Evaluating area D using polar coordinates

Question

Let $D$ be the region in the xy-plane bounded on the left by the line $x=2$ and on the right by the circle $x^2 + y^2 = 16$. Evaluate $$\iint (x^2 + y^2)^{-3/2}dA$$

Answer 1

I'll first remind you of polar coordinates: $$x=r\cos(\theta)\\ y=r\sin(\theta)$$ That means that $x^2+y^2=r^2.$ First, sketch your region. You will get something like the following (excuse my bad drawing):
img http://puu.sh/lRm2G/17d858184a.png
When you are integrating, your $r$ bounds will be starting from $x=2$ and ending at $x^2+y^2=16$. Let's convert to polar coordinates to get these as functions of $\theta$: $x=2\rightarrow r\cos(\theta)=2\rightarrow r=\frac{2}{\cos(\theta)}$ $x^2+y^2=16\rightarrow r=4$
$\theta$ bounds are a little more tricky. Plug $x=2$ into the circle equation to get $4+y^2=16$, therefore the intersections of $x=2$ and $x^2+y^2=16$ are $(2, \pm\sqrt{12})$ To find the angle made, make a right triangle to get: $$\theta=\arctan\frac{\sqrt{12}}{2}=\pm\frac{\pi}{3}$$ So your theta bounds are $\theta\in\big[-\frac{\pi}{3}, \frac{\pi}{3}\big]$. Turning the function you're integrating into polar coordinates, we get our set up integral: $$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int_{\frac{2}{\cos(\theta)}}^{4}\big(r^2\big)^{-\frac{3}{2}}r\,dr\,d\theta=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int_{\frac{2}{\cos(\theta)}}^{4}\bigg(\frac{1}{r^3}\bigg)r\,dr\,d\theta.$$ I assume you can go from here.

Answer 2

Each arc of radius $r$ has length $2r\arccos\!\left(\frac2r\right)$

Since $\mathrm{d}A=2r\arccos\!\left(\frac2r\right)\mathrm{d}r$, the integral is $$ \int_2^4\frac1{r^3}\left[2r\arccos\!\left(\frac2r\right)\right]\mathrm{d}r $$ Substitute $\frac2r=\cos(\theta)$ and integrate by parts.