[微積] 汗水期末考古

看板NTUBIME100HW作者benimut時間17年前 (2008/06/13 20:52)推噓5(5推 0噓 34→)

留言39則, 4人參與討論串1/3 (看更多)

一 Find a point on the surface z = 8 - 3x^2 - 2y^2 at which the tangent plate is perpendicular to the line x = 7 - 3t, y = 12 + 8t, z = 10 - t.(20%) 二 Let D={(x,y,z) | x^2 + y^2 + z^2 <= 90} be a solid and the temperature at the point (x,y,z) on the D is T(x,y,z) = 3xy + 4x^2 + z^2. 三 Let D = {(x,y,z) | x^2 + y^2 + z^2 <= 49, z >= 2} be a solid with the density den(x,y,z) = 2z. (i)Find the volume of D.(7%) (ii)Find the mass of D.(8%) (iii)Find the center of mass of D.(10%) 3 √(9+x^2) 四 Find the integral ∫ [∫ √(x^2 + y^2) dy] dx.(10%) 0 0 [Hint: Use polor coordinates] 五 Let D be the triangular region bounded by 3 lines x = 0, y = 2, x - 2y = 0, and C = (偏微分)D be the bounary of D. Find the line integral ∫ (x + 2xy)dx + (x^2 + 3x)dy.(15%) c 其實我是為了求解答才發文的我相信大家手上都有了我期末考考古題第一題題目就看不懂聽了華均的說明還是不會做第二題我也不會做為什麼我的z自動消掉ꐨ汗) 第三題我用 E={(ρ,φ,θ) | 2secφ <= ρ <= 7, 0 <= φ <= arccos(2/7), 0 <= θ <= 2π}解 (算式很長 PTT很糟跳下面問題看比較快) ----------------------------------------------------------------- arccos(2/7) 7 2π ∫ ∫ [∫ ρ^2*sinφ dθ] dρ dφ 0 2secφ 0 ----------------------------------------------------------------- arccos(2/7) 7 =2π∫ [∫ ρ^2*sinφ dρ] dφ 0 2secφ ----------------------------------------------------------------- arccos(2/7) =2π∫ [(1/3)*7^3*sinφ - (1/3)*(2secφ)^3*sinφ] dφ 0 ----------------------------------------------------------------- arccos(2/7) =2π∫ [(1/3)*7^3*sinφ] dφ 0 arccos(2/7) - 2π∫ [(1/3)*(2secφ)^3*sinφ] dφ 0 ----------------------------------------------------------------- arccos(2/7) =2π*(sin(arccos(2/7)))*7^3 - 2π*(2/3)∫ [(secφ)^3*sinφ] dφ 0 ----------------------------------------------------------------- arccos(2/7) =2π*3√5*7^3 - 2π*(2/3)∫ [(secφ)^3*sinφ] dφ 0 ----------------------------------------------------------------- 其中(上面只是說明為什麼我會有這個問題，這裡才是問題) ----------------------------------------------------------------- ∫(secφ)^3*sinφ] dφ ----------------------------------------------------------------- =∫tanφ(secφ)^2 dφ ----------------------------------------------------------------- 而d(tanφ)=(secφ)^2 dφ ----------------------------------------------------------------- 故 ∫tanφ(secφ)^2 dφ ----------------------------------------------------------------- =∫tanφd(tanφ) ----------------------------------------------------------------- =(1/2)(tanφ)^2 ----------------------------------------------------------------- ----------------------------------------------------------------- 代回原式 arccos(2/7) 2π*3√5*7^3 - 2π*(2/3)∫ (secφ)^3*sinφ] dφ 0 -----------------------------------------------------------------] =2π*3√5*7^3 - 2π*(2/3)[(1/2)*(tan(arccos(2/7)))^2 - (1/2)*(tan0)^2] ----------------------------------------------------------------- =2π*3√5*7^3 - 2π*(2/3)[(1/2)*(3√5/2)^2 - (1/2)*(tan0)^2] ----------------------------------------------------------------- 天啊! 為什麼會出現tan0啊(暈第五題教過嗎? 誰來救救我啊!(吶喊) (發現換回windows就有積分符號可以打了不過還是沒有偏微分...) ※ 編輯: benimut 來自: 61.231.8.26 (06/13 22:36)

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chaochienyao

06/13 22:47, , 1^F

06/13 22:47, 1^F

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hasheesh

06/13 23:20, , 2^F

06/13 23:20, 2^F

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hygeiasmiuk

06/14 11:34, , 3^F

06/14 11:34, 3^F

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