Results 1 to 5 of 5
Like Tree3Thanks
  • 1 Post By HallsofIvy
  • 1 Post By hollywood
  • 1 Post By HallsofIvy

Math Help - double integral help, polar coordinates

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    613

    double integral help, polar coordinates

    I get to this bit

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

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

    =

     r^{2} + r ??
    Attached Thumbnails Attached Thumbnails double integral help, polar coordinates-screen-shot-2014-05-16-12.37.24.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,041
    Thanks
    1675

    Re: double integral help, polar coordinates

    The r in " r drd\theta" is outside the square root. Use the substitution u= r^2+ 1.
    Thanks from Tweety
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2008
    Posts
    613

    Re: double integral help, polar coordinates

    Quote Originally Posted by HallsofIvy View Post
    The r in " r drd\theta" is outside the square root. Use the substitution 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

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

    than multiply that through by r? why is this wrong ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    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 \sqrt{3^2+1} is \sqrt{10}, not 4. Or looking the other way, (r+1)^2 is r^2+2r+1, not r^2+1.

    When you make HallsofIvy's substitution, you get

    \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
    Thanks from Tweety
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,041
    Thanks
    1675

    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

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

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

    Or (r+ 1)^2= r^2+ 2r+ 1 not r^2+ 1.
    Thanks from Tweety
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: November 10th 2012, 11:47 AM
  2. Double Integral with Polar Coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 8th 2011, 07:43 AM
  3. Double Integral In Polar Coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 13th 2010, 01:48 AM
  4. double integral using polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 18th 2009, 01:14 AM
  5. Double Integral Over Polar Coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 9th 2007, 06:26 AM

Search Tags


/mathhelpforum @mathhelpforum