Re: [理工] [工數]-線積分
※ 引述《doom8199 (~口卡口卡 修~)》之銘言:
: ※ 引述《magiciano (magiciano)》之銘言:
: : 求∫G(x,y)ds
: : G(x,y)=x^2/y^3
: : 2y=3x^(2/3)
: : < <
: : 1 = x = 8
: : 我算到ds= √1+x^(-1/3) dx
: : 然後接下來的積分就卡住了
: : 希望各位幫幫忙~
: ---
: 2y = 3x^(2/3) → y' = x^(-1/3)
: ds = √[1+ (y')^2] dx = √[1+ x^(-2/3)] dx
: 8
: then ∫ x^2/y^3 ds = 8/27∫ √[1+ x^(-2/3)] dx
: c 1
: (set x^(-1/3) = tanθ
: → -1/3*[x^(-1/3)]^4 dx = (secθ)^2 dθ )
: = -8/9∫ secθ * (secθ)^2 / (tanθ)^4 dθ
: θ
: = -8/9∫ (sinθ)^(-4) d(sinθ)
: θ
= 8/27 * (sinθ)^(-3) |
θ
x=8
= 8/27 * [x+x^(1/3)]*√[1+ x^(-2/3)] |
~~~~~~~~~~~ x=1
這邊怪怪的...?
: = 8/27 *( 5√5 - 2√2 )
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