An exercise asks to integrate
$$f(x,y) = \sqrt{4-x^2-y^2}$$
inside circle with radius 2 and center $(2,0)$.
When I set up the double integral I get
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{4cos(\theta)} \sqrt{4-r^2} r dr d\theta$$
Is that the correct way to set up the integral?
Should I really evaluate that as
$$ - \frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{4-16cos^2(\theta)} u^{1/2} du d\theta = - \frac{1}{3} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} u^{3/2}\bigg|_{0}^{4-16cos^2(\theta)} d\theta = \cdots $$
which seems like a dead-end?