I am trying to evaluate the integral:
$$\iint_D \sqrt{1-y^2} dx dy$$
Where D is the disk centered at $(1, 0)$ and with radius 1.
Notice that this integral clearly converges: in the disk, $y^2<1$, so $|\sqrt{1-y^2}|<1$. This also means the integral satisfies Fubini.
One change of variables $(x, y) \mapsto (x+1, y)$ moves the circle to the origin, and leaves the integrand unchanged. I then change to polar coordinates: $(\rho, \theta) \mapsto (\rho\cos \theta, \rho \sin \theta)$ obtaining:
$$ \int_0^{2\pi} \int_0^1 (1-\rho^2\sin^2\theta)^\frac12\rho \ d\rho d\theta $$
I found an antiderivative of the integrand as a function of $\rho$:
$$ -\frac 1 {3\sin^2\theta} (1-\rho^2\sin^2\theta)^\frac32 $$
Evaluating it at $0$ and $1$ reduces the inner integral to the following:
$$ \int_0^{2\pi} -\frac 1 {3\sin^2\theta} \left [ (1-\sin^2\theta)^\frac32-1 \right ] d\theta$$
This simplifies to:
$$ -\frac {1} 3 \int_0^{2\pi} \frac{\cos^3\theta-1}{\sin^2\theta}d\theta$$
But Wolframalpha says this does not converge.
Where did I go wrong?
Thanks to the helpful comments above, this is the problem:
$$ (\cos^2\theta)^\frac32 = | cos^3\theta|$$
Because $ \cos^2\theta$ is always positive.
The rest of the integral is as follows:
First, notice that both $\sin^2\theta$ and $|\cos\theta|$ have period $\pi$ and not just $2\pi$. Then further notice that they are even around $\pi/2$. With this, $\int_0^{2\pi} f(\theta)d\theta = 4\int_0^\frac\pi2 f(\theta)d\theta$. More specifically, the integral becomes:
$$ -\frac43 \int_0^\frac\pi2 \frac {\cos^3\theta -1}{\sin^2\theta}d\theta $$
Where I dropped the absolute value because $\cos(\theta)$ is always positive on this new domain anyways.
I have an antiderivative of the integrand:
$$ \frac {-1-\sin^2\theta+\cos\theta} {\sin\theta} $$
Which evaluates as $-2$ at $\frac\pi2$ and as $0$ at $0$ (taking the limit).
Therefore the integral evaluates as $-\frac43[-2-0]=\frac83$.