※ 引述《dayeh (dayeh)》之銘言：
: Determine the curve integral
: ∫√(x^2+y^2)ds, Γis the curve x^2+y^2=ax
: Γ
: 請問這題要怎麼算？
: 謝謝
x^2 + y^2 = ax
=> 2x + 2y(dy/dx) = a
=> dy/dx = (a-2x)/2y
=> (dy/dx)^2 = (a-2x)^2/4y^2
=> 1 + (dy/dx)^2 = 1 + [(a^2-4ax+4x^2)/4y^2]
= (a^2-4ax+4x^2+4y^2)/4y^2
= a^2/4y^2 (∵ x^2 + y^2 = ax ∴ 4x^2 + 4y^2 = 4ax)
=> ds = sqrt[1+(dy/dx)^2]dx = a/2y dx
x^2 + y^2 = ax
=> [x-(a/2)]^2 + y^2 = a^2/4
let x = (a/2) + (a/2)cost and y = (a/2)sint 0≦t≦2π
dx = -(a/2)sint dt
=> ds = a/2y dx = a/2[(a/2)sint] [-(a/2)sint]dt
= -(a/2)dt
=> ∫ sqrt(x^2+y^2) ds = ∫ sqrt(ax) ds
Γ Γ
2π
= ∫ sqrt(a)sqrt[(a/2)+(a/2)cost][-(a/2)dt]
0
2π
= [-a^(3/2)/2]∫ sqrt[(a/2)+(a/2)cost]dt
0
2π
= [-a^2/2^(3/2)]∫ sqrt(1+cost)dt
0
= [-a^2/2^(3/2)]*4sqrt(2)
= -2a^2
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◆ From: 140.113.66.158
※ 編輯: sulanpa 來自: 140.113.66.158 (02/16 22:48)