1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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!)

$$\int_0^{2\sqrt2}\int_{\pi/4}^{\pi/2}r^4\,dtdr=\frac\pi4\int_0^{2\sqrt2}r^4\,dr=\frac\pi{20}\cdot128\sqrt2=\frac{32\sqrt2\;\pi}5$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.