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阿仁 asked in 科學數學 · 1 decade ago

急~高等微積分請大師幫忙

Let Cr denote the circle of radius r about the origin in the xy-plane, oriented counterclockwise as viewed from the positive y-axis. Suppose F is a C1 vector field on the complement of the y-axis in R3 such that ∫C1 F‧dx = 5 and curl F(x,y,z) = 3j + (zi-xk) / (x^2+z^2)^2 .Compute ∫Cr F‧dx for every r .

Update:

1.題目打錯 是 in the xz-plane

Update 2:

r > 1時 ρ 的上下界是r~1?

還是應該是 1~r?

1 Answer

Rating
  • 天助
    Lv 7
    1 decade ago
    Favorite Answer

    1. xy-plane?

    2. Suppose F is a C1 vector field on the complement of the y-axis in R3 such that ∫C1 F‧dx = 5 and curl F(x,y,z) = 3j + (zi-xk) / (x^2+z^2)^2

    是否有誤?

    2010-05-11 11:00:41 補充:

    圖片參考:http://imgcld.yimg.com/8/n/AD04686329/o/1510051008...

    如圖,設C=C1+Cr,A為環形區域, nda=(0,1,0)dzdx=(0,1,0) ρdρ dθ

    若r<1,則C1順時,Cr逆時,C順時

    若r>1,則C1順時,Cr逆時,C逆時

    (設r<1) (以下線積分均為逆時方向,順時方向差一個負號)

    F在A內為diff函數(已知),so,可用Stoke定理

    ∫_Cr F‧dl-∫_C1 F‧dl=-∫_C F‧dl=-∫∫_A curl(F)‧nda

    =-∫[0~2π]∫[r~1] ( ρsinθ, 3ρ^4, -ρcosθ)/ρ^4 * (0,1,0)ρdρdθ

    =-∫[0~2π]∫[r~1] 3ρdρdθ=3π(r^2-1)

    so, ∫_Cr F‧dl-∫_C1 F‧dl=3π(r^2-1),

    hence, ∫_Cr F‧dl=5+3π(r^2-1)

    (設r>1) (以下線積分均為逆時方向,順時方向差一個負號)

    F在A內為diff函數(己知),so,可用Stoke定理

    ∫_Cr F‧dl-∫_C1 F‧dl=∫_C F‧dl=∫∫_A curl(F)‧nda

    =∫[0~2π]∫[r~1] ( ρsinθ, 3ρ^4, -ρcosθ)/ρ^4 * (0,1,0)ρdρdθ

    =∫[0~2π]∫[r~1] 3ρdρdθ=3π(1-r^2)

    so, ∫_Cr F‧dl-∫_C1 F‧dl=3π(1-r^2),

    hence, ∫_Cr F‧dl=5+3π(1-r^2)

    Ans:∫_Cr F‧dl=5-3π |1-r^2|

    Note: C順時,逆時針方向時,nda差一個負號

    2010-05-11 11:41:32 補充:

    謝謝更正!

    r>1時應是1~r沒錯,積分結果是5+3π (r^2-1)

    so, the answer is 5+3π(r^2-1) for any r not equal 0.

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