Integration Over a Complicated RegionDate: 11/1/96 at 19:12:28 From: brenchi Subject: Integration The region R in the Cartesian plane consists of the points whose coordinates simultaneously satisfy the five relations: 0 <= x <= (pi/2), y >= 0, y >= sinx, y <= cosx, y <= tanx Prove that the area of R is: (1/2)*{2*2^(1/2) - 1 - 5^(1/2) - log {[5^(1/2)-1]/2}} units square From Billy Date: 11/2/96 at 16:39:1 From: Doctor Anthony Subject: Re: Integration Given that the region R in the Cartesian plane consists of the points whose coordinates simultaneously satisfy the five relations: 0 <= x <= (pi/2), y >= 0, y >= sinx, y <= cosx, y <= tanx, sketch the graphs and note that if A is the point where the tan(x) and cos(x) curves cut, we need the areas: INT(0 to A)[tan(x)-sin(x)]dx + INT(A to pi/4)[cos(x)-sin(x)]dx The point A has tan(x) = cos(x) so sin(x) = cos^2(x) sin(x) = 1 - sin^2(x) sin^2(x) + sin(x) - 1 = 0 Using the quadratic formula, sin(x) = [sqrt(5)-1]/2 x = arcsin{[sqrt(5)-1]/2} This is the x coordinate of point A, where cos(x) = sqrt(sin(x)). Integrating the expressions given above: [-ln(cos(x)) + cos(x) from 0 to A] + [sin(x) + cos(x) from A to pi/4] = [-ln(sqrt(sin(x)))+sqrt(sin(x)-0-1] + [1/sqrt(2) + 1/sqrt(2) - sin(x)-sqrt(sin(x)] where sin(x) = [sqrt(5)-1]/2 = [-(1/2)ln{(sqrt(5)-1)/2} -1 + 2/sqrt(2) - (sqrt(5)-1)/2 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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