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

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SUMMARY

The discussion focuses on evaluating the surface integral ∫∫ydS over the paraboloid defined by y = x^2 + z^2, constrained within the cylinder x^2 + z^2 = 4. The integral is transformed into polar coordinates, resulting in ∫∫r^2√(4r^2 + 1)rdrdθ, with r ranging from 0 to 2 and θ from 0 to 2π. A suggested method for integration involves substituting u = 4r^2 + 1, simplifying the expression to facilitate calculation. This approach provides a clear pathway to evaluate the integral effectively.

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bodensee9
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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.
 
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bodensee9 said:
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.
 
Many thanks.
 

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