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Math Help - Surface area of revolution using a parametrization

  1. #1
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    Surface area of revolution using a parametrization

    Hi, the question is attached below.

    Attempt at solution

    Surface area = 2*pi*r*ds = 2*pi*integral (from 1 to 2) of x^3*sqrt(1+9x^4)dx

    I ended up with 199.4804797 as my answer.

    However, I'm having trouble finding the area of revolution using the parametrization method. What is r(t)?
    Attached Thumbnails Attached Thumbnails Surface area of revolution using a parametrization-untitled.jpg  
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  2. #2
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    Since x= t and y= x^3, r(t)= t^3 so the parametric equations are x= t, y= t^3 cos(\theta), z= t^3 sin(\theta)

    As a vector equation, that is \vec{r}(t,\theta)= t\vec{i}+ t^3 cos(\theta)\vec{j}+ t^3 sin(\theta)\vec{k}.

    The derivatives of that function, with respect to t and \theta are
    \vec{r}_t= \vec{i}+ 3t^2 cos(\theta)\vec{j}+ 3t^2 sin(\theta)\vec{k} and
    \vec{r}_\theta= -t^3 sin(\theta)\vec{j}+ t^3 cos(\theta)\vec{k}

    The cross product of those two derivatives (the "fundamental vector product" for this surface) is
    3t^5\vec{i}- t^3 cos(\theta)\vec{j}- t^3 sin(\theta)\vec{k}

    Its length is \sqrt{9t^4+ t^2}= t^2\sqrt{9+ t^2} so the "differential of surface area" for this surface is
    t^2\sqrt{9+ t^2} dtd\theta

    The surface area is given by
    \int_{\theta= 0}^{2\pi}\int_{t=1}^2 t^2\sqrt{9+ t^2}dtd\theta
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