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?