calculus
posted by lssa .
Sketch the region in the first quadrant enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
y=10cosx, y=10sin2x,x=0

first solve the two to find where they cross
10sin2x = 10cosx
sin2x  cosx = 0
2sinxcox  cosx = 0
cosx(2sinx  1) = 0
cosx = 0 or sinx = 1/2
x = π/2 or x = π/4 in the first quadrant
so I would do it in two integrals
1. integral [ 10sin2x] from 0 to π/4
2. integral [10cosx] from π/4 to π/2
Area = 20cos2x  from 0 to π/4 + 10sinx  from π/4 to π/2
= (20cos2(π/4) + 20cos 0) + (10sinπ/2  (10sinπ/4))
= 0 + 20 + 0  10/√2
= 20  10/√2
= (20√2  10)/√2 = appr. 12.929