evaluate the double integral $\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$ evaluate the double integral
$\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$
Hi all, could someone give me a hint on this question?
I've actually tried converting to polar coordinates but i cant seem to get the limits. But if polar coordinates are the way to go i'll just keep working on it. Thanks in advance.
edit: so the trick is to change the order of integration.
$\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$
=$\int_0^2 \int_0^{x^2} \sqrt{x^2+y}\, dydx$
=$ \frac{2}{3}\int_0^2 (2x^2)^{3/2}-x^3\,dx$
=$ \frac{2}{3}(2\sqrt2-1) \int_0^2 x^3\,dx$
=$\frac{8}{3}(2\sqrt2-1)$
 A: Your best bet here would be to reverse the order of integration so you can integrate over $y$ first.  Just draw a picture of the integration region; you'll see you can rewrite the integral as follows:
$$\int_0^2 dx \int_0^{x^2} dy \, \sqrt{x^2+y} $$
Now the integration over $y$ is less messy; we then get a single integral:

$$\frac{2}{3} \int_0^2 dx \left [\left ( 2 x^2\right )^{3/2} - \left (x^2\right )^{3/2} \right ]  = \frac{2}{3} \left ( 2 \sqrt{2}-1 \right ) \int_0^{2} dx \, x^3 = \frac{8}{3} \left ( 2 \sqrt{2}-1 \right )$$

A: Here is a polar treatment. 
First get rid of the square root by $z=\sqrt{y}$ with $dy=2zdz$: $$I=\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy=2\int_0^2 \int_{z}^2 \sqrt{x^2+z^2}z\, dxdz$$
Now draw a graph to see the integration regions:

Therefore with $z=r\sin \theta$, $x=r \cos \theta$ and $dxdz=r dr d\theta$ we have
\begin{align}
I&=2\int_0^{\frac{\pi}{4}}\int_0^{\frac{2}{\cos \theta}}r^3 \sin \theta dr d\theta\\
&=8\int_0^{\frac{\pi}{4}}\frac{\sin \theta}{\cos^4 \theta} d\theta\\
&=8\times \Big(\frac13 (2\sqrt{2}-1)\Big)
\end{align}
Note: $$\int\frac{\sin \theta}{\cos^4 \theta} d\theta=\frac{1}{3\cos^3 \theta}$$
