<s>[INST] Evaluate double integral (Need help picking $u$ and $v$) Evaluate
$$
\iint_R 8(x^2+y^2)(x^2-y^2)dA
$$
where $R$ is a region in the first quadrant of $xy$-plane bounded by the circles $x^2+y^2=1,\ x^2+y^2=4$, and the line $y=x+1,\ y=x-1$.
Ok I need help picking $u$ and $v$. I have tried $u^2=x^2+y^2$, and $v^2=x^2-y^2$ but I'm having a hard time simplifying the Jacobian and the integral.
 [/INST] This integral is easily evaluated in polar coordinates.  Obviously, $r \in [1,2]$.  The bounds on the angle $\theta$ depend on $r$ and are derived from the lines:
$$y-x=\pm 1 \implies r(\sin{\theta} - \cos{\theta}) = \pm 1$$
Using $$\sin{\theta}-\cos{\theta} = 2 \cos{\left(\frac{\pi}{4}\right)} \sin{\left(\theta-\frac{\pi}{4}\right)}$$
we get the bounds on $\theta$ as
$$\theta \in \left [ \frac{\pi}{4}-\arcsin{\left(\frac{1}{\sqrt{2} r}\right)},  \frac{\pi}{4}+\arcsin{\left(\frac{1}{\sqrt{2} r}\right)}\right ]$$
The integral is now
$$8 \int_1^2 dr \, r^5 \: \int_{\pi/4 - \arcsin{(1/(\sqrt{2} r))}}^{\pi/4 + \arcsin{(1/(\sqrt{2} r))}} d\theta \, \cos{2 \theta}$$
Note that I used $x^2-y^2 = r^2 (\cos^2{\theta} - \sin^2{\theta}) = r^2 \cos{2 \theta}$.
The integral over $\theta$ is  simple and turns out to be zero:
$$\int_{\pi/4 - \arcsin{(1/(\sqrt{2} r))}}^{\pi/4 + \arcsin{(1/(\sqrt{2} r))}} d\theta \, \cos{2 \theta} = \frac12 \sin{\left ( \frac{\pi}{2} + 2 \arcsin{\left(\frac{1}{\sqrt{2} r}\right)}\right )} - \frac12 \sin{\left ( \frac{\pi}{2} -2 \arcsin{\left(\frac{1}{\sqrt{2} r}\right)}\right )}$$
which you should see is zero by expanding the sines (noting that $\cos{(\pi/2)} = 0$).  Therefore the integral is zero.
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