20230224

1. Draw a bear in polar coordinates, and then Cartesian coordinates.....

   

The definition of radian measure is given by $$\text{radian measure} = \frac{\text{arc length}}{\text{radius of circle}}\quad \text{or}\quad \theta = \frac{s}{r}.$$ We've traveled (arc length) $s=2\pi$ units along a circle with radius $r=6$. The radian measure is $$\theta = \frac{s}{r} = \frac{2\pi}{6} = \frac{\pi}{3}.$$

Note that $s = r\theta$. In particular, note that if the angle we traverse is a little angle $d\theta$, then the little distance traveled is $$ds = rd\theta.$$

With Mathematica, we can quickly get the answer using either of the following. Notice the order for bounds.

Integrate[Integrate[r, {r, 0, 7}], {theta, 0, 2 Pi}]
Integrate[r, {theta, 0, 2 Pi}, {r, 0, 7}]

By hand, we compute $$\ds \int_0^{2\pi}\int_{0}^{7} r\, dr d\theta = \int_0^{2\pi}\frac{r^2}{2}\bigg|_0^{7} d\theta = \int_0^{2\pi}\frac{7^2}{2} d\theta = \frac{7^2}{2}\theta\bigg|_0^{2\pi} = \pi7^2. $$ This shows that that area inside a circle of radius $7$ is $\pi 7^2$. Changing the 7 to an an arbitrary radius $a$, the above work shows that the area inside a circle of radius $a$ is $\pi a^2$.

Differentials give $du = 1dx+1dy$ and $dv = 1dx-1dy$, which we can write as the linear combination $$\begin{pmatrix}du\\dv\end{pmatrix} = \begin{pmatrix}1\\1\end{pmatrix}dx + \begin{pmatrix}1\\-1\end{pmatrix}dy.$$ We have $u$ and $v$ in terms of $x$ and $y$, which means the Jacobian of $u$ and $v$ with respect to $x$ and $y$ is $$\frac{\partial(u,v)}{\partial(x,y)} = |(1)(-1)-(1)(1)| = 2.$$

How do we find $\ds \frac{\partial (x,y)}{\partial (u,v)}$? This notation means we need $x$ and $y$ in terms of $u$ and $v$. We just have to solve $u=x+y$ and $v=x-y$ for $x$ and $y$. Summing the equations gives $u+v = 2x$. Subtraction yields $u-v = 2y$. This means $$x = \frac{u+v}{2}, y = \frac{u-v}{2}.$$ Differentials give $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix}1/2\\1/2\end{pmatrix}du + \begin{pmatrix}1/2\\-1/2\end{pmatrix}dv.$$ The area of the parallelogram formed by the partial derivatives gives $$\frac{\partial(x,y)}{\partial(u,v)} = |(1/2)(-1/2)-(1/2)(1/2)| = 1/2.$$

Note that $\frac{\partial(u,v)}{\partial(x,y)} = (\frac{\partial(x,y)}{\partial(u,v)})^{-1}$. This is not a coincidence.

The area is $\ds\iint_RdA =\iint_Rr drd\theta = \int_{\pi/2}^{\pi}\int_0^{5}rdrd\theta$.

Let's use Mathematica to compute the integral and plot the region.

outerB = {theta, Pi/2, Pi};
innerB = {r, 0, 5};

Integrate[r, outerB, innerB]

coordinates = {r Cos[theta], r Sin[theta]};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]

The centroid $(\bar x, \bar y)$ is the point that gives the average $(x,y)$ coordinates of the points in the region. It's the geometric center, and variable densities are ignored. These are just average $x$ and $y$ values, so we use the average value formula $\bar f = \frac{\iint_R f dA}{\iint_R dA}$, with $f$ equal to $x$ and $y$ respectively, to obtain

• $\ds\bar x = \frac{\iint_RxdA}{\iint_RdA} =\frac{\iint_R(r\cos\theta)rdrd\theta}{\iint_Rrdrd\theta} = \frac{\int_{\pi/2}^{\pi}\int_0^{5}(r\cos\theta)rdrd\theta}{\int_{\pi/2}^{\pi}\int_0^{5}rdrd\theta}$ and
• $\ds\bar y = \frac{\iint_RxdA}{\iint_RdA} =\frac{\iint_R(r\sin\theta)rdrd\theta}{\iint_Rrdrd\theta} = \frac{\int_{\pi/2}^{\pi}\int_0^{5}(r\sin\theta)rdrd\theta}{\int_{\pi/2}^{\pi}\int_0^{5}rdrd\theta}$.

With Mathematica, we can quickly compute these, with the last line below drawing the region to help us verify we've got the correct region.

outerB = {theta, Pi/2, Pi};
InnerBounds = {r, 0, 5};

Integrate[(r Cos[theta]) r, outerB, innerB]/
 Integrate[r, outerB, innerB]
Integrate[(r Sin[theta]) r, outerB, innerB]/
 Integrate[r, outerB, innerB]

coordinates = {r Cos[theta], r Sin[theta]};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]