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Evaluate

$$\int_{0}^{2}\int_{0}^{\sqrt{2x-x^2}} \sqrt{x^2+y^2}dydx$$ 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: $$\int_{0}^{\pi}\int_{0}^{2} (r) d(r) d(\theta)$$

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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.

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  • $\begingroup$ 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

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