I am stuck on the following problem
Evaluate :
$$I=\iint x^2 y^2 z dS $$ where S is the positive side of lower half of the sphere $x^2 + y^2 + z^2 = a^2$
I tried using spherical coordinates and their jacobians but cannot seem to find the answer which is $$I= \frac {2\pi a^7} {105}$$
Substitute $z=-\sqrt{a^2-x^2-y^2}$ and $dS=\sqrt{1+z_x^2+z_y^2}dA$. which turn $$I=-a\iint_{R} x^2y^2dA.$$ Use polar co-ordinate.$x=r\cos\theta, y=r\sin \theta$. $dxdy=rdrd\theta$. $r$ varies from $0$ to $a$ and $\theta$ varies from $0$ to $2\pi$. Which turn $I=-a\int_{0}^{2\pi}\sin^2\theta \cos^2\theta d\theta \int_{0}^a r^5 dr=\frac{-a^7}{6}\int_{0}^{2\pi}\sin^2\theta \cos^2\theta d\theta =\frac{-a^7}{24}\int_{0}^{2\pi}\sin^2 2\theta d\theta=\frac{-a^7}{48}\int_{0}^{2\pi}({1-\cos4\theta}) d\theta=\frac{-a^7\pi}{24}$
Consider the following transformation: \begin{equation} \begin{split} x&=a\sin\phi\cos\theta\\ y&=a\sin\phi\sin\theta\\ z&=a\cos\phi \end{split} \end{equation} where $\theta\in[0,2\pi]$ and $\phi\in[-\frac{\pi}{2},0]$, then $dS=a^2\sin\phi\;d\phi\;d\theta$. Now $$ \iint_Sx^2y^2z\:dS=a^7\int_{0}^{2\pi}\cos^2\theta\sin^2\theta\;d\theta\int_{-\pi/2}^0\sin^5\phi\cos\phi\;d\phi=a^7\bigg[\frac{\pi}{4}\bigg]\cdot\bigg[-\frac{1}{6}\bigg]=-\frac{\pi a^7}{24}. $$ Note that $$\int_{-\pi/2}^0\sin^5\phi\cos\phi\;d\phi=\int_{-\pi/2}^0\sin^5\phi(\sin\phi)'\;d\phi=\bigg[\frac{1}{6}\sin^6\phi\bigg]_{-\pi/2}^0=-\frac{1}{6} $$
