Presentations
5.7, 14 (part 3), 16-21Pacing Tracker
- The quizzes have included questions for 22 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 through 4. Have you started your self-directed learning project for each unit?
- The 6th project can be over any topic from the entire semester. Feel free to get started on this one as soon as you have an idea.
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- 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.
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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