Find the area of the region enclosed by one loop of the curve.
r = 2sin(7θ)
r = 2sin(7θ)
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one loop , r=0
sin 7T=0
7T = 0+kpi , k=0,1
T=0
T= pi/7
dA= rdrdT
A=INT INT rdr dT
A= INT r^2/2 dT
0
A= (1/2) INT 4sin^2(7T) dT
A= 2INT sin^2(7T) dT
0
If u=7T , sin^2u = (1-cos2u)/2
A=2 INT (1-cos14T) /2 dT
A= T -(1/14) sin14T
A= pi/7 -(1/14) sin 2pi - (0- (1/14) sin0)
A= pi/7
sin 7T=0
7T = 0+kpi , k=0,1
T=0
T= pi/7
dA= rdrdT
A=INT INT rdr dT
A= INT r^2/2 dT
0
A= 2INT sin^2(7T) dT
0
A=2 INT (1-cos14T) /2 dT
A= T -(1/14) sin14T
A= pi/7 -(1/14) sin 2pi - (0- (1/14) sin0)
A= pi/7
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2sin(7θ) = 0 -----> θ = 0, π/7, 2π/7, . . .
So for one loop we integrate from 0 to π/7
To find area using polar coordinates, we calculate as follows:
A = ∫ₐᵇ 1/2 r² dθ
A = ∫ [0 to π/7] 1/2 (2 sin(7θ))² dθ
A = 2 ∫ [0 to π/7] sin²(7θ) dθ
A = 2 ∫ [0 to π/7] (1/2 − 1/2 cos(14θ)) dθ
A = 2 (1/2 θ − 1/28 sin(14θ)) | [0 to π/7]
A = (θ − 1/14 sin(14θ)) | [0 to π/7]
A = π/7 − 1/14 (sin2π) + 0 − sin(0)
A = π/7
So for one loop we integrate from 0 to π/7
To find area using polar coordinates, we calculate as follows:
A = ∫ₐᵇ 1/2 r² dθ
A = ∫ [0 to π/7] 1/2 (2 sin(7θ))² dθ
A = 2 ∫ [0 to π/7] sin²(7θ) dθ
A = 2 ∫ [0 to π/7] (1/2 − 1/2 cos(14θ)) dθ
A = 2 (1/2 θ − 1/28 sin(14θ)) | [0 to π/7]
A = (θ − 1/14 sin(14θ)) | [0 to π/7]
A = π/7 − 1/14 (sin2π) + 0 − sin(0)
A = π/7
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2 sin 7θ = 0
sin 7θ = 0
7θ = 0, π, 2π, ...
θ = π/7
A = ∫ {0, π/7} 2 sin 7θ dθ = 2/7 (-cos 7θ) = 2/7 [-cos (7 • π/7) + cos (7 • 0)] = 2/7 (-cos π + cos 0) = 2/7 (1 + 1) = 4/7
sin 7θ = 0
7θ = 0, π, 2π, ...
θ = π/7
A = ∫ {0, π/7} 2 sin 7θ dθ = 2/7 (-cos 7θ) = 2/7 [-cos (7 • π/7) + cos (7 • 0)] = 2/7 (-cos π + cos 0) = 2/7 (1 + 1) = 4/7