推 magiciano:感謝~~ ^^ 09/01 13:06
※ 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~
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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
= 8/27 *( 5√5 - 2√2 )
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