per the usual formula,
A = ∫ 1/2 r^2 θ dθ
but the boundary changes at Ø = arctan(1/2) so
A1 = ∫[0,arctan(1/2)] 1/2 (4sinθ)^2 dθ = 4Ø - 8/5
A2 = ∫[arctan(1/2),π/2] 1/2 (2cosθ)^2 dθ = π/2 - 2/5 - Ø
A = A1 + A2 = π/2 - 2 + 3Ø = 0.9617
check using rectangular coordinates; we have two circles
(x-1)^2 + y^2 = 1
x^2 + (y-2)^2 = 4
They intersect at (8/5, 4/5) so the area is
∫[0,8/5] √(1 - (x-1)^2) - (2-√(4-x^2)) dx = 0.9617
Find all points of intersection (r,θ) of the curves r=2cos(θ), r=4sin(θ).
Note: In this problem the curves intersect at the pole and one other point. Only enter the answer for nonzero r in the form (r,θ) with θ measured in radians.
POINT OF INTERSECTION: (4/sqrt5, atan(1/2))
Given the point of intersection above what is the area?? I have been stuck working on this for awhile and many tutors can not get the right area...
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1 answer