I'm stuck with the following problem.
Evaluate the surface-integral $\int_Y xyz dS$. where $Y$ is the part of the sphere $x^2 + y^2 + z^2 =1$, that lies above the area $\{ y\leq x, x\leq 0, 0\leq x^2 + y^2 \leq 1 \}$.
I think some kind of parametric equations is necessary to solve this by hand ... but I don't know how to find the integral limits. Please help!
Let $z=\sqrt{1-x^2-y^2}$ then (recall the definition of surface integral) $$I=\int_Yxyz\, dS=\int_D xy\sqrt{1-x^2-y^2} \sqrt{1+z_x^2+z_y^2}\,dx dy$$ where $D=\{(x,y)\in\mathbb{R}^2\::\; y\leq x, x≤0, 0≤ x^2 + y^2 ≤1\}$. Since $$\sqrt{1+z_x^2+z_y^2}=\sqrt{\frac{1}{1-x^2 -y^2}}$$ after changing to polar coordinates we get $$I=\int_D xy\,dx dy=\int_{\rho=0}^1\int_{\theta=-3\pi/4}^{-\pi/2}\rho^2\cos\theta \sin\theta \, \rho d\theta d\rho\\ =\left[\frac{\rho^4}{4}\right]_0^1\cdot \left[\frac{\sin^2\theta}{2}\right]_{-3\pi/4}^{-\pi/2}=\frac{1}{4}\cdot (\frac{1}{2}- \frac{1}{4})= \frac{1}{16}.$$