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.

$\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$

=$\int_0^2 \int_0^{x^2} \sqrt{x^2+y}\, dydx$

=$\frac{2}{3}\int_0^2 (2x^2)^{3/2}-x^3\,dx$

=$\frac{2}{3}(2\sqrt2-1) \int_0^2 x^3\,dx$

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

• Consider changing order of integration $$\int\limits_0^4 \int\limits_{\sqrt{y}}^2 \sqrt{x^2+y} \color{purple}{dx}\color{blue}{dy} =\int\limits_{?}^{?} \int\limits_{?}^{?} \sqrt{x^2+y}\color{blue}{dy}\color{purple}{dx}$$ – ganeshie8 Mar 13 '15 at 7:20
• i actually didnt think of that. i've been substituting to polar coordinates for the last few exercises i actually forgot i could change the order of integration. Thank you. – Helpisneeded Mar 13 '15 at 7:28

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:

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

• Hi Ron, i've completed my working and edited the post. But my answer is different from yours. Did i do something wrong? – Helpisneeded Mar 13 '15 at 7:45

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: $$\int\frac{\sin \theta}{\cos^4 \theta} d\theta=\frac{1}{3\cos^3 \theta}$$