I have the area defined by
$$A = \sqrt{x/4-(x/2)^2} < y < \sqrt{1-x^2)} \text{ and }0 < x < 1$$
and I'm supposed to find the integral of the function bound by these limits;
$$I = \int_0^1\int_{\sqrt{x/4-(x/2)^2}}^{\sqrt{1-x^2}}f(x,y)\mathrm{d}x\mathrm{d}x$$
So I think it would be better to convert these to polar coordinates;
$$I = \int_0^1\left(\int_0^{\sqrt{1-x^2}}f(x,y)\mathrm{d}y-\int_0^{\sqrt{x/4-(x/2)^2}}f(x,y)\mathrm{d}y\right)\mathrm{d}x\\ I = \int_0^1\left(\int_0^{\tau/4}g(r,\theta)\mathrm{d}\theta-\int_?^?g(r,\theta)\mathrm{d}\theta\right)r\mathrm{d}r$$
But I'm not really sure how to proceed from here. Any help would be appreciated!
$\displaystyle y^2 = \frac{x}{4}-\frac{x^2}{4}$
so
$\displaystyle 4r^2 \sin^2\theta = r\cos\theta-r^2\cos^2\theta$
and
$\displaystyle 4r = \cos\theta+3r\cos^2\theta$
You could try solving this for $r$: $\displaystyle r=\frac{\cos\theta}{4-3\cos^2\theta}$, then let $r$ be your inner integral and integrate over $r$ from zero to that limit. You might also try to solve for $\cos\theta$, let your inner integral be over $\theta$ from 0 to that limit, but that is looking kind of hairy to me. Unless $f(x,y)$ has some particularly nice form, I don't see any clean way to do this.
EDIT: Just saw the comments. Matthew Pereira has the right idea.