求圓方程在坐標上的部份面積問題

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圖片參考：https://lh6.googleusercontent.com/-FFIWajc6T2c/Te3Snm1ahBI/AAAAAAAAF8U/uP1rjLbz5G4/7D55.JPG

樓上的答案似乎錯了.

x² + y² - 6x + 4y - 12 = 0
y² + 4y + 4 = - x² + 6x + 12 + 4
(y + 2)² = - x² + 6x + 16
y = √( - x² + 6x + 16) - 2 ......(1)
y = 1 ......(2)

We integrate (1) - (2) from x = -1 = x = 4
Answer
=∫ [ √( - x² + 6x + 16) - 2 - 1 ] dx
=∫ [ √( 25 - x² + 6x - 9) - 3 ] dx
=∫ [ √( 25 - (x - 3)² ) ] dx - 15
Now let x - 3 = 5sinθ
dx = 5cosθ dθ
x = -1 ⇒ θ = sin-1(-4/5), sinθ = -4/5, cosθ = 3/5
x = 4 ⇒ θ = sin-1(1/5), sinθ = 1/5, cosθ = (2√6)/5
So we the definite integration from θ = sin-1(-4/5) to θ = sin-1(1/5)
Answer
=∫ [ √( 25 - 5sin²θ ) ] 5cosθ dθ - 15
=∫ [ 5cosθ ] 5cosθ dθ - 15
=25 ∫ cos²θ dθ - 15
=25 ∫ (cos2θ + 1)/2 dθ - 15
=(25/2) ∫ (cos2θ + 1) dθ - 15
=(25/2) [(1/2)sin2θ + θ] - 15
=(25/2) [sinθcosθ + θ] - 15
=(25/2) [ (1/5)(2√6)/5 + sin-1(1/5) - (-4/5)(3/5) - sin-1(-4/5)] - 15
=(25/2) [ (2√6)/25 + sin-1(1/5) + 12/25 + sin-1(4/5)] - 15
= √6 + 6 + (25/2)sin-1(1/5) + (25/2)sin-1(4/5) - 15
= √6 - 9 + (25/2)sin-1(1/5) + (25/2)sin-1(4/5)

The numerical answer is 7.55765....

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