Evaluating The Surface Integral With The Part Of A Plane Inside A Cylinder

Question: how to calculate the surface integral of

∫xz dσp

where p is the part of the plane z+x-5=0 that is inside the cylinder x2+y2=1.

Im trying to understand how to do this exercise… I guess that i have to convert the surface to the cylindrical coordinates but i cant find any problem similar to this to understand how to do it.

Thanks in advance

Solution

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∫∫xzdσp = ∫∫Dx(5 – x)√1 +(- 1)2dxdy = √2∫∫(5x – x2)dxdy; where D = {x2 + y2 ≤ 1}.

Wedchange to polar coordinates: x = rcosθ; y = rsinθ; 0 ≤ r ≤ 1; 0 ≤ θ ≤ 2π

∫∫xzdσp = √2∫01∫02π(5rcosθ – r2cos2θ)rdrdθ = √2 ∫01(5r2sinθ – 1/2r3θ – 1/4r3sin2θ)02√πdr =

√2∫01(- πr3)dr = – π√2/4