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Thread: double integral help, polar coordinates

  1. #1
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    double integral help, polar coordinates

    I get to this bit

    $\displaystyle \int^{\frac{\pi}{2}}_{0} \int^{1}_{0} \sqrt{r^{2} +1} r dr d\theta $

    from here, can I not simplify the expression $\displaystyle \sqrt{r^{2}+1} r $

    =

    $\displaystyle r^{2} + r $ ??
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  2. #2
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    Re: double integral help, polar coordinates

    The r in "$\displaystyle r drd\theta$" is outside the square root. Use the substitution $\displaystyle u= r^2+ 1$.
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  3. #3
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    Re: double integral help, polar coordinates

    Quote Originally Posted by HallsofIvy View Post
    The r in "$\displaystyle r drd\theta$" is outside the square root. Use the substitution $\displaystyle u= r^2+ 1$.
    I know that it is outside the square root, but why can i not just take the square root or r^2 and 1 that gives

    $\displaystyle \sqrt{r^{2} + 1} $ = r+ 1

    than multiply that through by r? why is this wrong ?
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  4. #4
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    Re: double integral help, polar coordinates

    Back to algebra class for you: you can't take the square root of each term like that. For example $\displaystyle \sqrt{3^2+1}$ is $\displaystyle \sqrt{10}$, not 4. Or looking the other way, $\displaystyle (r+1)^2$ is $\displaystyle r^2+2r+1$, not $\displaystyle r^2+1$.

    When you make HallsofIvy's substitution, you get

    $\displaystyle \int^{\frac{\pi}{2}}_{0} \int^{2}_{1} \frac{1}{2} u^{1/2}\, du \,d\theta$

    which you should be able to integrate directly.

    - Hollywood
    Last edited by hollywood; May 16th 2014 at 05:50 AM. Reason: LaTeX
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  5. #5
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    Re: double integral help, polar coordinates

    Quote Originally Posted by Tweety View Post
    I know that it is outside the square root, but why can i not just take the square root or r^2 and 1 that gives

    $\displaystyle \sqrt{r^{2} + 1} $ = r+ 1

    than multiply that through by r? why is this wrong ?
    Is $\displaystyle \sqrt{25+ 1}= \sqrt{26}= 6$?

    Or $\displaystyle (r+ 1)^2= r^2+ 2r+ 1$ not $\displaystyle r^2+ 1$.
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