Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.