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please help me! Integration
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: 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?
最佳解答:
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
其他解答:
這裏可以幫到你 http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536|||||http://img100.imageshack.us/img100/5934/10770502.png|||||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?
please help me! Integration
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發問: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: 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?
最佳解答:
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
其他解答:
這裏可以幫到你 http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536|||||http://img100.imageshack.us/img100/5934/10770502.png|||||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?
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