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


Do the usual and draw the region, and check that (fill in details)

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\;\;,\;\;\frac\pi4\le t\le \frac\pi2\;,\;\;0\le r\le 2\sqrt2$$

so your integral becomes (don't forget the Jacobian!)


  • $\begingroup$ I dont understand \pi/4 , for the draw? $\endgroup$ – Daniel ORTIZ May 2 '16 at 19:17
  • $\begingroup$ @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. $\endgroup$ – DonAntonio May 2 '16 at 19:20
  • $\begingroup$ Yes, thanks for your help. $\endgroup$ – Daniel ORTIZ May 2 '16 at 19:27

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