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Math Help - double integral polar

  1. #1
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    double integral polar

    Any help would be appreciated/ This is my first attempt at Latex so please bear with me.
    I have done the following integration

    \int_{0}^{h}\int_{0}^{y} x.dx.dy


    and have come up with h^3/6 which I'm pretty sure about.

    I am now trying to do this by converting to polar form, but am having difficulty finding the limits of integration to use. Any helpers?
    Last edited by Debsta; April 15th 2010 at 06:54 PM.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Debsta View Post
    Any help would be appreciated/ This is my first attempt at Latex so please bear with me.
    I have done the following integration

    \int_{0}^{h}\int_{0}^{y} x.dx.dy


    and have come up with h^3/6 which I'm pretty sure about.

    I am now trying to do this by converting to polar form, but am having difficulty finding the limits of integration to use. Any helpers?
    This integral is bounded by

     x=0 and x=y

    Also by

    y=0 and  y=h

    If you graph the domain you will see we are bounded by the line y=x and the roof h. Of course, the line y=x is associated with \frac{\theta}{4} and this would then be our start theta value. We are then bounded by x=0 or the y-axis.

    \frac{\theta}{4} \le \theta \le \frac{\theta}{2}

    We are not bounded by a cylinder, so we must find an equation of r that relates to theta. This can be done by looking at your domain and realising that

    sin ( \theta ) = \frac{H}{R}

    We then have

    R=Hcsc( \theta )

    Of course

    x=rcos(\theta)

    Why in gods name would you want to do this question in polar I have no idea. But this should be good
    Last edited by AllanCuz; April 15th 2010 at 08:59 PM.
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  3. #3
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    Quote Originally Posted by AllanCuz View Post
    This integral is bounded by

     x=0 and x=y

    Also by

    y=0 and  y=h

    If you graph the domain you will see we are bounded by the line y=x and the roof h. Of course, the line y=x is associated with \frac{\theta}{4} and this would then be our start theta value. We are then bounded by x=0 or the y-axis.

    \frac{\theta}{4} \le \theta \le \frac{\theta}{2}

    We are not bounded by a cylinder, so we must find an equation of r that relates to theta. This can be done by looking at your domain and realising that

    sin ( \theta ) = \frac{H}{R}

    We then have

    R=Hcsc( \theta )

    Of course

    x=rcos(\theta)

    Why in gods name would you want to do this question in polar I have no idea. But this should be good
    Thankyou very much. I see now that theta is between pi/4 and pi/2. I was thinking 0 and pi/4 which was where my mistake was.
    I used 0 < r < h cosec theta for the limits for r which I'm sure is right.
    Thanks again...yes I know it is a ridiculous way to do the problem, but that was what the question asked...just doing what I'm told. It's helped me to learn about the polar approach though so some good has come out of it!
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