※ 引述《admini (我是小ｐ阿~~)》之銘言：
: 題目是:find the area of the surface generated by
: revolving about the x-axis the curve
: y^2 - 2lny = 4x from y = 1 to 2 =_______
1/4 (y^2 - 2lny) = x 1/4 (2y - 2/y ) =dx/dy
∫2πydL = ∫2πy√[(dx)^2 + (dy)^2] = ∫2πy√[(dx/dy)^2 + 1]*dy
= ∫2πy√{[1/4( y^2 - 2 + 1/y^2]+1} * dy
= ∫2πy * 1/2( y + 1/y ) dy
= π/3 * y^3 + π * y │y=1 to 2
= 7π/3 + π = 10π/3
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※ 編輯: Elfiend 來自: 220.138.221.177 (06/12 19:23)