作者LuisSantos ( )
看板trans_math
标题Re: [积分] 请问一题积分问题
时间Tue Jun 24 10:00:12 2008
※ 引述《flygey (努力达成目标)》之铭言：
: 2 2
: 1.Find∫∫(x +y )dA,where P is the parallelogram bounded by the lines
: x+y=1, x+y=2, 3x+4y=5, and 3x+4y=6
: 请问一下这题如何解
: 感谢
令 u = x + y ------(1)
v = 3x + 4y ------(2)
(2) - (1)*3 => y = -3u + v 代入(1)得 x = u - y = 4u - v
| δx δx |
|----- -----|
| δu δv | | 4 -1|
J = | | = | | = 4 - 3 = 1 => |J| = 1
| δy δy | |-3 1|
|----- -----|
| δu δv |
1 ≦ x + y ≦ 2 => 1 ≦ u ≦ 2
5 ≦ 3x + 4y ≦ 6 => 5 ≦ v ≦ 6
∫∫x^2 + y^2 dA
2 6
= ∫ ∫ ((4u - v)^2 + (-3u + v)^2)*(|J|) dvdu
1 5
2 6
= ∫ ∫ (25)(u^2) - 14uv + (2)(v^2) dvdu
1 5
2 2 |v = 6
= ∫ (25)(u^2)(v) - (7)(u)(v^2) + (---)(v^3) | du
1 3 |v = 5
2 182
= ∫ (25)(u^2) - 77u + ----- du
1 3
25 77 182 |2
= (----)(u^3) - (----)(u^2) + (-----)(u) |
3 2 3 |1
175 231 182
= ----- - ----- + -----
3 2 3
357 231 231 7
= ----- - ----- = 119 - ----- = ---
3 2 2 2
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 140.119.27.58
1F：推 flygey:感谢版友解答 140.137.49.4 06/24 11:51