$$\int_{0}^{2}\int_{0}^{\sqrt{2x-x^2}} \sqrt{x^2+y^2}dydx$$ by converting to polar coordinates.

I sketch the region which is a half circle from $0$ to $2$ on the $x$-axis and $0$ to $1$ on the y-axis

$\theta \in [0 , \pi] $ $ r\in [0 , 2] $ $$\sqrt{x^2+y^2} = r$$

Then in Polar Coordinates would be: $$\int_{0}^{\pi}\int_{0}^{2} (r) d(r) d(\theta)$$


You drew the region correctly, but the bounds on your integral are incorrect.

If you draw a picture, you will see that $\theta \in [0,\frac{\pi}{2}]$ since the half-disk is entirely in the first quadrant and any radial line in the first quadrant intersects the half-disk.

For any fixed $\theta$, the minimum value of $r$ is $0$, and the maximum value of $r$ satisfies:

$(x-1)^2+y^2 = 1$

$(r\cos\theta - 1)^2 + (r\sin \theta)^2 = 1$

$r^2\cos^2\theta - 2r\cos\theta + 1 + r^2 \sin^2\theta = 1$

$r^2 - 2r\cos\theta = 0$

$r = 2\cos\theta$

Hence, the bounds for $r$ should be $r \in [0,2\cos\theta]$.

Also, don't forget the Jacobian of the polar transformation. Specifically, $\,dx\,dy = r\,dr\,d\theta$

Then, the double integral becomes $\displaystyle\int_{0}^{\pi/2}\int_{0}^{2\cos \theta}r \cdot r\,dr\,d\theta$. This should be easy to evaluate.

| cite | improve this answer | |
  • $\begingroup$ So for theta do you just look at the quadrant and not the circle? And for r how do you know to use that equation? $\endgroup$ – user177780 Oct 11 '14 at 20:25

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.