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Homework Help: Stuck on the integration bound

  1. Nov 30, 2017 #1
    1. The problem statement, all variables and given/known data
    Trying to help a friend with a problem. We are supposed to solve the below using polar coordinates. The actual answer is supposed to be π/16. Solving the integral is not the issue, just converting it.
    EKDTWGI.png

    2. The attempt at a solution
    What I got sort of worked, but it is only half of the answer given.
    92fa8XY.png

    The actual answer was this, but putting it into the calculator gives a completely different answer from what it's supposed to be.
    fnl2peT.png


    This is how I was reading the situation:
    The hatch is the area that it seems we need to integrate over, and it also seems as though my answer shouldn't even be close given how it is a rectangular area.
    swQt1ek.png


    Is there something I'm not seeing, or is the provided answer wrong?
     
  2. jcsd
  3. Nov 30, 2017 #2

    andrewkirk

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    How did you get those integration limits of ##[\sqrt 2,2]## for r? ##r## is the distance from the origin to:
    • the line ##y=1## for the lower limit and
    • the circle boundary for the upper limit
    Looking at the diagram, we see that, at ##\theta=0## the integration is for ##r\in[1,2]## and for ##\theta=\pi/4## it is for ##r\in [\sqrt 2,\sqrt 2]##. Those limits match the limits in the solution formula, but not the ones in the integral above it.
     
  4. Dec 1, 2017 #3
    From some of the examples I found you base r off of the origin to some point. From the origin to the top of the circle is a distance of √2. Having r ∈ [√2, 2cos(Θ)] Didn't work, but replacing it with its max at 2 made it a little bit closer to the answer.
     
  5. Dec 1, 2017 #4

    andrewkirk

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    Draw a line segment from the origin, at an angle of ##\theta## above the x axis, to the point where it hits the curve.
    The lower limit for the inner integral is the distance along that segment from the origin (call it O) to the point where the segment hits the vertical line ##x=1## (call that point Q).
    The upper limit for the inner integral is the distance from O to where the line segment hits the curve.
    Use geometry to calculate those two lengths as formulas in terms of ##\theta##.
     
  6. Dec 1, 2017 #5

    Ray Vickson

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    Let the shaded shape in your diagram be ABCA, where AB and CA are horizontal and vertical line segments and BC is the circular arc. The origin is O.

    One way to solve it using polar coordinates is to evaluate the integral over OBCO in polar coordinates, then subtract the integral over OACO.
     
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