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 | 1 | initial version | posted 2014-07-31 06:36:23 -0600  |
If you bring the 4 to the other side you get $x^2 + y^2 = 4$, which is a circle of radius 2. For this reason, I set this up as follows: $$\int_0^\frac{\pi}{2}\int_0^2(4-r^2)rdud\theta$$ $$\int_0^\frac{\pi}{2}(2r^2-\frac{r^4}{4})\bigg|_0^2d\theta$$ $$\frac{\pi}{2}(8-4)$$ $$=2\pi$$
Don't know if this is correct though.
