $\bar x$ and $\bar y$.
Solution
The centroid is $$\bar x = \frac{\int_a^b\int_{g(x)}^{f(x)}x dy dx}{\int_a^b\int_{g(x)}^{f(x)} dy dx} \quad\text{and}\quad \bar y = \frac{\int_a^b\int_{g(x)}^{f(x)}y dy dx}{\int_a^b\int_{g(x)}^{f(x)} dy dx}.$$ Note that if we had a density function $\delta (x,y)$, then the center-of-mass would be $$\bar x = \frac{\int_a^b\int_{g(x)}^{f(x)}x \delta(x,y) dy dx}{\int_a^b\int_{g(x)}^{f(x)} \delta(x,y)dy dx} \quad\text{and}\quad \bar y = \frac{\int_a^b\int_{g(x)}^{f(x)}y \delta(x,y) dy dx}{\int_a^b\int_{g(x)}^{f(x)} \delta(x,y) dy dx}.$$
Let's look at 39 now.
- Set up an integral to find the volume of the region in space above the $xy$-plane that is bounded above by the plane $z=8$ (so $\rho \cos\phi = 8$) and below by the cone $z^2=x^2+y^2$ (so $\phi = \pi/4$).
Solution
$$\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8\sec\phi}\rho^2\sin\phi d\rho d\phi d\theta$$
- Set up an integral to find $\bar z$ for the center-of-mass of the region above if the density is $\delta = x^2+y^2+z^2$. If you forgot, recall $z=\rho\cos\phi$ and the Jacobian is $\rho^2\sin\phi$.
Solution
The bounds don't change at all, rather we just have to add the correct pieces from the center-of-mass formulas. This gives $$\bar z = \frac{\iiint_D z\delta dV}{\iiint_D\delta dV}=\frac{\ds\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8\sec\phi}\overbrace{(\rho\cos\phi)}^{z}(\rho^2) (\rho^2\sin\phi) d\rho d\phi d\theta}{\ds\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8\sec\phi}\underbrace{(\rho^2)}_{\delta} \underbrace{(\rho^2\sin\phi) d\rho d\phi d\theta}_{dV}}$$
- 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 $$\bar f = \frac{\int_{-3}^{2}\int_{x+2}^{8-x^2}(x+y^2)dydx}{ \int_{-3}^{2}\int_{x+2}^{8-x^2}dydx}.$$
- The length of a wire that lies along the helix $\vec r(t) = (3\cos t,3\sin t, 4t)$ for $0\leq t\leq 4\pi$ is $$\int_{0}^{4\pi} \sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt.$$ Set up an integral to compute the average charge density, provided the charge density at each point on the wire is given by $\sigma(x,y,z) = x^2+y^2+z$.
Solution
The average charge density is $$\bar \sigma = \frac{ \int_{0}^{4\pi} [(3\cos t)^2+(3\sin t)^2 +4t]\sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt}{ \int_{0}^{4\pi} \sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt}.$$
- 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 $$\bar P = \frac{\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3}(10+\rho\sin\phi\cos\theta)\rho^2\sin\phi d\rho d\phi d\theta}{ \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta}.$$
Group problems
- Draw the solid whose volume is given by the integral $\ds\int_{0}^{\pi}\int_{\pi/3}^{\pi/2}\int_{0}^{3}\rho^2\sin\phi \,d\rho \,d\phi \,d\theta$. Check with Integration.nb.
- Set up an integral formula to compute the $z$-coordinate of the center of mass of the solid above, provided the density is given by $\delta = x^2+y^2+z^2$.
- Let $f = xy^2+3x$.
- Compute $\vec F = \vec \nabla f$. (We'll use capital $F$ for vector fields, and lower case $f$ for real-valued functions, i.e. the output is a single real number, rather than a vector.)
- Compute $D^2f$ (so $D\vec F$). It's a square matrix.
- Compute $\int_{ (2,1) }^{ (1,3) } df$ (total change in $f$ from $(2,1)$ to $(1,3)$). [Check: Plugging the points into $f$ yields $12-8 = 4$.]
- Let $\vec F = (2xy+4, x^2+2y)$.
- Compute $D\vec F$. It's a square matrix.
- Find a real-valued function $f$ so that $\vec \nabla f = \vec F $. In particular, this means $df = \vec F \cdot d\vec r$. [Check: $f = x^2y+4x+y^2$.]
- Find the work done by $\vec F$ (so $\int_C \vec F\cdot d\vec r$) to get from $(2,0)$ to $(0,3)$. (Or simpler, just compute $\int_C df$, the total change in $f$ from $(2,0)$ to $(0,3)$, which you can do because $df = \vec F \cdot d\vec r$. ) [Check: $9-4=5$.]
- Let $\vec F = (2x+3y, 4x+5y)$.
- Compute $D\vec F$.
- Why it is impossible to find a function $f$ so that $\vec F = \vec \nabla f$.
- Given a vector field $\vec F$, what condition must be true about $D\vec F$ for there to be a function $f$ such that $\vec\nabla f = \vec F$? We call such function $f$ a potential for $\vec F$. When $\vec F$ has a potential, we say that $\vec F$ is a gradient field.
- Compute the derivative of each vector field $\vec F$ below (obtaining a square matrix). Then find a potential for $\vec F$ or explain why the vector field has no potential.
- $\vec F = (2x,3y)$ [Check: $D\vec F = \begin{bmatrix}2&0\\0&3\end{bmatrix}$ and $f = x^2+\frac{3}{2}y^2$ yields $\vec \nabla f = (2x,3y)$. We can quickly verify that $\vec\nabla f = \vec F$ by a direct computation. ]
- $\vec F = (2y,3x)$
- $\vec F = (3y,3x)$
- $\vec F = (4x,5y,6z)$
- $\vec F = (4x,5z,6y)$
- $\vec F = (4x,5z,5y)$
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
- (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 $$V = \int dV =\int_a^b \underbrace{\pi (\text{radius of disc at height $z$})^2}_{\text{area of disc at height $z$}} \underbrace{dz}_{\text{little height}}.$$
Textbook practice
If you want more integrals to work with. These all come from Thomas's calculus, the 14th edition. You can also get lots of practice with integration from chapter 5 of OpenStax's text. If you want to complete an SDL project that involves working through many problems from these sections, and then sharing a video or slide show of how you solve a few that helped you learn the most, I'll happily approve it. Here are the sections in Thomas's Calculus.
- 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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