## Volume of a Sphere

The volume of a solid can be found using the formula
$V= \int \int \int_V x \: dy \: dz$
. The equation of the surface of a surface of a sphere of radius
$a$
is
$a^2 =x^2 +y^2 +z^2$
.
Rearranging this gives
$x= \sqrt{a^2 - y^2 -z^2}$

The volume in the first quadrant,
$\: 0 \leq x, \: y, \: z \leq a$
is one eighth the volume of the sphere. The integral becomes
\begin{aligned} V &= 8 \int^a_0 \int^{\sqrt{a^2 -z^2}}_0 \sqrt{a^2 - y^2 - z^2} dy \: dz \\ &= 8 \int^a_0 \int^{\sqrt{a^2 -z^2}}_0 \sqrt{(a^2 - z^2) - y^2} dy \: dz \end{aligned}

Substitute
$y=\sqrt{a^2-z^2} sin \theta \rightarrow dy = \sqrt{a^2-z^2} cos \theta d \theta$

The limits become
$0, \:\frac{\pi}{2}$

\begin{aligned}V &= 8 \int^a_0 \int^{\pi / 2}_0 \sqrt{(a^2 -z^2) - (a^2 -z^2) sin^2 \theta} \sqrt{a^2-z^2} cos \theta d \theta \: dz \\ &= 8 \int^a_0 \int^{\pi / 2}_0 (z^2 -a^2 ) cos^2 \theta d \theta \: dz \\ &= 8 \int^a_0 \int^{\pi / 2}_0 (z^2 -a^2 ) (\frac{1}{2} + cos 2 \theta) d \theta \: dz \\ &= 8 \int^a_0 (z^2 -a^2 ) [ \frac{\theta}{2} + \frac{sin 2 \theta}{2}]^{\pi / 2}_0 dz \\ &= \int^a_0 (a^2 -z^2) \frac{\pi}{4} dz \\ &= 8 \frac{\pi}{4} [a^2 z - \frac{z^3}{3}]^a_0 \\ &= 2 \pi (a^3 - \frac{a^3}{3} ) \\ &= \frac{4 \pi a^3}{3} \end{aligned}