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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.
- 5.08 - Cecillia
- 5.09 - Zack, Jordan
- 5.11 - Ethan, Gavin
- 5.12 - Tanner
- 5.13 -
- 5.14 -
- 5.15 -
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.
- 5.07 - Cheyenne
- 5.08 -
- 5.09 -
- 5.11 -
- 5.12 -
- 5.13 -
- 5.14 -
- 5.15 -
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 triangular region $R$ in the plane that is bounded by the curves $y=2x$, $y=4$, and $x=0$. Set up a double integral that compute the area of this region.
Solution
There are two possible answers, namely $$ \int_{0}^{2}\int_{2x}^{4}dydx \text{ or } \int_{0}^{4}\int_{0}^{y/2}dxdy. $$ Both are equally correct.
- Using the same region $R$ as above, set up a double integral formula to compute $\bar x$, the $x$-coordinate of the centroid of $R$ (the centroid is the geometric center-of-mass assuming uniform density).
Solution
There are two possible answers, namely $$ \frac{\int_{0}^{2}\int_{2x}^{4}xdydx} {\int_{0}^{2}\int_{2x}^{4}dydx} \text{ or } \frac{\int_{0}^{4}\int_{0}^{y/2}xdxdy} {\int_{0}^{4}\int_{0}^{y/2}dxdy}. $$ Both are equally correct. The bounds are completely independent of the formula $\bar x = \dfrac{\iint_R xdA}{\iint_R dA}$.
- Using the same region $R$ as above but adding a varying density of $\delta(x,y) = x^2y$, set up a double integral formula to compute $\bar y$, the $y$-coordinate of the center-of-mass of $R$.
Solution
There are two possible answers, namely $$ \frac{\int_{0}^{2}\int_{2x}^{4}y(x^2y)dydx} {\int_{0}^{2}\int_{2x}^{4}(x^2y)dydx} \text{ or } \frac{\int_{0}^{4}\int_{0}^{y/2}y(x^2y)dxdy} {\int_{0}^{4}\int_{0}^{y/2}(x^2y)dxdy}. $$ Both are equally correct. The bounds are completely independent of the formula $\bar x = \dfrac{\iint_R xdm}{\iint_R dm}$, where $dm=\delta dA$.
- If $\vec u$ and $\vec v$ are both 3D vectors, what physical quantity does $|\vec u\times \vec v|$ compute?
Group problems
- Consider the region $R$ that is bounded by the lines $x=0$, $y=6$, and $x=y/2$. The density (mass per area) is given by $\delta(x,y)$.
- Set up a double integral to compute the area of $R$.
- Set up a double integral to compute the mass of $R$.
- Set up a double integral formula to compute $\bar x$ and $\bar y$ for the centroid of $R$.
- Set up a double integral formula to compute $\bar x$ and $\bar y$ for the center-of-mass of $R$.
- Consider $\int_{0}^{4}\int_{x}^{4}\cos(y^2)dydx$.
- Draw the region described by the bounds.
- Swap the order of the bounds on the integral (use $dxdy$ instead of $dydx$).
- Compute the integral from the last step.
- Draw the region described the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3\pi/2}\int_{0}^{2+2\cos\theta}rdrd\theta$
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- $\ds\int_{0}^{3}\int_{0}^{\pi}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- 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 $z$-coordinate of the center-of-mass (so $\bar z$) of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$.
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (t^2+1, 3t, t^3)$ for $-1\leq t\leq 2$.
- Compute $ds$. (Remember - a little distance equals the product of the speed and a little time.)
- Set up an integral to find $\bar x$, then $\bar y$, then $\bar z$, for the centroid of $C$.
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