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I have to evaluate $$\iint_TxdS$$
Where $T$ is the part of the sphere $x^2+y^2+z^2=a^2$ which lies in the first octant $x,y,z\geq0$. So I used polar coordinates with $dS=rdrd\theta$ where $0\leq r\leq a$ and $0\leq \theta\leq \pi/2$. Hence
$$I= \frac{a^3}{3}$$
So my question is, is this correct? and is there any other way to solve this i.e spherical coordinates?
1 Answer
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Check your answer and I think something is wrong. The sphere in the first octant can be expressed as $$ x=a\sin\phi\cos\theta,y=\sin\phi\sin\theta,z=a\cos\theta $$ where $\phi,\theta\in[0,\pi/2]$. Then $$ dS=a^2\sin\phi d\phi d\theta$$ and hence $$\iint_TxdS=\int_0^{\pi/2}\int_0^{\pi/2}a\sin\phi\cos\theta a^2\sin\phi d\phi d\theta=\frac{a^3\pi}{4}.$$
