Derivation (Proof) of Volume of a Sphere

Before we start, remember the Pythagorean Theorem: $a^2+b^2=c^2$ in a right triangle with legs $a,b$ and hypotenuse $c$. We'll use this down below.

Suppose that we have a sphere with center $O$ and radius $r$. We would like to find the volume $V$ of this sphere.

Take a line that passes through $O$, and let $X$ be a point on the line that lies inside or on the sphere. Thus, the directed distance $OX$ ranges from $-r$ to $r$. Let $x$ equal the directed distance $OX$, and let $S(x)$ denote the cross-sectional area when a plane cuts the sphere at a distance $x$ from the center.

Using the integral formula for volume,

$$V=\int_{-r}^{r}S(x)dx$$

Using the symmetry of the sphere,

$$V=2\int_{0}^{r}S(x)dx$$

Using the Pythagorean Theorem, we may find the radius of the cross-section: $$r_x=\sqrt{r^2-x^2}$$ Since all of the cross-sections are circles, we can find the area of the cross-section: $$S(x)=\pi r_x^2$$ $$S(x)=\pi(r^2-x^2)$$

Now, a simple integration gives us the volume $V$: $$V=2\int_{0}^{r}S(x)dx$$ $$V=2\int_{0}^{r}\pi(r^2-x^2)dx$$ $$V=2\pi(r\cdot r^2 -\frac{1}{3}r^3 -0\cdot r^2+\frac{1}{3}0^3)$$ $$V=\frac{4}{3}\pi r^3$$

Thus, we see that the volume of a sphere is $\frac{4}{3}\pi$ times the radius cubed.