Find the area of the surface of revolution generated by revolving the curve
y=√x,0≦x≦4 , about the x -axis
y=√x,0≦x≦4繞x軸的表面積
2014-05-21 10:47 pm
回答 (10)
2014-05-25 12:45 am
✔ 最佳答案
繞 x軸 的 旋轉體, 其側表面積公式是S = ∫_[a,b] 2πy ds
y = √x, 則 ds = √{1+[1/(2√x)]^2} dx = √{1+1/(4x)} dx
故
S = ∫_[0,4] 2π√x √{1+1/(4x)} dx = ∫_[0,4] 2π √ (x+1/4) dx
簡單的 u = √(x+1/4) 代換, 得
s = u^2 - 1/4, dx = 2u du, x=0～4 則 u=1/2～(√17)/2
故
S = ∫_[1/2,√17/2] 2π u (2u) du = (4π/3){(√17/2)^3-(1/2)^3}
2014-06-21 10:35 pm
2014-06-20 9:48 pm
2014-06-11 1:02 am
2014-06-03 3:00 pm
2014-05-30 4:17 pm
2014-05-29 4:05 pm
2014-05-28 3:59 pm
2014-05-25 10:37 am
可參考下面的 旋转曲面面积之计算
wenku.baidu.com/view/f18b91c46137ee06eff91883.html
wenku.baidu.com/view/f18b91c46137ee06eff91883.html
2014-05-21 11:01 pm
∫﹝0,4﹞2π√xdx
=2π( 1/(3/2))X^(3/2)，x=﹝0,4﹞
=(4π/3)(8-0)
=32π/3
=2π( 1/(3/2))X^(3/2)，x=﹝0,4﹞
=(4π/3)(8-0)
=32π/3
收錄日期: 2021-05-04 01:55:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140521000010KK08844