Evaluating a double integral 
Using a formula from geometry evaluate $\displaystyle \int_0^3 \! \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sqrt{9-x^2-y^2}\,dy\,dx$.

Attempt. What formula? Any hints please. I know I could use some trigonometric equations for integrals, but I suppose there is a much easier way to evaluate these integrals.
 A: Essentially you want to integrate $$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} z dy dx$$ where $z = \sqrt{9 - y^2 - x^2}$. What geometric object is this? It might be of help to rewrite it as $$x^2 + y^2 + z^2 = 9$$
Move your mouse over the gray area below if you want further hints and the solution.

 The equation $$x^2 + y^2 + z^2 = 9$$ represents a sphere in $3$D. The limits of the integral will give you the portion of the sphere over which you are integrating and hence the integral will give you the volume of the portion of the sphere. In this case, $z$ is always positive. This restricts the object to a hemi-sphere. The $y$ takes negative and positive values. The $x$ takes only positive values which again restricts the object to half of the hemi-sphere. Hence, the portion of the sphere you are integrating is just $1/4$ of the total sphere. The total volume of the sphere is $\dfrac43 \times \pi \times 3^3$. Hence, the integral is $$\dfrac{\dfrac43 \times \pi \times 3^3}{4} = 9 \pi$$

