Evaluating the surface integral with the part of a plane inside a cylinder

Question

Question: how to calculate the surface integral of

∫xz dσ_p

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

Answer 1

∫∫xzdσ_p = ∫∫_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

∫∫xzdσ_p = √2∫₀¹∫₀^2π(5rcosθ - r²cos²θ)rdrdθ = √2 ∫₀¹(5r²sinθ - 1/2r³θ - 1/4r³sin2θ)₀^2√πdr =

√2∫₀¹(- πr³)dr = - π√2/4