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.
 A: 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.
A: Your region looks like so:

The inner and outer circles can be written as 
$$g(x,y)=1 \mbox{ or } g(x,y)=4,$$ 
where $g(x,y)=x^2+y^2$.  The lines can be written as 
$$f(x,y)=-1 \mbox{ or } f(x,y)=1,$$ 
where $f(x,y)=x-y$.  
So, set $u=x^2+y^2$ and $v=x-y$, because these expressions are constant on the boundary of the region.  The bounds for $u$ will be 1 and 4 and the bounds for $v$ will be -1 and 1.  You can then solve that system for $x$ and $y$.  The determinant of your Jacobian matrix should look something like 
$$\frac{1}{2 \sqrt{2 u-v^2}}.$$
A little more detail
If you set $u=x^2+y^2$ and $v=x-y$ and solve for $x$ and $y$ in terms of $u$ and $v$, one solution should be
$$x=\frac{1}{2} \left(v+\sqrt{2 u-v^2}\right)$$
and
$$y=\frac{1}{2} \left(v+\sqrt{2 u-v^2}\right).$$
I solved these equations with Mathematica.  One way to do it by hand would be to write $y=x-v$, plug that into the $u$ equation to get $u=x^2+(x-v)^2$ and solve that quadratic for $x$.  You can then get $y$.  However you do it, plugging these into your integrand and simplifying, you get
$$8\left(x^2+y^2\right)\left(x^2-y^2\right) =8 u v \sqrt{2 u-v^2}.$$
The $\sqrt{2u-v^2}$ looks intimidating but, happily the determinant of the Jacobian of this transformation is
$$-\frac{1}{2 \sqrt{2 u-v^2}}$$
Thus, you end up with the integral
$$-\int _{-1}^1\int _1^4 8uv\,dudv$$
which is zero. In hindsight, that's totally obvious, as your original integrand has an odd symmetry about the line $y=x$!
