# Thread: Calculate the volume of a part of a sphere inside a cylinder

1. Originally Posted by Jhevon
how exactly did you end up with that integral? that guy never showed up when i did it
$V=4 \int_0^{\frac{\pi}{2}} \int_0^{a\sin (\theta)} \int_0^{\sqrt{a^2-r^2}} rdzdrd\theta$.
I call $I$ the $dz$ integral. I got that $I=r\sqrt{a^2-r^2}$
Thus $V=4 \int_0^{\frac{\pi}{2}} \int_0^{a\sin (\theta)} r\sqrt{a^2-r^2}drd\theta$.
I call $J$ the $dr$ integral. I got that $J=-\frac{1}{3}(a\sin (\theta))^{\frac{3}{2}}$ via a $u$ substitution.
Hence $V=-\frac{4}{3} a^{\frac{3}{2}} \int _0^{\frac{\pi}{2}} \sin ^{\frac{3}{2}} (\theta) d\theta$.

2. Originally Posted by arbolis
$V=4 \int_0^{\frac{\pi}{2}} \int_0^{a\sin (\theta)} \int_0^{\sqrt{a^2-r^2}} rdzdrd\theta$.
I call $I$ the $dz$ integral. I got that $I=r\sqrt{a^2-r^2}$
Thus $V=4 \int_0^{\frac{\pi}{2}} \int_0^{a\sin (\theta)} r\sqrt{a^2-r^2}drd\theta$.
I call $J$ the $dr$ integral. I got that $J=-\frac{1}{3}(a\sin (\theta))^{\frac{3}{2}}$ via a $u$ substitution.
Hence $V=-\frac{4}{3} a^{\frac{3}{2}} \int _0^{\frac{\pi}{2}} \sin ^{\frac{3}{2}} (\theta) d\theta$.
well, that's your problem! $J = - \frac 13 (a^2 - a^2 \sin^2 \theta)^{3/2} - a^3$

check again, you'll see that's what you should have

3. Originally Posted by Jhevon
well, that's your problem! $J = - \frac 13 (a^2 - a^2 \sin^2 \theta)^{3/2} - a^3$

check again, you'll see that's what you should have
As yesterday I forgot to change the limits of the integral when making the substitution.
$u=a^2-r^2 \Rightarrow du=-2rdr$ so $J=-\frac{1}{2} \int_{a^2}^{a^2-a^2\sin ^2 (\theta)} \sqrt u du=-\frac{1}{2} \int_{a^2}^{a^2(1-\sin ^2 (\theta))} \sqrt u du=-\frac{1}{2} \int_{a^2}^{a^2(\cos^2 (\theta))} \sqrt u du$.
$\Rightarrow J= - \frac{1}{3} u^{\frac{3}{2}} \big | _{a^2}^{a^2\cos^2 (\theta))}=-\frac{1}{3} (a^3\cos ^3 (\theta)-a^3)$.

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