The Volume of a Sphere with Calculus

This post is second in a series of 3 beginning with “Coordinates in 3-space”. We have been taught that the volume of a sphere V = \frac{4}{3}\pi r^3. We will go through two derivations of this, beginning with the Calculus version. We will make use of a theorem in calculus.

Theorem: \int \int \int dV = \int \int \int r^2 sin \phi dr d\theta d\phi.

sphcoordel

Suppose we have a sphere of radius r. Then from the previous post involving spherical coordinates, we see that every point in the sphere is made of (\rho, \theta, \phi where 0 \leq \rho \leq r, 0 \leq \theta \leq 2 \pi, and 0 \leq \phi \leq \pi. These are exactly our limits of integration hence we just have to compute the following integral!

\int_0^\pi \int_0^{2\pi} \int_0^r \rho^2 sin\phi d\rho d\theta d\phi

= \int_0^\pi \int_0^{2\pi} \frac{r^3}{3} sin\phi d\theta d\phi

= \int_0^\pi 2\pi \frac{r^3}{3} sin\phi d\phi

= \frac{2\pi r^3}{3} \int_0^\pi sin\phi d\phi

= \frac{2\pi r^3}{3} cos\phi |_0^\pi

= \frac{4\pi r^3}{3}.

 

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