※ 引述《r833123 (whales186)》之銘言：
: 1.http://lib.ncue.edu.tw/exam/trans/10001/phy/0201.pdf
: '想請問第五題這種收斂發散的解答要怎麼寫
: 難道就直接寫出我所背的公式嗎 p數列啊 交錯級數啊 ratio test啊
: 這樣會得分嗎????這是第一個問題
: 2.第二個問題是同一份考卷的第六題和第七題 我想破頭實在不知道要怎麼解
: 可能功力還差一大截
: 請各位高手指點一下!!
7.
x_i = {x, y, z}
1 2 3
_
x_i = ∫x_i dm / ∫dm
∫dm = ∫ρdxdydz
2 2-x 2-x-y
= ∫∫ ∫ 2x dz dy dx
0 0 0
= ∫∫2x(2-x-y)dydx
= ∫[2x(2-x)^2 - x(2-x)^2]dx
= ∫{x(2-x)^2]}dx
= ∫{-(2-x)^3 + 2(2-x)^2}dx
= (-16/4) + (16/3) = 4/3
∫xdm = ∫2x^2 dxdydz
2 2-x 2-x-y
= ∫∫ ∫ 2x^2 dz dy dx
0 0 0
= ∫∫2x^2 (2-x-y) dy dx
= ∫2x^2 (2-x)^2 - x^2 (2-x)^2 dx
= ∫[(2-x)^4 - 4(2-x)^3 + 4(2-x)^2] dx
= -[(-32/5) + 16 - (32/3) ] = 16/15
∫ydm = ∫2x y dzdydx
2 2-x 2-x-y
= ∫∫ ∫ 2x y dz dy dx
0 0 0
= ∫∫ 2xy(2-x-y) dydx
= ∫ x(2-x)^3 - (2/3)x(2-x)^3 dx
= ∫ (1/3)x(2-x)^3 dx
= ∫ (1/3)[-(2-x)^4 +2(2-x)^3] dx
= (-1/3)[(-1/5)(-32) + (1/2)(-16)]
= 8/15
= ∫zdm
_
x = ∫x dm / ∫dm = (16/15)/(4/3) = 4/5
_ _
y = ∫y dm / ∫dm = (8/15)/(4/3) = 2/5 = z