# Surface integral over hemisphere $z = \sqrt{R^2 - x^2 - y^2}$

Evaluate: $$\iint_S y\,dS,$$ where $S$ is the hemisphere defined by $z = \sqrt{R^2 -x^2 - y^2}.$

Attempt:I found two tangents, a normal and said $$dS = \frac{R}{\sqrt{R^2 -x^2 - y^2}} dx\,dy$$ In polars, $y = r\sin\theta,$ so I believe I should compute$$\int_0^{2\pi} \int_0^R \frac{r\sin\theta \cdot R}{\sqrt{R^2 - r^2}} r\,dr\,d\theta$$ Is this okay?

-

The surface element on a spherical surface is given by $dS = r^2 \sin\theta d\theta d\phi$ in spherical coordinates $(r, \theta, \phi)$. Thus your surface integral can be evaluated as follows:

$$\iint_S y \,dS = \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} R \sin\theta \sin\phi \cdot R^2 \sin\theta \,d\theta \,d\phi =$$ $$= R^3 \cdot \left[-\cos\phi\right]_{\phi=0}^{2\pi} \cdot \left[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}\right]_{\theta=0}^{\pi/2} = R^3 \cdot 0 \cdot \frac{\pi}{4} = 0$$

However, by noting that the integral of an odd function over a symmetric interval is always zero, it is possible to obtain the same result without any calculations.

-
Thanks. The reason I opted for cylindrical coords was the fact that we had symmetry with respect to the z axis. Why is my answer not correct? – CAF Dec 21 '12 at 7:18
Your answer is correct; try to carry out the integration! Sorry for the late reply. – Librecoin Feb 4 '13 at 11:35