Evaluate $\iint (ax^2+by^2+cz^2)dS$, where $S$ is surface of unit sphere $x^2+y^2+z^2 = 1$.

My attempt. Using Gauss divergence theorem, $F.n = (ax^2+by^2+cz^2) \implies F=\langle ax,by,cz\rangle$

$\text{div} F = (a+b+c)$. by gauss divergence theorem, $$\iint_S F.n ds = \iiint_V \text{div} F dV = \iiint (a+b+c)dV = \frac{4}{3}\pi(a+b+c)$$

But if I calculate by surface integral I am getting wrong answer.

My try:

Changing to spherical coordinates $x=\sin \phi \cos \theta, y = \sin \phi \sin \theta, z= \cos \phi$ (I am taking radius $r = 1$ over sphere)

$$\iint_{S: \phi: 0 - \pi,\; \theta:0-2\pi} (a\sin^3 \phi \cos^2 \theta+b\sin^3 \phi \sin^2 \theta + c\cos^2\phi \sin \phi)\ d\phi d\theta = \Big(2\pi a+2\pi b +\frac{2}{3}c\Big).$$ Both the answers are different. I know that I made some mistake in latter part. Please help.

  • $\begingroup$ can any 1 pls have a look at this? $\endgroup$ – Magneto Nov 8 '17 at 2:40

Check your last evaluation. Note that $$\int_{\phi=0}^{\pi}\int_{\theta=0}^{2\pi} \sin^3 \phi \cos^2 \theta d\theta d\phi=\int_{\phi=0}^{\pi} \sin^3 \phi d\phi\int_{\theta=0}^{2\pi}\cos^2 \theta d\theta = \frac{4}{3}\cdot \pi$$ (the same result if you replace $\cos^2 \theta$ with $\sin^2 \theta$) and $$\int_{\phi=0}^{\pi}\int_{\theta=0}^{2\pi} \cos^2 \phi \sin \phi d\theta d\phi=2\pi\int_{\phi=0}^{\pi} \cos^2 \phi \sin \phi d\phi = 2\pi\cdot\frac{2}{3}.$$

  • $\begingroup$ @anirudh Is it clear now? $\endgroup$ – Robert Z Nov 8 '17 at 10:56
  • $\begingroup$ got it . i was checking my calculation. thanks a lot $\endgroup$ – Magneto Nov 8 '17 at 11:03

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