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Math Help - Simplify the integral (is my version correct?)

  1. #1
    Member Pranas's Avatar
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    Simplify the integral (is my version correct?)

    Basically I am asked to simplify the integral by going to polar coordinates. But professor's summary is not available at this moment, so I'm afraid I might be doing something wrong, because this material is very new to me.

    \displaystyle \[I = \int\limits_0^{\frac{R}{{\sqrt {1 + {R^2}} }}} {\left( {\int\limits_0^{Rx} {f\left( {\frac{y}{x}} \right)dy} } \right)} dx + \int\limits_{\frac{R}{{\sqrt {1 + {R^2}} }}}^R {\left( {\int\limits_0^{\sqrt {{R^2} - {x^2}} } {f\left( {\frac{y}{x}} \right)dy} } \right)} dx\]

    I assume it to be a sector of a circle. My angle increases until it reaches line Rx (that's why I wrote first upper limit to be arctan(R)). Then for every angle I go from 0 to R with my rho as well. Only then I go to usual polar substitutions.

    \displaystyle \[I = \int\limits_0^{\arctan (R)} {\left( {\int\limits_0^R {\left( {f\left( {\frac{{y(\rho ,\varphi )}}{{x(\rho ,\varphi )}}} \right) \cdot \rho } \right)d\rho } } \right)} d\varphi  = \int\limits_0^{\arctan (R)} {\left( {\int\limits_0^R {\left( {f\left( {\tan (\varphi )} \right) \cdot \rho } \right)d\rho } } \right)} d\varphi  = \]

    \displaystyle \[ = \frac{{{R^2}}}{2}\int\limits_0^{\arctan (R)} {f\left( {\tan (\varphi )} \right)} d\varphi \]

    Does it seem like a reasonable enough idea? What logic should I use to check if the outcome is likely to be good?

    P.S. That was based on my previous and first overall (yet don't know if it's correct) solution of this kind:

    \displaystyle \[I = \iint\limits_D {2x \cdot dS}\]

    where D is area within

    \displaystyle \[{x^2} + {y^2} = 2x \Rightarrow {(x - 1)^2} + {y^2} = 1\]

    I came up with

    \displaystyle \[I = 2 \cdot \int\limits_0^{\frac{\pi }{2}} {\left( {\int\limits_0^{2\cos (\varphi )} {2{\rho ^2} \cdot \cos (\varphi )d\rho } } \right)} d\varphi  = 4\int\limits_0^{\frac{\pi }{2}} {\frac{8}{3}} {\left( {\cos (\varphi )} \right)^4}d\varphi  = \frac{{32}}{3} \cdot \frac{3}{8} \cdot \frac{\pi }{2} = 2\pi \]

    But how should I check (using WolframAlpha of similar stuff) the answer? I mean how to calculate the answer using purely the first given form?


    Thanks a lot and sorry for possibly quite bad English.
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  2. #2
    MHF Contributor
    Joined
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    Hi

    Everything you have written is OK for me
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