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?