I have a problem that involves computing the vector surface of z=e^(1-x^2-y^2) where z is greater then or equal to 1, for a Vector Field F=xi+yj+(2-2*z)k. This is to be done using Gauss' theorem. I got an answer of 0, which doesn't seem right I just want to ask if anyone can confirm or deny the result. Here is how I did it. So this is a region sort of like a hump. The normal on the hump points upwards, and the hump and the plane z=1 I believe forms a closed surface, cutting out a circle if you look at down on it. So I calculate the Divergence of the Field and find that it equals to 0. x dx + y dy +(2-2*z)dz = 1+1-2. So then since the triple integral over the solid region is 0 I should be able to say the surface integral over the hump is equal to the sum of the double integral over the intersection of the hump and the double integral of the plane circle, which I think isof x^2+y^2=1. Therefore I calculate simply the surface integral of the circular region with a downward pointing normal (0,0,-1). Therefore F dot product n = 2-2*z. Here is the part where things get confusing for me. I try and subsitute z=e^(1-x^2-y^2). Then converting to polar coordinates so that I get e^(1-cos^2(a)-sin^2(a))=e^-0=1. There that makes the inner term I am integrating 2*1-2. Which equals 0 and making the whole thing 0. I don't think this is correct, can someone tell me where I am going wrong?