Surface integrals of vectors: Need check of my work- answer doesn't look right

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Homework Statement


Find surface integral of vector field F=<x,y,x+y> over the surface z=x^2+y^2 where x^2+y^2 less than 1. Use outward pointing normals


Homework Equations





The Attempt at a Solution


So I did the whole thing and got a zero which doesn't look right to me. My algebra seems right so can you please validate the approach? Thanks:
I parametrize the surface in polar coordinates: x=r cos(theta) and y = r sin (theta). So my surface is (rcos(theta), rsin(theta), r^2).
I take partial derivatives of r and theta, find the determinant and my normal to the surface is < 2r^2cos(theta), 2r^2sin(theta), -r >
then F in terms of r and theta = < rcos(theta), rsin(theta), r(cos(theta)+sin(theta))>
dot product that with the normal to get (2r^3-r^2)(cos(theta)+sin(theta)) and integrate that for r= 0 to 1 and theta= 0 to 2pi
 
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Never mind- it was a calculation error... Answer is pi.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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