- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
This is day 5 of Unit 3
Brain Gains (Rapid Recall, Jivin' Generation)
- Set up an iterated double integral formula that gives the area of the region in the $xy$-plane that lies inside the curve $r=2-2\cos\theta$.
Solution
The curve starts to repeat after $\theta= 2\pi$. This gives the solution as $$ \iint_R dA = \int_{0}^{2\pi}\int_{0}^{2-2\cos\theta} r dr d\theta. $$
We can use Mathematica to quickly plot the curve.
PolarPlot[2 - 2 Cos[theta], {theta, 0, 2 Pi}]
- Set up an iterated double integral formula that gives the area of the region in the $xy$-plane that lies inside the curve $r=2\cos\theta$.
Solution
For this curve, note that after $\theta = \pi$, the curve begins wrapping around itself. The area is given by $$ \iint_R dA = \int_{0}^{\pi}\int_{0}^{2\cos\theta} r dr d\theta. $$
We can use Mathematica to quickly plot the curve.
PolarPlot[2 Cos[theta], {theta, 0, Pi}]
- Set up an iterated double integral formula that gives the area of the region in the first quadrant of the $xy$-plane that lies both inside the curve $r=2-2\cos\theta$ and inside the curve $r=2\cos\theta$.
Solution
We can use Mathematica to quickly plot the curves.
PolarPlot[{2 - 2 Cos[theta], 2 Cos[theta]}, {theta, 0, Pi/2}]
Notice that rays starting from the origin, and heading out to the curves, hit the cardioid for $0\leq \theta\leq \pi/3$, and then swap to hitting the circle for $\pi/3\leq \theta\leq \pi/2$. This means we'll need two different integrals. One solution is $$ \int_{0}^{\pi/3}\int_{0}^{2-2\cos\theta} r dr d\theta + \int_{\pi/3}^{\pi/2}\int_{0}^{2\cos\theta} r dr d\theta. $$
- Set up an iterated double integral formula that gives the $x$-coordinate of the center-of-mass of the region in the $xy$-plane that lies inside the curve $r=2\cos\theta$, provided the density is given by $\delta(x,y) = x^2+y^2$.
Solution
We already found the area is given by $$ \iint_R dA = \int_{0}^{\pi}\int_{0}^{2\cos\theta} r dr d\theta. $$ Note that the density, in polar coordinates, is $\delta = r^2$. The center-of-mass is then $$ \bar x = \frac{ \iint_R x dm }{ \iint_R dm } = \frac{ \iint_R x \delta dA }{ \iint_R \delta dA } = \frac{ \int_{0}^{\pi}\int_{0}^{2\cos\theta} (r\cos\theta) r^2 r dr d\theta }{ \int_{0}^{\pi}\int_{0}^{2\cos\theta} (r\cos\theta) r^2 r dr d\theta }. $$
Group Problems
- Consider a metal disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$.
- Set up an iterated integral to compute the mass of the disc.
- Set up an iterated integral formula to compute the $x$-coordinate of the center of mass (so $\bar x$) of the disc.
- Draw the region whose area is given by $\ds\int_0^{\pi/2}\int_0^{3\sin(2\theta)}rdrd\theta$.
- A metal plate lies over the right most petal of the rose $r = 4\sin(3\theta)$. The density at points on the plate is given by $\delta(x,y) = xy^2$. Set up a double integral in polar coordinates to give the average density.
- A metal plate lies over one the right most petal of the rose $r = 3\cos(2\theta)$. The density at points on the plate is given by $\delta(x,y) = x$. Set up a double integral in polar coordinates to give the $y$-coordinate of the center of mass.
- Draw the region described by the bounds of each integral.
- $\ds\int_{0}^{2}\int_{2x}^{4}dydx$
- $\ds\int_{0}^{4}\int_{0}^{y/2}dxdy$
- $\ds\int_{0}^{3\pi/2}\int_{0}^{2+2\cos\theta}rdrd\theta$
- $\ds\int_{-3}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
Here's a reminder of code snippets from the prep about how to draw the regions.
ClearAll[x, y, z, r, theta]
(*Cartesian 2D coordinates*)
outerB = {x, 0, 5}
innerB = {y, 2, x + 2}
ParametricPlot[{x, y}, outerB, innerB, Mesh -> {10, 0}]
(*Polar coordinates*)
outerB = {theta, 0, Pi};
innerB = {r, 1, 3+2 Cos[theta]};
ParametricPlot[{r Cos[theta], r Sin[theta]}, outerB, innerB, Mesh -> {10, 0}]
(*Cartesian 3D coordinates*)
outerB = {x, 0, 5};
middleB = {y, 0, 7};
innerB = {z, 0, 10 - 2 x};
R = ParametricRegion[{x, y, z}, Evaluate[{outerB, middleB, innerB}]];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
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