[Math] evaluate the double integral $\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$

Question

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$

=$\frac{8}{3}(2\sqrt2-1)$

Answer 1

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:

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