F.6 數學定積分應用1

圖片參考：http://img137.imageshack.us/img137/7444/integration.png

y^2 + 4x = 0 => x = -y^2/4 ... (1)
x^2 + 4y = 0
(-y^2/4)^2 + 4y = 0
y^4/16 + 4y = 0
y^4 + 64y = 0
y(y^3 + 64) = 0
y(y + 4)(y^2 + 4y + 16) = 0
y = 0 或y = -4
(1) => y = 0 => x = 0
y = -4 => x = -4
兩曲線交點為 (0,0), (-4,-4)
因y^2 + 4x = 0 和x^2 + 4y = 0對稱於x = y
曲線(y^2) + 4x = 0 以及 (x^2) + 4y = 0 所圍成的圖形的面積
= 2∫(-x^2/4 - x ) dx 由-4到0
= 2[-x^3/12 - x^2/2]由-4到0
= 2[-64/12 + 8]
= 16/3