computation of moment of spherical shell

In using the formula for area integration over a sphere derived in the http://planetmath.org/node/6668last example, we need to keep in mind that to every point in $xy$ plane, there correspond two points on the sphere, which are obtained by taking the two signs of the square root. The importance of this fact in obtaining a correct answer is illustrated by our next example, the calculation of the moment of inertia of a spherical shell.

The moment of a spherical shell is given by the integral

$$I={\int }_S{x}^2{d}^{2}A.$$

While we could compute this by first converting to spherical coordinates and then using the result of http://planetmath.org/node/6664example 1, we can avoid the trouble of changing coordinates by treating the sphere as a graph. Using the result of the previous example, our integral becomes

where the factor of 2 takes into account the observation of the preceding paragraph that two points of the sphere correspond to each point of the $xy$ plane. Computing this integral, we find

$$2{\int }_{-r}^{+r}{\int }_{-\sqrt{r^2-y^2}}^{+\sqrt{r^2-y^2}}\frac{r{x}^2}{\sqrt{r^2-x^2-y^2}}𝑑x𝑑y=$$

$${2r{\int }_{-r}^{+r}\left(-\frac{1}{2}x\sqrt{r^2-x^2-y^2}+\frac{1}{2}(r^2-y^2)\arcsin \frac{x}{\sqrt{r^2-y^2}}\right)|}_{-\sqrt{r^2-y^2}}^{+\sqrt{r^2-y^2}}dy=$$

$$2r{\int }_{-r}^{+r}\frac{\pi }{2}(r^2-y^2)𝑑y=\frac{4}{3}\pi r^4$$

Quick links:

• http://planetmath.org/node/6660main entry

• http://planetmath.org/node/6668previous example