How to evaluate a surface integral involving a paraboloid and a cylinder?

[bodensee9]

I am wondering if someone could help me evaluate the following:
I am asked to find the surface integral ∫∫ydS where S is part of the paraboloid y = x^2+z^2 that lies inside the cylinder x^2+z^2 = 4.
The double integral could be rewritten as ∫∫y*√(4(x^2+z^2)+1)dS, or ∫∫(x^2+z^2)*√(4(x^2+z^2))dxdz. But this seems very difficult to integrate, so if I convert to polar coordinates, I should have ∫∫r^2*√(4r^2+1)rdrdθ, where r is between 0 and 2 and θ is between 0 and 2π. But I’m not really sure how to integrate r^3*√(4r^2+1)? Or did I set this up incorrectly? Thanks.

 

[HallsofIvy]

I am wondering if someone could help me evaluate the following:
I am asked to find the surface integral ∫∫ydS where S is part of the paraboloid y = x^2+z^2 that lies inside the cylinder x^2+z^2 = 4.
The double integral could be rewritten as ∫∫y*√(4(x^2+z^2)+1)dS, or ∫∫(x^2+z^2)*√(4(x^2+z^2))dxdz. But this seems very difficult to integrate, so if I convert to polar coordinates, I should have ∫∫r^2*√(4r^2+1)rdrdθ, where r is between 0 and 2 and θ is between 0 and 2π. But I’m not really sure how to integrate r^3*√(4r^2+1)? Or did I set this up incorrectly? Thanks.

Write $r^3\sqrt{4r^2+ 1}dr$ as $r^2\sqrt{4r^2+ 1}(rdr)$ and let $u= 4r^2+1$.

 

[bodensee9]

Many thanks.