Show that $\int_{S}x^2 d\sigma = \frac{a^{4} \pi}{4}$ where $S$ is a disk of radius $a$ centered at the origin in the plane $z=0$.

Question

Show that $$\int_{S}x^2 d\sigma = \frac{a^{4} \pi}{4}$$ where $S$ is a disk of radius $a$ centered at the origin in the plane $z=0$.

Deduce that, $$\int_{S}(x^2 + y^2)d\sigma = \frac{a^4 \pi}{2},$$ And, $$ \int_{S}(x^2 + z^2)d\sigma = \int_{S}(y^2 + z^2)d\sigma = \frac{a^4 \pi}{4}$$

Know where to go with this problem after the initial setup, but as there is no height given, do you assume a height?

Also would i be correct in saying you use polar coordinates?

Thank you in advance.

Answer 1

With polar coordinates, the first integral becomes

$$\int_0^a\int_0^{2\pi}(r\cos (t))^2 (r)drdt $$

$$=\int_0^ar^3dr \int_0^{2\pi}\cos^2 (t)dt $$

$$=\frac {a^4}{4}\Bigl [\frac {t}{2}+\frac {1}{4}\sin(2t)\Bigr]_0^{2\pi} $$

$$\frac{\pi a^4}{4} $$

for the second, by symetry, $$\int_Sx^2d\sigma=\int_Sy^2d\sigma $$

for the third $$\int_Sz^2d\sigma =0$$