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Thread: Surface area integral

  1. #1
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    Surface area integral

    If you revolve the function $\displaystyle y=x^3$, $\displaystyle 1\leq x\leq 2$ around the x-axis how do you calculate the surface area by using the parametrization
    $\displaystyle x=t$
    $\displaystyle y=r(t)\cos\theta$
    $\displaystyle z=r(t)\sin\theta$
    for a suitable $\displaystyle r(t)$
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  2. #2
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    Quote Originally Posted by qwesl View Post
    If you revolve the function $\displaystyle y=x^3$, $\displaystyle 1\leq x\leq 2$ around the x-axis how do you calculate the surface area by using the parametrization
    $\displaystyle x=t$
    $\displaystyle y=r(t)\cos\theta$
    $\displaystyle z=r(t)\sin\theta$
    for a suitable $\displaystyle r(t)$
    When you have a surface defined in terms of two parameters, x= f(u,v), y= g(u,v), z= h(u,v), you can write each point as the vector equation, $\displaystyle \vec{r}(u,v)= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}$. The two partial derivatives, with respect to u and v are $\displaystyle \vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}$ and $\displaystyle \vec{t}= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}$. The "fundamental vector product" for the surface is $\displaystyle \vec{r}_u\times\vec{r}_v$.

    Finally, the "differential of surface area is the magnitude of that product time drdt: $\displaystyle \left|\vec{r}_u\times\vec{r}_v\right|dudv$

    Here, since this is rotated around the x-axis, x= t, as you say, and so r(t), perpendicular to the x, axis, is $\displaystyle r(t)= t^3$. $\displaystyle x= t$, $\displaystyle y= t^3cos(\theta)$, and $\displaystyle z= t^3sin(\theta)$. $\displaystyle \vec{r}(t,\theta)= t\vec{i}+ t^3cos(\theta)\vec{j}+ t^3 sin(\theta)\vec{k}$.
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