calculus
posted by zoe on .
find the centroid of the plane region bounded by the curves y = cos x, y=sinx, x=0,

Double check my math, because it gets messy...
Break it into two intervals: [0,pi/4] and [pi/4,pi/2]
Int(sin x)[0,pi/4] + Int(cos x)[pi/4,pi/2]
= (cos x)[0,pi/4] + (sin x)[pi/4,pi/2]
= 2√2
This is the denominator in the formulas for xbar and ybar: D = 2√2
Now for the numerators:
xbarn = Int(x sin x)[0,pi/4] + Int(x cos x)[pi/4,pi/2]
Recall that using integration by parts,
Int(x sin x) = x cos x + sin x)
Int(x cos x) = x sin x + cos x)
If my math is right, xbarn = pi/4 * (2√2)
So, xbar = xbarn/D = pi/4
Makes sense, since the area is symmetric about the line x = pi/4.
Now for ybar
ybarn = 1/2 Int(sin^2 x)[0,pi/4] + 1/2 Int(cos^2 x)[pi/4,pi/2]
Recall that
sin^2 x = (1  cos(2x))/2
cos^2 x = (1 + cos(2x))/2
ybarn = 1/4 Int(1  cos 2x)[0,pi/4] + 1/4 Int(1 + cos 2x)[pi/4,pi/2]
= 1/4 (x  1/2 sin 2x)[0,pi/4] + 1/4 (x + 1/2 sin 2x)[pi/4,pi/2]
= 1/8 (pi2)
So, ybar = ybarn/D = (pi2)/(8*(2√2))