# 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?

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