# Triple integral cylindrical coordinates, cylinder and sphere

A drill with radius $$r_1$$ is drilled through the center of a sphere with radius $$r_2$$How do i find the volume of the remaining ring-shaped body using cylindrical coordinates? What i have so far is the volume of the cylinder using triple integration but i'm not sure what to do about the sphere. $$\int_0^{2\pi}\int_0^{2r_2}\int_0^{r_1} r_1 drdzd\theta$$ The answer should be $$V=\pi h^3/6$$

If I got the problem right: $$V_{drilled \; sphere} = \int_{-\sqrt{r_2^2 - r_1^2}}^{\sqrt{r_2^2 - r_1^2}} \int_{0}^{2\pi} \int_{r_1}^{\sqrt{r_2^2 - z^2}} r \;\; dr \; d\theta \; dz = 2\pi \int_{-\sqrt{r_2^2 - r_1^2}}^{\sqrt{r_2^2 - r_1^2}} dz \int_{r_1}^{\sqrt{r_2^2 - z^2}} r \; dr = 2\pi \int_{-\sqrt{r_2^2 - r_1^2}}^{\sqrt{r_2^2 - r_1^2}} dz \; \frac{r^2}{2} \bigg\vert _{r = r_1}^{\sqrt{r_2^2 - z^2}} = \pi \int_{-\sqrt{r_2^2 - r_1^2}}^{\sqrt{r_2^2 - r_1^2}} dz \left( r_2^2 - r_1^2 - z^2 \right) = \pi \left[ \left( r_2^2 - r_1^2 \right) z \big\vert_{z=-\sqrt{r_2^2 - r_1^2}}^{\sqrt{r_2^2 - r_1^2}} - \frac{z^3}{3} \bigg\vert_{z=-\sqrt{r_2^2 - r_1^2}}^{\sqrt{r_2^2 - r_1^2}} \right] = \pi \frac{4 \left(r_2^2 - r_1^2 \right)^{3/2}}{3}$$

By the way, what is $h$ in your answer? And where are $r_1$ and $r_2$?

• h is the height of the ring. The answer should be in terms of h.
– Aes
Aug 17, 2017 at 11:29
• @Aesthetic well, in my variables, $h = 2 \sqrt{r_2^2 - r_1^2}$. Aug 17, 2017 at 11:41
• @Aesthetic Now my answer is right Aug 17, 2017 at 11:47
• How do you get h=2sqrt(r2^2-r1^2)?
– Aes
Aug 17, 2017 at 11:48
• just draw the picture, try to see :) It is easy, I can not attach photo now. Aug 17, 2017 at 11:52