# Convert Integral Rectangular to Polar

How can convert this problem

$$\int_0^2 \int_x^\sqrt{8-x^2} \left(x^2+y^2\right)^{3/2} dydx$$

I convert limits and funtion to polar cordinates as follows: $$\begin{split} r^2 &= x^2+y^2\\ x = 2 &\to r \cos \theta = 2\\ x = 2 &\to r \cos \theta = 0\\ x = y &\to r \cos \theta = r \sin \theta \to 1 = \tan \theta\\ y = \sqrt{8 - x^2} &\to y^2 + x^2 = 8 \to r^2 = 8 \end{split}$$ But i dont know how to rebuild de integral with your limits

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\;\;,\;\;\frac\pi4\le t\le \frac\pi2\;,\;\;0\le r\le 2\sqrt2$$
$$\int_0^{2\sqrt2}\int_{\pi/4}^{\pi/2}r^4\,dtdr=\frac\pi4\int_0^{2\sqrt2}r^4\,dr=\frac\pi{20}\cdot128\sqrt2=\frac{32\sqrt2\;\pi}5$$
• @DanielORTIZ The variable $\;y\;$ begins from below on $\;y=x\;$ (this is $\;\pi/4\;$ radians with the positive $\;x\,-$ axis), and all the way up to the y axis, which is $\;\pi/2\;$ radians. – DonAntonio May 2 '16 at 19:20