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Math Help - Double integral set up #2

  1. #1
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    Double integral set up #2

    so I am given

    \int_{0}^{2}\int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx

    Could I do this as

    2\int_{0}^{2}\int_{0}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx

    I remember doing some integrals in calculus 2 where you could do that. I think they were stuff dealing with the polar coordinates though
    Last edited by Jonroberts74; July 22nd 2014 at 06:09 PM.
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  2. #2
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    Re: Double integral set up #2

    Quote Originally Posted by Jonroberts74 View Post
    so I am given

    \int_{0}^{2}\int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx

    Could I do this as

    2\int_{0}^{2}\int_{0}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx

    I remember doing some integrals in calculus 2 where you could do that. I think they were stuff dealing with the polar coordinates though
    You need to do more than just guess.

    $\displaystyle{\int_0^2 \int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2}}\left(\dfrac{5}{\sqrt{2+x}}+y^3\right)~dy~dx=$

    $\displaystyle{\int_0^2}\dfrac{5}{\sqrt{2+x}} \left( \displaystyle{\int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2}}y^3~dy\right) ~dx$

    Now the integral over $dy$ is that of an odd function over limits symmetric about zero.

    What does that tell you about it's value?
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    Re: Double integral set up #2

    it goes to zero when an odd function is integrated over the symmetric about zero
    Last edited by Jonroberts74; July 22nd 2014 at 07:19 PM.
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  4. #4
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    Re: Double integral set up #2

    Correct, so what is the value of the double integral?
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    Re: Double integral set up #2

    after integrating for y

    \int_{0}^{2} 15\sqrt{2-x} dx = -10(2-x)^{3/2} \Bigg|_{0}^{2} = 20\sqrt{2}
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    Re: Double integral set up #2

    Quote Originally Posted by Jonroberts74 View Post
    after integrating for y

    \int_{0}^{2} 15\sqrt{2-x} dx = -10(2-x)^{3/2} \Bigg|_{0}^{2} = 20\sqrt{2}
    correct.
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