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rkahn rkahn
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Posts: 47
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10 years ago
Evaluate the given integral by making an appropriate change of variables, where R is the trapezoidal region with vertices (2, 0), (10, 0), (0, 10), and (0, 2).
L=double integral of 5(cos(3(y-x)/(y+x))) dA
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10 years ago
"First of all, note that the edges of the trapezoid have equations
x = 0, y = 0, x+y = 2, and x+y = 10.

This and the integrand suggests the substitution
u = y - x, and v = y + x.
So, x = (-u+v)/2 and y = (u+v)/2.

The boundaries of R transform as follows:
x = 0 ==> u = v, and y = 0 ==> u = -v, while
x+y = 2 and x+y = 10 ==> v = 2, v = 10.

Moreover, ∂(x,y)/∂(u,v) =
|-1/2 1/2|
|1/2 1/2| = -1/2.

So, ∫∫R 5 cos(3(y-x)/(y+x))) dx dy
= ∫(v = 2 to 10) ∫(u = -v to v) 5 cos(3u/v) * |-1/2| du dv
= ∫(v = 2 to 10) (5/2) * (v/3) sin(3u/v) {for u = -v to v} dv
= ∫(v = 2 to 10) (5/6) * v (sin 3 - sin(-3)) dv
= ∫(v = 2 to 10) (5/6) sin 3 * 2v dv
= (5/6) sin 3 * v^2 {for v = 2 to 10}
= 80 sin 3."
If you are reading this, you are awesome, and I wish you a happy and prosperous day!
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