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Homework Help: Evaluate Double integral

  1. Apr 21, 2008 #1

    tony873004

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    1. The problem statement, all variables and given/known data
    Evaluate the integral [tex]\int {\int\limits_R {\left( {x + y} \right)\,dA} } [/tex] where R is the region that lies to the left of the y-axis between the circles [tex]x^2 + y^2 = 1[/tex] and [tex]x^2 + y^2 = 4[/tex] by changing to polar coordinates.



    2. Relevant equations
    x=r cos theta
    y=r sin theta


    3. The attempt at a solution
    my effort:

    [tex]\begin{array}{l}
    r_{inner} = \sqrt 1 = 1,\,\,r_{outer} = \sqrt 4 = 2 \\
    R = \left\{ {\left( {r,\theta |1 \le r \le 4,\,\frac{{3\pi }}{2}\,\theta \le \pi } \right)} \right\} \\
    x = r\cos \left( \theta \right),\,\,\,y = r\sin \left( \theta \right) \\
    \int\limits_1^2 {\int\limits_{\pi /2}^{3\pi /2} {\left( {r\cos \left( \theta \right) + r\sin \left( \theta \right)} \right)\,d\theta \,dr} } \\
    \end{array}[/tex]

    The solution shows an extra instance of r in the integral. If the original question is for (x+y) then why isn't it simply r cos + r sin, rather than r(r cos + r sin)?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 21, 2008 #2
    Assuming everything else is correct:

    there is an additional r because of Jacobian
    you will learn co-od transformations soon
    when you transform, area changes, and that
    r takes into account of that
     
  4. Apr 21, 2008 #3
    the area element in polar coordinates is :

    [tex] dA = rdrd\theta[/tex]
    that is where the extra r comes from.
     
  5. Apr 21, 2008 #4

    tony873004

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    Thanks!

    I was about to say never mind because I just found the formula in the book. It gives the extra r.

    But I'm glad I asked, because now I understand where it comes from.
     
  6. Apr 21, 2008 #5
    indeed, you can see this also if you draw out the picture of the infinitesimal pieces: rdtheta will be a small slice of length. multiplying by dr will give you approximately a small square. so dxdy =~ rdrdtheta
     
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