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.

d S
= √(1 + z_{x}'^{2}
+ z_{y}'^{2})
dx dy, where z = √(x^{2}
+ y^{2});
z_{x}'
= x/√(x^{2}
+ y^{2})
; z_{y}'
= y/√(x^{2}
+ y^{2})
, so that

dS
= √(2)
dxdy

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

∫∫ (xy
+ y √(x^{2}
+ y^{2})
+ x√(x^{2}
+ y^{2}))
√(2)
dxdy taken over the region: x^{2}
+ y^{2}
<= 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)
∫∫ r^{2}
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}
r^{3}dr
= √(2)a^4/4
∫ _{-π/2}^{π/2}
cos^{5}
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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