I've looked at tons of videos and read all chapters in my book and can't seem to be able to solve this task.
Evaluate the surface integral $$\int\int_{Y} (x+y)z dS$$ where the surface Y is the part of the sphere $1=x^2+y^2+z^2$, that is located above the area $y\leq -x, x\leq y, 0\leq x^2+y^2\leq 1$.
I tried parametrization and thinking symmetry and I believe I got some kind of answer but can't seem to be able to fit in the limits.
Edit: The conclusion I came up with was: $$\int\int_{R}\dfrac{1}{\sqrt{-x^2-y^2+1}}dA$$ and: $$\int\int sin(\phi)d\phi d\theta$$
I'm using geographical coordinates $(\phi,\theta)$. Then the upper half of $S^2$ is produced by the map $$(\phi,\theta)\mapsto{\bf r}(\phi,\theta)=\bigl(\cos\phi\cos\theta,\sin\phi\cos\theta,\sin\theta\bigr)\qquad\bigl(0\leq\phi\leq2\pi, \ 0\leq\theta\leq{\pi\over2}\bigr)\ .$$ One computes $|{\bf r}_\phi\times{\bf r}_\theta|=\cos\theta$, hence $${\rm d}S=\cos\theta\ {\rm d}(\phi,\theta)\ .$$ The constraints $x\leq y\leq -x$ define the sector ${3\pi\over4}\leq\phi\leq{5\pi\over4}$. It follows that $$\eqalign{\int_Y (x+y)z\>{\rm d}S&=\int_0^{\pi/2}\int_{3\pi/4}^{5\pi/4}(\cos\phi+\sin\phi)\cos^2\theta\sin\theta\>d\phi\>d\theta\cr &=\int_0^{\pi/2}\cos^2\theta\sin\theta\>d\theta\ \int_{3\pi/4}^{5\pi/4}(\cos\phi+\sin\phi)\>d\phi\ .\cr}$$