Proving volume of a sphere

Question

I randomly decided to derive the volume of a sphere.

The area of a circle is $\pi r^2$.

So the volume, I thought, should be $\int \pi r^2 dr = \frac{\pi r^3}{3} $, summing up the area of many discs.

Shouldn't there be a $4$ in there? Why isn't there?

Answer 1

The method you tried to apply actually works like this: $$ V = \int_{-r}^r \pi y^2\ dx\qquad (1) $$ where $x^2+y^2=r^2$. Plugging the Pythagorean identity in $(1)$ gives $$ V = \int_{-r}^r \pi (r^2-x^2)\ dx = \pi\left[r^2x-\frac{x^3}{3}\right]_{x=-r}^r=\frac{4}{3}\pi r^3 $$ For more details see this derivation.

Answer 2

The missing $4$ actually comes from the area of spherical shell with radius $r$.

Let $(x,y)=r(\cos t, \sin t)$. Then $(\dot{x},\dot{y})=r(-\sin t, \cos t)$.

\begin{align*} S &= \int_{-r}^{r} 2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx \\ &= 2\pi\int_{-\pi}^{\pi} y\sqrt{\dot{x}^{2}+\dot{y}^{2}} dt \\ &= 2\pi \int_{-\pi}^{\pi} r^{2} \sin t \, dt \\ &= 4\pi r^{2} \\ V &= \int_{0}^{R} 4\pi r^{2} \, dr \\ &= \frac{4}{3} \pi R^{2} \end{align*}