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Math Help - Surface Area Question

  1. #1
    Newbie jojoyce's Avatar
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    Question Surface Area Question

    Find the surface area of the cone z=4sqrt(x^2+y^2) and above a region in the xy-plane with area 3.
    I tried this problem and here's what i've done------
    r=sqrt(x^2+y^2)
    The slope s is s=sqrt(r^2+16r^2)=sqrt(17)r

    A surface=pir^2+pi r s=pi r^2(1+sqrt17)

    3=
    pi r^2(1+sqrt17) ===> r=0.4317

    this is as far as i went, but what do I do next to get the surface area?

    Thanks in advance!!!
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  2. #2
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    Re: Surface Area Question

    Have you learnt how to use cylindrical coordinates?
    If so consider using
    x=rcos\theta
    y=rsin\theta
    z=z

    dA= dxdy= |J|drd\theta
    where |J| is the determinant of the Jacobian

    integrate dA between r=0 and r=0.4317 and between \theta=0 and \theta=2\pi
    Last edited by Shakarri; March 19th 2013 at 12:26 PM.
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  3. #3
    Newbie jojoyce's Avatar
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    Re: Surface Area Question

    Thanks! but I still don't get it is the determinant equals r? or 4r? I tried both and neither worked
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  4. #4
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    Re: Surface Area Question

    The jacobian is
    \begin{bmatrix}\frac{dx}{dr}&\frac{dx}{d\theta}\\ \frac{dy}{dr}&\frac{dy}{d\theta}\end{bmatrix}

    Which becomes
    \begin{bmatrix}cos\theta&-rsin\theta\\sin\theta&rcos\theta\end{bmatrix}

    The determinant is r, perhaps you are making a mistake elsewhere.
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