Surface integral of xyz on a sphere above a cone

[mit_hacker]

Homework Statement

Compute the surface integral:

g = xyz on x^2+y^2+z^2 = 1 above z^2=x^2+y^2.

Homework Equations

The Attempt at a Solution

I'm only doubtful about the parameterization. Under normal circumstances, since x^2+y^2+z^2 = 1 is a sphere, we can write:

r = (SinCos[v])i + (SinSin[v])j + (Cos)k.

However, how do you account for the "above z^2=x^2+y^2."

Do I simply sum the square of the x and y components and write:

r = (SinCos[v])i + (SinSin[v])j + (Sin^2)k.

Is this correct?

 

[HallsofIvy]

No, of course not! z^2= x^2+ y^2 has nothing to do with the sphere- it is below the sphere!

z^2= x^2+ y^2 is a cone with axis the positive z-axis. Since you are only concerned with the surface area, you would only use that to determine the limits of integration. In spherical (angular) coordinates, x= SinCos[v], y= SinSin[v] and z= Cos. Obviously, x^2+ y^2= Sin^2(u) and z^2= Cos^2(u). The equaton of the cone is just Sin^2(u)= Cos^2(u) which results in (since we are talking about z> 0 here) u= $\pi/4$. v ($\theta$ in spherical coordinates) goes from 0 to 2$\pi$ and u ($\phi$ in spherical coordinates) goes from 0 to $\pi/4$.

 

[mit_hacker]

Thanks a lot once again

Thanks very much again HallsofIvy!