# 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 $z=0$.

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?

• Forget $z$, it is useless. You are asked to compute $$\iint_{x^2+y^2\leq a^2}x^2\,dx\,dy\qquad\text{and}\qquad \iint_{x^2+y^2\leq a^2}(x^2+y^2)\,dx\,dy$$ and they are clearly related by symmetry. Jun 8, 2017 at 0:45
• You may easily solve them through polar coordinates or just by Cavalieri's principle. Jun 8, 2017 at 0:46
• @JackD'Aurizio What would my boundaries for each integral be then in this case. I understand that the one will be $[0,2\pi]$. Jun 8, 2017 at 0:50
• The boundaries depend on the parametrization you choose for the disk $x^2+y^2\leq a^2$. Jun 8, 2017 at 1:11
• For instance in $$\iint_{x^2+y^2\leq a^2}x^2\,dx\,dy = \int_{-a}^{a}2\sqrt{a^2-x^2}x^2\,dx = 4a^4\int_{0}^{1}u^2\sqrt{1-u^2}\,du = \frac{\pi}{4}a^4$$ there is no angle ranging over $[0,2\pi]$. Jun 8, 2017 at 1:14

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 $$\int_Sz^2d\sigma =0$$