4
$\begingroup$

Evaluate the double integral by converting to polar coordinates: $$\int_0^2 \int_0^x \sqrt{x^2 + y^2} \, dy \, dx + \int_2^{2\sqrt{2}} \int_o^{\sqrt{8 - x^2}} \sqrt{x^2 + y^2} \, dy \, dx $$

What I have done: First, I tackled the left hand integral. Using the fact that $x = r \cos\theta$ and $y = r\sin\theta$, I converted the left hand integral to $\int r^2 \, dr \, d\theta$. I drew the graph of the region to get a right triangle. I am having trouble determining the limits of integration for $r$ and $\theta$

$\endgroup$
  • $\begingroup$ @E.H.E: I don't think there is. $\endgroup$ – joriki Oct 17 '15 at 20:35
4
$\begingroup$

The two regions together form a sector of a circle with radius $\sqrt8$ centred at the origin, so the corresponding integral in polar coordinates is

$$ \int_0^{\frac\pi4} \, \mathrm d\phi\int_0^\sqrt8r \, \mathrm dr \, r\;. $$

$\endgroup$
  • $\begingroup$ Wolfram Alpha also gave me ~5.91 which I obtained with this method $\checkmark$ $\endgroup$ – WAS Oct 18 '15 at 16:54

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.