How to calculate a surface integral

Calculus
Tutor: None Selected Time limit: 1 Day

Calculate the surface integral (xy + yz + zx)dS, where S is the part of the cone z = sqrt(x^2 + y^2) inside the surface x^2 + y^2 = 2ax, where a is a positive parameter.

Apr 9th, 2015

∫∫ (xy + yz + zx) dS =

d S = √(1 + zx'2 + zy'2) dx dy, where z = √(x2 + y2); zx' = x/√(x2 + y2) ; zy' = y/√(x2 + y2) , so that

dS = √(2) dxdy

Thus, the surface integral is reduced to the following double integral

∫∫ (xy + y √(x2 + y2) + x√(x2 + y2)) √(2) dxdy taken over the region: x2 + y2 <= 2ax.

Since the region is symmetric with respect to x-axis and the first two terms inside the integral are odd functions with respect to y, their values are zero. To calculate the rest of the integral use the polar coordinates:

√(2) ∫∫ r2 cos θ rdr dθ taken over -π/2 < θ < π/2; 0 < r < a cos θ

equals to the integral √(2) ∫ -π/2 π/2 cos θ dθ ∫0 acos θ r3dr = √(2)a^4/4 ∫ -π/2 π/2 cos5 θ dθ = √(2)a^4 /4 *(16/15) = 4a^4 √(2)/15 (the last integral was calculated by substitution u = sinθ).

Answer: 4a^4√(2)/15 

Apr 9th, 2015

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