Spherical coordinates in surface integrals 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 $I=-a\int_{0}^{2\pi}\sin^2\theta \cos^2\theta d\theta \int_{0}^a r^5 dr=\frac{-a^7}{6}\int_{0}^{2\pi}\sin^2\theta \cos^2\theta d\theta  =\frac{-a^7}{24}\int_{0}^{2\pi}\sin^2 2\theta d\theta=\frac{-a^7}{48}\int_{0}^{2\pi}({1-\cos4\theta}) d\theta=\frac{-a^7\pi}{24}$
