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Integrate over area enclosed by curve

  1. Nov 8, 2006 #1
    Hi, i've been asked to use polar coordinates and to integrate

    [tex]\int\int_{A}^{} x^2 + y^2 + \sqrt{x^2 + y^2}dxdy[/tex]

    over the area A enclosed by the curve [tex]x^2 + y^2 = 4[/tex]

    I know

    [tex]x=rcos(\theta)[/tex]
    [tex]y=rsin(\theta)[/tex]
    [tex]dxdy=rdrd\theta[/tex]

    So i think the integral can be written as

    [tex]\int\int_{A}^{} r^3 \pm r^2 drd\theta[/tex]

    and i think the limits might be:

    [tex]\int_{\theta=0}^{2\pi}\int_{r=-2}^{2} r^3 \pm r^2 drd\theta[/tex]

    But i think these are probably wrong and i'm not sure how to account for the negative areas and such because i can't use symmetry if it's a function like that can i?

    If i do that i think the area comes out as [tex]64\pi/3[/tex]
     
    Last edited: Nov 8, 2006
  2. jcsd
  3. Nov 8, 2006 #2

    OlderDan

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    r is never negative in polar coordintes. The lower limit of the r integral is zero. The theta variable is what takes you all the way around the curve. When a square root is written as you have been given, the positve square root is implied. The integrand is just r^3 + r.
     
  4. Nov 8, 2006 #3
    Ok i think i understand would that mean it should be [tex]\int_{\theta=0}^{2\pi}\int_{r=0}^{2} r^3 + r^2 drd\theta[/tex]


    Which i think comes out with an answer of [tex]40\pi/3[/tex]

    If the inetgrand is [tex]r^3 + r[/tex] does that mean the r from the [tex]rdrd\theta[/tex] is only multiplied to one of them? I wasn't sure when i was writing it and just guessed it'd be multiplied to both.

    Thanks for your reply btw :)
     
    Last edited: Nov 8, 2006
  5. Nov 8, 2006 #4

    OlderDan

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    In any coordinates you use, dA multiplies the whole integrand. The r instead of r^2 was just from careless typing. Sorry. It should be r^2. I fixed it in my earlier post. Your latest expression looks good.
     
    Last edited: Nov 8, 2006
  6. Nov 9, 2006 #5
    Thanks again my main problem with these types of questions is visualising what it actually means.
     
  7. Nov 9, 2006 #6
    Didn't think i should post this in a different thread as it's sort of the same topic but could someone please check this for me:

    Question:

    Integrate the integral below

    [tex]\int\sqrt{1+x^2}ds[/tex]

    along the curve [itex]y=\frac{x^2}{2}[/itex] between [tex]x=0[/tex] and [tex]x=1[/tex]. Remember that [itex]ds=\sqrt{dx^2 + dy^2}[/itex]

    What i did:

    [tex]\frac{dy}{dx}=x[/tex]


    => [tex]dy=xdx[/tex]


    => [tex]dy^2 = x^2 dx^2[/tex]


    [tex]\int\sqrt{1+x^2}\sqrt{dx^2 + dy^2}[/tex]


    => [tex]\int\sqrt{1+x^2}\sqrt{dx^2 + x^2 dx^2}[/tex]


    => [tex]\int\sqrt{1+x^2}\sqrt{dx^2 (1+x^2)}[/tex]


    => [tex]\int\sqrt{1+x^2}\sqrt{1+x^2}dx[/tex]


    => [tex]\int1+x^2dx[/tex]


    => [tex]\int_{0}^{1}1+x^2dx[/tex]


    => [tex][1+\frac{1^3}{3}] = \frac{4}{3}[/tex]
     
    Last edited: Nov 9, 2006
  8. Nov 9, 2006 #7

    OlderDan

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    That looks good.
     
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