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Math Help - Surface Integrals

  1. #1
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    Surface Integrals



    I have to evaluate the given surface integrals \iint s {z}^{2 } d\sigma where S is the portion of the cone z = \sqrt{{x}^{2 } +{y}^{2 } } between the planes z = 1 and z = 2

    I dont know where to start
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  2. #2
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    Quote Originally Posted by adam_leeds View Post


    I have to evaluate the given surface integrals \iint s {z}^{2 } d\sigma where S is the portion of the cone z = \sqrt{{x}^{2 } +{y}^{2 } } between the planes z = 1 and z = 2

    I dont know where to start
    The top and bottom are circles or radius 2 and 1 respectivily and can be parametrized using cylindrical polar coordinates.

    Note that on the side the azimuthal angle is fixed at

    \theta=\frac{\pi}{4}

    So the surface can be parameterized as follows

    x=r\sin\left(\frac{\pi}{4} \right)\cos(\phi) \quad y=r\sin\left(\frac{\pi}{4} \right)\sin(\phi) \quad z=r\cos\left(\frac{\pi}{4} \right)

    So as a vector we get

    \mathbf{r}(r,\phi)=[r\sin\left(\frac{\pi}{4} \right)\cos(\phi)]\mathbf{i}+[r\sin\left(\frac{\pi}{4} \right)\sin(\phi)]\mathbf{j}+[r\cos\left(\frac{\pi}{4} \right)]\mathbf{k}

    Can you finish from here?
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  3. #3
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    I'll have a go, whats the formula for getting theta?

    Is the next step F.r' ?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    You can use the standard method


    \displaystyle\iint_S F(x,y,z)dS=\displaystyle\iint_DF(x,y,f(x,y))\sqrt{  1+\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}}\;dxdy

    In this case

    F(x,y,z)=z^2,\;z=f(x,y)=\sqrt{x^2+y^2},\;D\equiv 1\leq x^2+y^2 \leq 4


    Edited: Sorry, I didn't see the TheEmptySet's post.
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  5. #5
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    Quote Originally Posted by adam_leeds View Post
    I'll have a go, whats the formula for getting theta?

    Is the next step F.r' ?
    You have the cone

    z=\sqrt{x^2+y^2}=r

    From spherical coordinates we know that

    z=r\cos(\theta) \iff \theta =\cos^{-1}\left( \frac{z}{r}\right)

    but from the equation of the cone we know that

    z=r

    So we get

    \theta =\cos^{-1}\left( \frac{r}{r}\right) =\cos^{-1}\left( 1\right)=\frac{\pi}{4}

    If we have the vector form of a surface its surface area is given by

    \iint \bigg|\frac{\partial \mathbf{r}}{\partial r} \times  \frac{\partial \mathbf{r}}{\partial \theta} \bigg| dr d\theta
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  6. #6
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    Quote Originally Posted by FernandoRevilla View Post
    You can use the standard method


    \displaystyle\iint_S F(x,y,z)dS=\displaystyle\iint_DF(x,y,f(x,y))\sqrt{  1+\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}}\;dxdy

    In this case

    F(x,y,z)=z^2,\;z=f(x,y)=\sqrt{x^2+y^2},\;D\equiv 1\leq x^2+y^2 \leq 4


    Edited: Sorry, I didn't see the TheEmptySet's post.
    This method seems more familiar to me,

    F(x,y,f(x,y)) = {x}^{ 2} + {y}^{ 2} , is that correct?
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  7. #7
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    Yes
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