Need help with Double Definite Integration I need help in solving this double integral:
$$\int\limits^2_{-2}\;\;\int\limits^\sqrt{4-x^2}_{-\sqrt{4-x^2}}{(x^2+y^2)^{5/2}}dy\,dx$$
Maybe introducing polar coordinates might help?
 A: $$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(x^2+y^2)^{5/2}dy\,dx=\int_0^{2\pi}\int_0^2r^5\,rdr\,d\theta=\frac{256}{7}\pi$$
after transforming to polar coordinates with 
$$x=r\cos \theta\\\\
y=r\sin \theta$$
and the Jacobian equal to $r$ so that $dx\,dy\to r\,dr\,d\theta$.
A: You can express $x^2 + y^2$ in polar coordinates using:
$x = rcos\theta, \ y = rsin\theta$
$$x^2 + y^2 = r^2cos^2\theta + r^2sin^2\theta = r^2$$
$$(x^2 + y^2)^{\frac{5}{2}} = (r^2)^\frac{5}{2} = r^5$$
Essentially, the region is a circle with radius 2 and the angle goes from $0$ to $2\pi$
$$\frac{\partial (x, y)}{\partial (r, \theta)} = r$$
The integral now becomes:
$$ \int_0^{2\pi}\int_0^2 r^5.r \ dr\ d\theta$$
$$\int_0^{2\pi} \frac{r^7}{7} d\theta = \int_0^{2\pi}\frac{128}{7}d\theta$$
$$\left(\frac{128}{7}\theta\right)_0^{2\pi}$$
$$ \frac{256}{7}\pi$$
In general, you can transform an integral over a region into another integral over a second region using:
$$ \int \int_D f(x, y)\ dx \ dy =\int \int_{\Delta} f(g_1(u, v), g_2(u, v))\left|\frac{\partial (x, y)}{\partial (u, v)}\right| \  du\  dv  $$
