# Solid volume with Triple Integrals

I'm having some issues to get the volume of the solid below.

Calculate the solid volume bounded above by $z =\sqrt{25-x^2-y^2}$ , below the plane $xy$, and sideways by the cylinder $x^2+y^2=9$

Your integral to be computed is given by $I := \iint_\limits{B} \sqrt{25 - x^2 - y^2} d(x,y)$, where $B := \{ (x,y) \in \mathbb{R}^2 | x^2 + y^2 \leq 9 \}$, which is obviously a circle with radius $r = 3$ centered at the origin. In this case, it is always a good idea switching to polar coordinates: Choosing $\theta \in [0, 2 \pi]$ and $r \in [0, 3]$ leads to: $I = \int \limits_0^{2 \pi} \int \limits_{0}^{3} \sqrt{25 - r^2}\cdot r \, \mathrm{d}r \ \mathrm{d}\theta$, where the factor $r$ after the square root comes from the functional determinant. Now an anti-derivative of the integrand (with respect to $r$) is given by $-\frac{1}{3}(25 - r^2)^{3/2}$; Hence: $I = 2 \pi \cdot \left[ -\frac{1}{3}(25 - r^2)^{3/2} \right]_{0}^{3} = \frac{2\pi}{3}(25^{3/2} - 16^{3/2})$.

• Thanks for your explanation, now it's much more clear on how to solve this kind of exercise, by using polar coordinates! – Steve Setchus Sep 19 '16 at 19:26
• Funny how these exercises looks so simple now... I was trying to solve this with a close minded and couldn't figure out an alternative way. – Steve Setchus Sep 19 '16 at 19:30
• You're welcome, nice to hear that! As far as I'm concerned, this feeling of "oh my god, is it really that simple?" shows up after a lot of exercises of this kind ;-) – ComplexFlo Sep 20 '16 at 6:34

Hint

Your volume is defined as the intersection of a hemisphere with a cylinder. It will be simpler in polar coordinates.

$$V=\int_0^{2\pi}\int_0^3 \sqrt{25-r^2}r\mathrm dr\mathrm d\theta$$

See for instance this. Notice the third example is almost identical to your question.

Now compute the integral.

• Thanks for your hint, this conversion made this much easier. – Steve Setchus Sep 19 '16 at 19:26