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)