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Use polar coordinates to evaluate.

  1. Oct 15, 2009 #1

    Ral

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    1. The problem statement, all variables and given/known data
    http://img162.imageshack.us/img162/9831/97118623.jpg [Broken]


    2. Relevant equations



    3. The attempt at a solution
    I first drew R, and from the circle equation, I know the radius of the circle is 12.5. Since the region is in the first quadrant, that'll mean that my limits of integration for r and theta would be[tex]\int^{2\pi}_{0}\int^{12.5}_{0}[/tex]

    I'm getting confused on how to do the actual integration. My professor said to integrate by parts, but I'm still getting confused when doing it.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 15, 2009 #2

    lanedance

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    Homework Helper

    does that say
    [tex] e^\infty [/tex]
    what does that mean?

    the radius is not 12.5, try looking at when y=0, what is x? This will be the point of the circle on the x axis. Also it should be theta from 0 to pi/2

    now what is an area element dA in polar coordinates? Then set up you double integral with integrands, limits & the function in polar cooordinates, so we can see what the integral actually is
     
    Last edited by a moderator: May 4, 2017
  4. Oct 15, 2009 #3

    Mark44

    Staff: Mentor

    No, the radius is not 12.5.
    As already noted, the radius is not 12.5. Also, the region is not the entire circle, but just the part in the first quadrant.
    Show us your converted polar-form double integral and we can go from there.
     
    Last edited by a moderator: May 4, 2017
  5. Oct 15, 2009 #4

    Mark44

    Staff: Mentor

    I believe it is ex, but you're right, it is a bit hard to read.
     
  6. Oct 15, 2009 #5

    Ral

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    Geez, making such stupid mistakes. That part that's hard to read is e^x, sorry about that. I'll retype it here.

    [tex]\int\int_{R}ye^x[/tex]

    So when I go into polar coordinates, I replace x and y with [tex]rcos\vartheta[/tex] and [tex]rsin\vartheta[/tex] I get

    [tex]\int^{\pi/2}_{0}\int^{5}_{0}rsin\vartheta*e^{rcos\vartheta} rdrd\vartheta[/tex]

    From there, I'm getting lost. Just realized I was missing an r too.
     
    Last edited: Oct 15, 2009
  7. Oct 15, 2009 #6

    Mark44

    Staff: Mentor

    You're missing a factor of r. dx dy --> r dr d[itex]\theta[/itex].
    Your first integral is with respect to r. It essentially looks like this:
    [tex]A\int r^2 e^{Br}dr[/tex]
    where A = sin[itex]\theta[/itex] and B = cos[itex]\theta[/itex].

    Since you're going to integrate with respect to r first, the sine factor can be treated as a constant. So can the cosine factor in the exponent on e.

    This looks like a job for integration by parts to me. You'd like to get a new integral with a single factor r and the exponential factor in the integrand, and then do integration by parts one more time to get an integral with no factors of r, and just the exponential.
     
  8. Oct 15, 2009 #7

    lanedance

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    Homework Helper

    makes a bit more sense
     
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