Setup the following integral in polar coordinates: $\int_1^2 \int_0 ^{x} (x^2 + y^2)^{-3/2} dy dx$

Question

Setup the following integral in polar coordinates:

$\int_1^2 \int_0 ^{x} (x^2 + y^2)^{-3/2} dy dx$

I am unsure where to start (particularly in terms of changing the limits).

I know the graph of R looks like this:

Any help would be greatly appreciated!

Answer 1

Okay you know the limits: $x\in[1,2]$ and $y\in[0,x]$. Polar coordinates: \begin{align} x= r \cos \theta, \hspace{10pt} y=r\sin\theta\end{align} What we can see is:

$0\leq y\leq x$ means $0\leq r\sin\theta \leq r\cos\theta$, that is true for $\theta\in[0,\pi/4]$, and $1\leq x \leq 2$ means $1 \leq r\cos\theta \leq 2$ and that is true for $r\in[\frac{1}{\cos\theta}, \frac{2}{\cos\theta}]$. So we know the limits in polar coordinates, now we can write the integral: \begin{align} \int^2_1\int^x_0 (x^2+y^2)^{-3/2}\mathrm d y\mathrm d x=\int^{\frac{\pi}{4}}_0 \int^{\frac{2}{\cos\theta}}_\frac{1}{\cos\theta}(r^2\cos^2\theta+r^2\sin^2\theta)^{-3/2} \color{red}{r\mathrm d r \mathrm d \theta} \end{align} Simplifying this a bit: \begin{align}\int^{\frac{\pi}{4}}_0 \int^{\frac{2}{\cos\theta}}_\frac{1}{\cos\theta}r^{-2} \mathrm d r \mathrm d \theta\end{align} You can take it from here?