求曲線圍成的面積(2)
回答 (2)
2010-10-03 8:30 pm
✔ 最佳答案
======================================================圖片參考：http://img20.imageshack.us/img20/8406/17140746.png
2010-10-03 12:30:37 補充：
http://img20.imageshack.us/img20/8406/17140746.png
2010-10-03 14:17:28 補充：
對,那上限應該是sqrt(2),是錯手打錯了.
但面積好像沒有sqrt(2),讓我再對一下.
2010-10-03 14:24:32 補充：
Integrate v du 的上限是sqrt(2),轉化t後上下限沒錯了.
2010-10-03 10:02 pm
To: Nelson
u^4+v^4=4面積的上下限不是0~1
2010-10-03 14:12:57 補充：
謹供參考:坐標變換(u,v)=(x-y, 2x+3y), 則Jacobian= 1/5
面積=∫∫_D dxdy=∫∫_D (1/5)dudv (D: u^4+v^4=4)
=(4/5)∫[0~√2]∫[0~(4-u^4)^(1/4)] dudv [u^2= 2sint ]
=(4/5)∫[0~π/2] (cost)^(3/2)/(√sint) dt
=(2/5)B(5/4, 1/4)=(2/5)Γ(5/4)Γ(1/4)/Γ(3/2)
=(2/5)*(1/4)[Γ(1/4)]^2/[(1/2)Γ(1/2)]=[Γ(1/4)]^2/(5√π)
u^4+v^4=4面積的上下限不是0~1
2010-10-03 14:12:57 補充：
謹供參考:坐標變換(u,v)=(x-y, 2x+3y), 則Jacobian= 1/5
面積=∫∫_D dxdy=∫∫_D (1/5)dudv (D: u^4+v^4=4)
=(4/5)∫[0~√2]∫[0~(4-u^4)^(1/4)] dudv [u^2= 2sint ]
=(4/5)∫[0~π/2] (cost)^(3/2)/(√sint) dt
=(2/5)B(5/4, 1/4)=(2/5)Γ(5/4)Γ(1/4)/Γ(3/2)
=(2/5)*(1/4)[Γ(1/4)]^2/[(1/2)Γ(1/2)]=[Γ(1/4)]^2/(5√π)
收錄日期: 2021-04-24 10:21:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100928000051KK00075