[微積] 汗水期末考古

[???]

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)
推 chaochienyao:你打得好辛苦喔!
推 hasheesh:有看有推！
推 hygeiasmiuk:為什麼不能有tan0
→ benimut:因為tan0不存在吧
推 hygeiasmiuk:tan0=0 tan(π/2)才不存在
→ benimut:說的也是 看來我耍白痴花了這麼多時間精力啊(汗)
→ benimut::那這個答案(2744√5π-45)/4似乎相當扯 有人算出來嗎？
→ benimut:或者說這種算法到底對還是錯뀠
→ benimut:我另外用
→ benimut:F={(x,y,z)|0 <=r<= √(7^2 - z^2),0 <=θ<= 2π,2<=z<=7}
→ benimut:------------------------------------------------------
→ benimut:    7
→ benimut:2π∫  √(7^2 - z^2)dz
→ benimut:      2
→ benimut:------------------------------------------------------
→ benimut:let z = 7sinφ
→ benimut:    √(7^2 - z^2) = 7cosφ
→ benimut:    φ = arcsin(z/7)
→ benimut:    dz = 7cosφ
→ benimut:代回原式
→ benimut:     π/2
→ benimut:2π∫           [(7cosφ)^2]dφ
→ benimut:     arcsin(z/7)
→ benimut:------------------------------------------------------
→ benimut:         π/2
→ benimut:2π*7^2∫           [(cos2φ + 1)/2]dφ
→ benimut:         arcsin(z/7)
→ benimut:------------------------------------------------------
→ benimut:                         π/2
→ benimut:=2π*7^2[1/2sin2φ + 1/2φ]
→ benimut:                         arcsin(z/7)
→ benimut:------------------------------------------------------
→ benimut:                         π/2
→ benimut:=2π*7^2[sinφcosφ + 1/2φ]
→ benimut:                         arcsin(z/7)
→ benimut:------------------------------------------------------
→ benimut:=-84√5 + 49π - arcsin(2/7)
→ benimut:答案竟然不同啊
推 chaochienyao:辛苦了!