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.
- 5.13 - Chase
- 5.14 - Parker
- 5.15 - Ralph, Jae, Tanner
- 5.16 - Zack
- 5.17 -
- 5.18 - Jeremy
- 5.19 - Ethan
- 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.
- 5.13 -
- 5.14 -
- 5.15 -
- 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 iterated integral that computes the area of the region $R$ that is inside a circle of radius $a$ centered at the origin, and left of the $y$-axis.
Solution
$$A = \int_{\pi/2}^{3\pi/2}\int_0^a r\,dr\,d\theta = \int_{-a}^a\int_{-\sqrt{a^2-y^2}}^0\,dx\,dy = \int_{-a}^0\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
- Suppose the region $R$ above describes a laminar object with density $\delta(x,y) = 5+x^2+y^2$. Set up a double integral formula to compute the mass of the object.
Solution
$$m = \int_{\pi/2}^{3\pi/2}\int_0^a (5+r^2)\, r\,dr\,d\theta = \int_{-a}^a\int_{-\sqrt{a^2-y^2}}^05+x^2+y^2\,dx\,dy = \int_{-a}^0\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}5+x^2+y^2\,dy\,dx$$
- Consider two vectors $\vec u = \left< a,b,0 \right>$ and $\vec v = \left< c,d,0 \right>$ in the $xy$-plane. They form a parallelogram in the $xy$ plane.
- Geometrically explain why the cross product either points directly up, or directly down (so is parallel to the $z$-axis).
- Give the magnitude of the cross product of $\vec u$ and $\vec v$.
- Use the cross product formula $\vec u \times \vec v = \left<u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1\right>$ to compute the cross product.
Solution
- The cross product gives a vector that is orthogonal to both. The $z$-axis meets any vector in the $xy$-plane at a 90 degree angle.
- Because the cross product has a magnitude equal to the area of the parallelogram formed, we can use $A = |ad-bc|$.
- Using the formula gives $$\vec u \times \vec v = \left<b 0 - 0 d, 0 c - a 0, a d - b c\right>= \left<0, 0, a d - b c\right>.$$ Note that the length of this vector is $|\vec u \times \vec v| = \sqrt{(ad-bc)^2}=|ad-bc|.$
Group problems
- Consider $\int_{0}^{4}\int_{x}^{4}e^{y^2}dydx$.
- Draw the region described by the bounds.
- Swap the order of the bounds on the integral (use $dxdy$ instead of $dydx$).
- Compute the integral.
- Draw the region described the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{3-x}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- $\ds\int_{0}^{\pi}\int_{0}^{3}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- 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 $z$-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 $2x+3y+6z=6$ (so $\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$).
- 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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