- 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)
- For the change-of-coordinates $x=2u+v$, $y=u-2v$, compute the differentials $dx$ and $dy$ and state them using the vector form (linear combination) $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} ?\\ ?\end{pmatrix}du+\begin{pmatrix}?\\?\end{pmatrix}dv.$$
Solution
We have $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} 2\\ 1\end{pmatrix}du+\begin{pmatrix}1\\-2\end{pmatrix}dv.$$
- The box in the $r\theta$ plane with $2\leq r\leq 3$ and $\frac{\pi}{6}\leq \theta\leq \frac{\pi}{4}$ corresponds to a region in the $xy$ plane. Shade this region in the $xy$ plane, and then state the length of the inner edge (the edge that lies on the circle of radius $2$). This is similar to Task 11.
Solution
We'll draw the region together, which you can check with Mathematica.
ParametricPlot[{r Cos[theta], r Sin[theta]}, {theta, Pi/6, Pi/4}, {r, 2, 3}, PlotRange -> 3]
The length of the inner edge is the radius $r=2$ multiplied by the angle traversed $d\theta = \pi/4-\pi/6 = \pi/12$, so $r d\theta = (2)(\pi/12)$. The key point is that the little length ($ds$) of the inner edge is $$ds=rd\theta.$$
- For the polar coordinate transformation given by the change-of-coordinates $x=r\cos\theta$ and $y=r\sin\theta$, compute the differential $(dx,dy)$, and then show that the Jacobian is $\ds\frac{\partial (x,y)}{\partial (r,\theta)} = |r|$. This is part of Task 12.
Solution
We compute $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} \cos\theta\\ \sin\theta\end{pmatrix}dr+\begin{pmatrix}-r\sin\theta\\r\cos theta\end{pmatrix}dr.$$ The Jacobian is then $$\dfrac{\partial(x,y) }{\partial(x,y) }=\left|(\cos\theta)(r\cos\theta)-(\sin\theta)(-r\sin\theta)\right| = |r|.$$
- Set up an iterated double integral formula that gives the mass of the region in the $xy$-plane that lies inside the curve $r=2-2\cos\theta$ and has density given by $\delta = x^2+y^2$. This is similar to Task 13.
Solution
Note first that $\delta = x^2+y^2=r^2$. The curve starts to repeat after $\theta= 2\pi$. This gives the solution as $$ \iint_R dA = \int_{0}^{2\pi}\int_{0}^{2-2\cos\theta} (r^2)|r| dr d\theta. $$ The absolute values on $|r|$ can be dropped as we know $r\geq 0$ from the lower bound of the inner integral.
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 $$ \iint_R dA = \int_{0}^{\pi}\int_{0}^{2\cos\theta} r dr d\theta. $$
Beware! In this problem, note that $2\cos\theta<0$ for $\pi/2<\theta\leq \pi$. We end up with the correct value for the integral because both $r$ is negative and $dr$ is negative for this region. A possibly better option that avoids any issues with a negative $r$ or $dr$ is $$ \iint_R dA = \int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta} r dr d\theta. $$
Note that we obtain $$ \int_{0}^{\pi}\int_{0}^{2\cos\theta} |r| dr d\theta=0, $$ as here only $dr$ is counted negative for $\pi/2<\theta\leq \pi$.
We can use Mathematica to quickly plot the curve, and evaluate various integrals.
PolarPlot[2 Cos[theta], {theta, 0, Pi}]
Integrate[r, {theta, 0, Pi}, {r, 0, 2 Cos[theta]}]
(*Incorrect*)Integrate[Abs[r], {theta, 0, Pi}, {r, 0, 2 Cos[theta]}]
Integrate[r, {theta, -Pi/2, Pi/2}, {r, 0, 2 Cos[theta]}]
Integrate[Abs[r], {theta, -Pi/2, Pi/2}, {r, 0, 2 Cos[theta]}]
Group Problems
- Draw the region whose area is given by $\ds\int_0^{\pi/2}\int_0^{3\sin(2\theta)}rdrd\theta$.
- 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.
- 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}]
|
Sun |
Mon |
Tue |
Wed |
Thu |
Fri |
Sat |