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Homework Help: Polar coordinates and multivariable integrals.

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Im righting this down for my roommates since he's having tons of trouble trying to figure this out and I can't answer it.
    also sorry for having to hotlink it.

    the equation is on the image since its very difficult to type it all out. Apologies about that.
    we understand everything up till the 2 is introduced and the integral goes from 0 to pi/6. We don't understand how you are able to just introduce a 2 and take out the -pi/6.
  2. jcsd
  3. Mar 15, 2012 #2


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    This combines a simple property of integrals along with the fact that the function of theta is an even function (meaning that it is symmetric around the y-axis).

    The property of integrals is that you can split the interval of integration up into a bunch of sub-intervals. The integral over the overall interval is then equal to the sum of the integrals over the subintervals. For instance, you can split the interval from -pi/6 to +pi/6 up into two sub-intervals. The first interval goes from -pi/6 to 0, and the second interval goes from 0 to +pi/6. According to the rule I just stated, then, we can write:[tex]\int_{-\pi/6}^{\pi/6} f(\theta)\,d\theta = \int_{-\pi/6}^{0} f(\theta)\,d\theta + \int_{0}^{\pi/6} f(\theta)\,d\theta [/tex]Now, if f(θ) is an even function, then it is symmetric around the y-axis. This means that the area under the curve from -pi/6 to 0 is exactly the same as the area under the curve from 0 to pi/6. Therefore, you can substitute one integral for the other (they are numerically equivalent):[tex]\int_{-\pi/6}^{\pi/6} f(\theta)\,d\theta = \int_{-\pi/6}^{0} f(\theta)\,d\theta + \int_{0}^{\pi/6} f(\theta)\,d\theta = \int_{0}^{\pi/6} f(\theta)\,d\theta + \int_{0}^{\pi/6} f(\theta)\,d\theta [/tex]In the last expression, I just replaced the integral from -pi/6 to 0 with an integral from 0 to pi/6, since they are the same. Obviously the last sum can then by written as[tex] = 2\int_0^{\pi/6} f(\theta)\,d\theta[/tex]
  4. Mar 15, 2012 #3
    thank you, this makes a lot more sense.
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