How to calculate the value of following triple integral?

Question

What is the volume of the region that lies under the sphere $x^2+y^2+z^2 = 9$, above the plane $z = 0$ and inside the cylinder $x^2+y^2 = 5$.

For me value should be $$\int_{-\sqrt{5}}^{\sqrt{5}}\int_{ -\sqrt{5-x^2}}^{\sqrt{5-x^2}} \int_{z = 0}^2 dz dydx$$

I am right or not?

Answer 1

I guess a picture would help you with finding the boundaries. Anyway note that the upper bound of the region is $x^2 + y^2 + z^2 = 9$. Therefore we must have $z = \sqrt{9 - x^2 - y^2}$. Hence the integral would be:

$$\int_{-\sqrt{5}}^{\sqrt{5}}\int_{ -\sqrt{5-x^2}}^{\sqrt{5-x^2}} \int_{0}^{\sqrt{9-x^2-y^2}} dz dydx$$

Answer 2

Why do you go only to $z=2$? You should go to $z=3$, but the limits of integration for $x$ and $y$ change when $3\gt z\gt2$. Below $z=2$ the radius in the $xy$ plane is $\sqrt{5}$, but above is $\sqrt{9-z^2}$