This seems like an introductory surface integral task but I have problems solving it nonetheless. I would like to calculate surface integral $$ \iint_S x dS $$ where S is this part of sphere $x^2 + y^2 + z^2 = R^2$ contained in the first octant. The answer should equal to $\frac{\pi R^3}{4}$.
I calculated the dS using known equations and obtained $dS = R^2\cos\psi$. What should I do next?
Surface integral was defined as $$ \iint_{\Sigma}f(x,y,z)dS $$
Is my $f(x,y,z) = x$ and I should just map $x = R \cos \varphi \cos \psi$ with $\varphi \in (0, \frac{\pi}{2})$ and $\phi \in (0,\frac{\pi}{2})$(to match first octant) and then just calculate: $$ \int_0^{\frac{\pi}{2}}d\varphi \int_0^{\frac{\pi}{2}} R \cos \varphi \cos \psi \cdot R^2 \cos \psi d\psi $$ ?
If not, what is the correct solution?
As Yalikesifulei pointed out in the comments, the solution seems correct: $$ \int_0^{\frac{\pi}{2}}d\varphi\int_0^{\frac{\pi}{2}}R\cos\varphi\cos\psi\cdot R^2 \cos \psi d\psi = \\ = R^3 \int_0^{\frac{\pi}{2}}\cos\varphi d\varphi \int_0^{\frac{\pi}{2}} \cos^2\psi d\psi = \\ = R^3 \int_0^{\frac{\pi}{2}} \frac{1+\cos2\psi}{2} d\psi = R^3 \frac{\pi}{4} $$