~瘋人~ |
2011-10-24 00:31 |
x^2 / a^2 + y^2 / b^2 = 1 (bx)^2 + (ay)^2 = (ab)^2 y=+/- sqrt [b^2 - (bx/a)^2] area =2*∫ y dx ( upper limit = a , lower limit = -a) =2*∫ sqrt [b^2 - (bx/a)^2] dx =2b ∫ sqrt [1-(x/a)^2] dx using substitution, x=a sin Θ =2b ∫ sqrt [1-(sinΘ)^2] d(a sinΘ) (upper limit = π/2 , lower limit = -π/2) =2ab ∫ (cosΘ)^2 dΘ =ab∫ (1+cos2Θ)dΘ =ab [Θ+(sin2Θ)/2](upper limit = π/2 , lower limit = -π/2) =ab(π/2 - (-π/2)) =πab as a reference, circle is a special case of an ellipse with area= πr^2 , since a=b=r in a circle.
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