I have this vector-field:
F(x; y; z) = (x*exp(xy)cos(exp(xy)) ; y*exp(xy)cos(exp(xy)) ; x^2 + y^2)
And I'm supposed to take the surface integral over half a torus (over the positive z-axis).
Now, my book says that for this surface integral, I won't have to do it over the torus,
if I take it over its projection on the xy-plane (an annlus), it will give the same answer.
So, my question is: how is that the same thing? I figured maybe the vector field didn't have any z-component or something, but obviously it does. My book does not give an explanation, so I thought I'd ask here.