Please Help, I'm not sure how to set up this integration;
Integrate the given function over the giver surface: h(x,y,z)=yz, over the part of the sphere x^2+y^2+z^2=4 that lies above the cone z=sqrt(x2+y2)
the surface is $\displaystyle S: \ z=\sqrt{4 - x^2 - y^2}$ and to find the region of integration, $\displaystyle D,$ we have to intersect $\displaystyle S$ and the cone, which gives us $\displaystyle x^2 + y^2 = 2.$ thus $\displaystyle D=\{(x,y): \ x^2 + y^2 \leq 2 \}.$
now we have $\displaystyle dS= \sqrt{z_x^2 + z_y^2 + 1} \ dx dy=\frac{2}{\sqrt{4-x^2-y^2}} \ dx dy=\frac{2}{z} \ dx dy.$ therefore the surface integral is: $\displaystyle \int \int_D h(x,y,z) \ dS=2 \int \int_{x^2 + y^2 \leq 2} y \ dx dy=0.$