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Math Help - double integral, mass using polar co-ordinates

  1. #1
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    double integral, mass using polar co-ordinates

    Can someone explain the solution for me very specifically and in simple terms, because im trying to understand the solution but just don't get it.
    i used cartesian- polar transforming but still dont get it, i have a test on monday too! help would be really nice
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    Quote Originally Posted by xibeleli View Post
    Can someone explain the solution for me very specifically and in simple terms, because im trying to understand the solution but just don't get it.
    i used cartesian- polar transforming but still dont get it, i have a test on monday too! help would be really nice
    Let's see: \iint\limits_R \delta(x,y)\,dA is just the definition of mass wrt a density function over a given region.

    Polar coordinates: x = r\cos \theta\,,\,\,y=r\sin \theta \Longrightarrow x^2+y^2=r^2 , and since the Jacobian of the change from cartesian to polar coordinates is r then we get that the above original double integral turns into

    \int\limits_0^\pi\int\limits_0^1 r^3\,drd\theta . As you can see, the limits for \theta\,\,are\,\,0\,\,and\,\,\pi , to cover the half upper unit circle, and r goes from 0 to 1 to cover all the interior of this half circle.

    The rest is just simple integration.

    Tonio
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    Mass:

    M=\int_{0}^{\pi}\int_{0}^{1}r^{3}drd{\theta}

    \overline{x}=0 because of the symmetry of the region.

    If you need it, the center of gravity:

    M_{x}=\int_{0}^{\pi}\int_{0}^{1}r^{4}sin{\theta}dr  d{\theta}

    \overline{y}=\frac{8}{5\pi}
    Last edited by galactus; October 24th 2009 at 05:07 AM. Reason: Oops, Tonio beat me. Well, we agree. so I reckon that's right :):)
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    Wow guys fast responses but i still don't get it, still puzzled on how they worked out the integral limits (pi integral 0) and (1 integral 0) can you please show the working out for how 1 of the limits is calculated using the formula as an example
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    You know the radius is 1. Look at the graph.

    Besides, y=\sqrt{1-x^{2}}

    Polar:

    rsin{\theta}=\sqrt{1-r^{2}cos^{2}{\theta}}

    r^{2}sin^{2}{\theta}=1-r^{2}cos^{2}{\theta}

    r^{2}sin^{2}{\theta}+r^{2}cos^{2}{\theta}=1

    r^{2}(\underbrace{sin^{2}{\theta}+cos^{2}{\theta}}  _{\text{this is 1}})=1

    r=\pm 1

    The upper half of the circle is integrated over 0 to Pi.
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    Quote Originally Posted by xibeleli View Post
    Wow guys fast responses but i still don't get it, still puzzled on how they worked out the integral limits (pi integral 0) and (1 integral 0) can you please show the working out for how 1 of the limits is calculated using the formula as an example

    To cover the half upper unit circle you have to "let" \theta to run from 0 to \pi, and the size of the radiuses must cover from 0 (in the origin (0,0)) to 1 (at the circle's perimeter).
    This way one gets to "sweep" the hole area between the half circle and the x-axis and that's why the coordinates belong to these ranges.

    Tonio
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