微積分的小問題很急幫幫我~~

這需要利用三角變換
∫√(4-x^2) dx, x=2sint-->dx=2cost dt
=4 ∫cos^2t dt
=4 ∫[1+cos(2t)]/2 dt
=2 ∫[1+cos(2t)] dt
=2t+sin(2t)
=2sin-1(x/2)+2(x/2)√(1-(x/2)^2)
=2sin-1(x/2)+x√(4-x^2)/2+C

2011-03-20 11:36:35 補充：
y=√x+1, x:[0,9] 繞x軸旋轉體
Vx=∫pi*y^2 dx
=pi∫ x+2√x+1 dx
={(1/2)x^2+(4/3)x^(3/2)+x|[0,9]}|pi
=(81/2+(4/3)*27+9)pi
=171pi/2

利用三角代換法
令x=asinu

令 x = 2sinθ﹐則dx = 2cosθ dθ

∫ √ (4 - x^2) dx

= ∫ √ (4 - 4sin^2 θ) (2cosθ) dθ

= 4 ∫ cos^2 θ dθ

= 2 ∫ 1 + cos2θ dθ

= 2θ + sin2θ/2 + C

= 2arcsin(x/2) + x√(4 - x^2)/2 + C

參考：http://integrals.wolfram.com/index.jsp

2011-03-20 10:44:27 補充：
sin2θ = 2sinθcosθ 再用三角比找出cosθ就可以了

2011-03-20 11:42:56 補充：
立體體積 V
= π ∫ y^2 dx
= π ∫ (x + 2√x + 1) dx
= π [(x^2/2 +(4/3)x^(3/2) +x] | [0,9]
= π (81/2 + (4/3)*27 + 9)
= 171π/2