Integrate (R/r)dr where r= (h^2+R^2)^1/2 with boundary from 0 to a.
Thanks
更新1:
sorry, it's dR
更新2:
The answer is correct, but must I use tan@ or sec^2@ to integrate it? Is there any simplier method?
please help me! Integration
2010-05-02 3:06 am
please help me integrate step by step.
Integrate (R/r)dr where r= (h^2+R^2)^1/2 with boundary from 0 to a.
Thanks
Integrate (R/r)dr where r= (h^2+R^2)^1/2 with boundary from 0 to a.
Thanks
回答 (4)
2010-05-02 4:36 am
✔ 最佳答案
S (0,a) R/r dR = S (0,a) R/(h^2 + R^2)^1/2 dRPut R = htan@, dR = hsec^2@d@
When R = 0, @ = 0, when R = a, @ = tan^-1(a/h)
So, the integral becomes:
S (0,tan^-1(a/h) h^2sec^2@tan@/hsec@ d@
= h S (0,tan^-1(a/h) sec@tan@ d@
= h [sec@](0,tan^-1(a/h))
= h[(1/h)sqrt(a^2 + h^2) - 1]
= sqrt(a^2 + h^2) - h
參考: Physics king
2010-05-05 11:43 pm
2010-05-02 7:10 am
2010-05-02 3:21 am
Integrate (R/r)dr, 0 to a
=R * Integrate (1/r) dr, 0 to a
=R * ln r, 0 to a
= R( ln a - ln 0)
=> undefined
wrong question?
do u meant dR?
=R * Integrate (1/r) dr, 0 to a
=R * ln r, 0 to a
= R( ln a - ln 0)
=> undefined
wrong question?
do u meant dR?
收錄日期: 2021-04-19 22:09:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100501000051KK01476