1. ## 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)

2. Originally Posted by s7b

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 $S: \ z=\sqrt{4 - x^2 - y^2}$ and to find the region of integration, $D,$ we have to intersect $S$ and the cone, which gives us $x^2 + y^2 = 2.$ thus $D=\{(x,y): \ x^2 + y^2 \leq 2 \}.$

now we have $dS= \sqrt{z_x^2 + z_y^2 + 1} \ dx dy=\frac{2}{\sqrt{4-x^2-y^2}} \ dx dy=\frac{2}{z} \ dx dy.$ therefore the surface integral is: $\int \int_D h(x,y,z) \ dS=2 \int \int_{x^2 + y^2 \leq 2} y \ dx dy=0.$

3. can you convert into polar coordinates once you have the integrals of y dxdy to find the limits easier??

4. Originally Posted by s7b

can you convert into polar coordinates once you have the integrals of y dxdy to find the limits easier??
sure you can or just use the symmetry and the fact that the integrand is an odd function.