1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluate Double integral

  1. Apr 21, 2008 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Evaluate Double integral
Loading...