Richard P. answered • 10/21/16

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for 0 < θ < π/2 , these curves cross at θ =arctan( 3

^{1/2}) = π/3The cosine curve is the upper one, so the desired area is

A = integral from 0 to π/3 of [ 3

^{1/2}cos(θ) - sin(θ) ]The anti-derivative of the expression in the square brackets is

[ 3

^{1/2}sin(θ) + cos(θ) ] So A = (3

^{1/2}sin(π/3) +cos(π/3) ) - (3^{1/2}sin(0) + cos(0) ) Remembering that π/3 radians is the same as 60 degrees, this evaluates to

(3/2 + 1/2) -1 = 1

This is actually the ares bounded by the two curves and the y axis. There is another

intersection of the two curves at θ = -2π/3

Plugging this value in for the lower limit results in A = 2 - (-3/2 - 1/2) = 4