Why is the volume of a sphere $\frac{4}{3}\pi r^3$? I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?

A complete answer using the disk method would be the following.
If you revolve $y=\sqrt{r^2-x^2}$ about the $x$-axis (and form a solid) you get the volume of a sphere.
 
Form a disk with height $f(x)$, and find its area.

The area of the red disk above is $\pi r^2$, or $\pi f(x)^2$, or we could say $\pi \sqrt{r^2-x^2}^2=\pi \left (r^2-x^2\right )$ at any point $x$ between $x=-r$ and $x=r$.
To find the volume of a sphere of radius $r$ then, just add up the areas of infinitesimally thin disks on as $x$ goes from $-r$ to $r$.
To work it out:\begin{align*}\int \limits _{-r}^r\pi \sqrt{r^2-x^2}^2\,dx & =2\pi \int \limits _0^rr^2-x^2\,dx \\
& =2\pi \left (\left .r^2x-\frac{1}{3}x^3\right )\right |_0^r \\
& =2\pi \left (r^3-\frac{1}{3}r^3\right ) \\
& =\frac{4\pi}{3}r^3.
\end{align*}
