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Math Help - Double integrals with polar co-ords

  1. #1
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    Double integrals with polar co-ords

    hello,

    I am having trouble deconstructing this problem given to me in class.

    i have been asked to sketch the region and evaluate the double integral of

    g(r,theta)=1, with the region bounded by r=1-cos(theta).

    I dont even really know what the region looks like so i cant begin to set bounds for my double integral.

    Any help would be greatly appreciated.
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  2. #2
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    Try converting your region back to Cartesians just so you can get a picture of what the region looks like...
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by olski1 View Post
    hello,

    I am having trouble deconstructing this problem given to me in class.

    i have been asked to sketch the region and evaluate the double integral of

    g(r,theta)=1, with the region bounded by r=1-cos(theta).

    I dont even really know what the region looks like so i cant begin to set bounds for my double integral.

    Any help would be greatly appreciated.
    The double integral can be written as...

    \displaystyle \int_{-\pi}^{\pi} d \theta\int_{0}^{1-\cos \theta} r\ d r

    ... so that it consists in two sucessive integrations...

    Kind regards

    \chi \sigma
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  4. #4
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    Okay, so i drew the cardoid described in my question, found my bounds for the inner integral to be from 0 to 1-cos(theta)

    but with my outter integral, why is it -pi to pi, does the cardoid not cover a region from 0 to 2pi?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by olski1 View Post
    Okay, so i drew the cardoid described in my question, found my bounds for the inner integral to be from 0 to 1-cos(theta)

    but with my outter integral, why is it -pi to pi, does the cardoid not cover a region from 0 to 2pi?
    If a function f(\theta) is periodic of period 2 \pi the integral \displaystyle \int_{\alpha}^{\alpha + 2 \pi} f(\theta)\ d \theta is independent from \alpha...

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by chisigma View Post
    If a function f(\theta) is periodic of period 2 \pi the integral \displaystyle \int_{\alpha}^{\alpha + 2 \pi} f(\theta)\ d \theta is independent from \alpha...

    Kind regards

    \chi \sigma
    I am not quite sure if i understand what you mean.
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  7. #7
    Member Pranas's Avatar
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    Quote Originally Posted by olski1 View Post
    I am not quite sure if i understand what you mean.
    Funny, it's exactly the same thing I am trying to prove at this very moment http://www.mathhelpforum.com/math-he...on-177414.html

    Don't know if it's enough of a clarification to you though
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