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Math Help - change of variable

  1. #1
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    change of variable

    Double integral \int \int_{D} (x^3 + xy^2) dx dy where D is the ring domain 1 \leq x^2 +y^2 \leq 2

    My solution:

    x=r cos\theta
    y=r sin \theta

    \int_{0}^{2 \pi} \int_{1}^{2} (r^3 cos^3 \theta +r cos \theta r^2 sin^2 \theta)  r dr d\theta =0<br />
    Last edited by colorful; May 17th 2009 at 03:54 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by colorful View Post
    Double integral \int \int_{D} (x^3 + y^3) dx dy where D is the ring domain 1 \leq x^2 +y^2 \leq 2

    My solution:

    x=r cos\theta
    y=r sin \theta

    \int_{0}^{2 \pi} \int_{1}^{2} (r^3 cos^3 \theta +r cos \theta r^2 sin^2 \theta) r dr d\theta =0<br />
    remember that x^3+y^3=(x+y)(x^2-xy+y^2)

    so you get

    \cos^3(x)+\sin^3(x)=(\cos(x)+\sin(x))(\cos^2(x)-\cos(x)\sin(x)+\sin^2(x))=

    (\cos(x)+\sin(x))(1-\sin(x)\cos(x))

    \int_{}^{}\int_{}^{} r^4[\sin(\theta)+\cos(\theta)-\sin^2(\theta)\cos(\theta)-\sin(\theta)\cos^2(\theta)]d\theta dr

    Good luck
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    remember that x^3+y^3=(x+y)(x^2-xy+y^2)

    Good luck
    OMG, I have typed the wrong integral, the question sould be \int \int_{D} (x^3 + xy^2) dx dy

    is my solution still holds?
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by colorful View Post
    OMG, I have typed the wrong integral, the question sould be \int \int_{D} (x^3 + xy^2) dx dy

    is my solution still holds?
    yes
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