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.
- 5.16 - Zack, Parker
- 5.17 - Jae, Kylar, Gavin
- 5.18 - Jeremy, Jordan, Lucy
- 5.19 - Next Time
- 5.20 - Spencer B
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.
- 5.16 -
- 5.17 -
- 5.18 -
- 5.19 -
- 5.20 -
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)
- Set up an integral formula to compute the $x$ coordinate of the center-of-mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = y+5$.
Solution
Recall that $dm = \delta dA$ and $dA = r dr d\theta$ in polar coordinates. Bounds for the region in polar coordinates are $0\leq \theta\leq 2\pi$ and $0\leq r\leq 2$. This gives $$\begin{align} \bar x &= \frac{\int\int_R x dm}{\int\int_Rdm}\\ &= \frac{\int\int_R x (y+5)dA}{\int\int_R(y+5) dA}\\ &= \frac{\int_0^{2\pi}\int_{0}^{3} (r\cos\theta) (r\sin\theta+5)rdrd\theta} {\int_0^{2\pi}\int_{0}^{3} (r\sin\theta+5)rdrd\theta} . \end{align}$$ You can instead use Cartesian coordinates to set up the integral, which gives $$\begin{align} \bar x &= \frac{\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x(y+5)dydx} {\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (y+5)dydx}\\ &= \frac{\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} x(y+5)dxdy} {\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} (y+5)dxdy} . \end{align}$$
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (3t, t^3)$ for $-1\leq t\leq 2$. Set up an integral formula to find $\bar y$ for the centroid of $C$.
Solution
Recall that a little length is given by $ds = v(t) dt$ (speed multiplied by a little time). Recall that the velocity $\vec v(t) = (3,3t^2)$ gives the speed $v(t) = \sqrt{9+9t^4}$. We then have $$\begin{align} \bar y &= \frac{\int_C y ds}{\int_C ds}\\ &= \frac{\int_{-1}^{2} t^3 \sqrt{9+9t^4} dt}{\int_{-1}^{2}\sqrt{9+9t^4} dt} . \end{align}$$
- Draw the region described the bounds of the integral $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$.
Solution
We'll draw this together, and use Integration.nb to check our work. We'll also discuss how changing the inner bounds will affect the integral with a few examples, such as $z=7-x$ instead of $z=5$, and $z=x$ or $z=-x^2$ instead of $z=0$.
- Draw the region described the bounds of the integral $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$.
Solution
We'll draw this together, and use Integration.nb to check our work.
This
is
space
for
drawing
Group problems
- Draw the region described the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{3-x}dzdydx$
- $\ds\int_{0}^{\pi}\int_{0}^{3}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- Set up an integral formula to compute each of the following:
- The mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$
- The $x$-coordinate of the center of mass (so $\bar x$) of the disc above.
- The volume of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$ (so $\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$).
- The $y$-coordinate of the center-of-mass (so $\bar y$) of previous object.
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