please help me! Integration

👨🏻‍🏫 學術台數學

[匿名]

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

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?

回答 (4)

Prof. Physics

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 dR

Put 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

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Nick

2010-05-05 11:43 pm

這裏可以幫到你
http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536

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