The definite integral which represents the area of the region in the fourth quadrant enclosed by r = 10 - cos is A = 210π - 80/8
Given:
the area of the region in the fourth quadrant enclosed by r = 10 - cos
we know that:
A = (limit a to b)∫ r²/2 dθ
we know that area of first quadrant = area of fourth quadrant (by symmetry)
A = (limit θ=0 to π/2)∫ (1-cosθ)²/2 dθ
= (limit θ=3π/2 to 2π)∫ (1-cosθ)²/2 dθ
A = 1/2(limit 0 to π/2)∫ (100 - 20cosθ + cos²θ)dθ
A = 1/2 (limit 0 to π/2)∫ (100 - 20cosθ + 1/2 + cos2θ/2)dθ
= 1/2 (limit 0 to π/2) ∫ (201/2 - 20cosθ + cos2θ/2)dθ
= 1/2 ( 201/2 θ- 20sinθ + sin2θ/4) limit π/2 to 0
= 1/2 ( 201π/4 - 20 + 0)
= 1/2 ( 210π - 80/4)
Hence we get the required answer.
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