問一道附加數學題(定積分)

我會以 "{{a, b}}" 來表達 "from a to b".

(i)
About x-axis: y=x-2 在外, y=(x-2)^(5/3) 在內.

Applying "V = ∫{a,b} [π y²] dx
Volume about x-axis
= ∫{{1, 3}} [π (x-2)²] dx - ∫{{1, 3}} [π [(x-2)^(5/3)]² ] dx
= ∫{{1, 3}} [π (x-2)²] dx - ∫{{1, 3}} [π [(x-2)^(10/3)] dx
= [(π/3) • (x-2)³] | {{1, 3}} - [(3π/13) • (x-2)^(13/3)] | {{1, 3}}
= 2π/3 - 6π/13
= 8π/39

(ii)
About y-axis: y由-1到0的部份, y=x-2 在外; y由0到1的部份, y=(x-2)^(5/3) 在外
y = x-2　<=>　x = y+2
y = (x-2)^(5/3)　<=>　x = y^(3/5) + 2

∫[π (y+2)²] dy
= (π/3) • (y+2)³ + C.

∫[π [y^(3/5) + 2]² ] dy
= ∫[π [y^(6/5) + 4 y^(3/5) + 4 ] dy
= π [(5/11) y^(11/5) + (5/2) y^(8/5) + 4y ] + C.

Applying "V = ∫{a,b} [π x²] dy
Volume about y-axis
= ∫{{-1, 0}} [π (y+2)²] dy
　- ∫{{-1, 0}} [π [y^(3/5) + 2]² ] dy
　+ ∫{{0, 1}} [π [y^(3/5) + 2]² ] dy
　- ∫{{0, 1}} [π (y+2)²] dy
= [(π/3) • (y+2)³ ] | {{-1, 0}}
　- π [(5/11) y^(11/5) + (5/2) y^(8/5) + 4y] | {{-1, 0}}
　+ π [(5/11) y^(11/5) + (5/2) y^(8/5) + 4y] | {{0, 1}}
　- [(π/3) • (y+2)³ ] | {{0, 1}}
= 7π/3
　- π (0 - [-(5/11) + (5/2) - 4] )
　+ π ( [(5/11) + (5/2) + 4] - 0)
　- 19π/3
= 7π/3 - 19π/3
　+ π ( [(5/11) + (5/2) + 4] - [-(5/11) + (5/2) - 4] )
= -4π + 5π
= π

2007-01-18 19:34:49 補充：
由於 我計算出的答案 跟 given answer 一樣, 我相信是 given answer 是對的.補充:y^(11/5) : 當y = -1, y^(11/5) = [(-1)^11] ^ (1/5) = (-1) ^ (1/5) = -1y^(8/5) : 當y = -1, y^(8/5) = [(-1)^8] ^ (1/5) = (+1) ^ (1/5) = +1後做開方, 是因為開單數次方.這是比較 tricky 的計算技巧.

想問你,你的答案是肯定對的?

2007-01-18 18:52:37 補充：
因為我計算到別的答案