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Thread: Surface rotation

  1. #1
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    Surface rotation

    A surface S is obtained by rotation, around the axis OZ, the arc of parabola $\displaystyle \alpha : z = \frac{x^2}{2}, y = 0$ with $\displaystyle 0 \leq x \leq 1$. Determine the area of the surface obtained
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    How do I calculate surface rotation? Must use the matrix?
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    Quote Originally Posted by Apprentice123 View Post
    How do I calculate surface rotation? Must use the matrix?

    It's a standard formula in Calc II $\displaystyle
    SA = 2 \pi \int_a^b y \sqrt{1+y'^2}\,dx
    $
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  4. #4
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    Quote Originally Posted by Danny View Post
    It's a standard formula in Calc II $\displaystyle
    SA = 2 \pi \int_a^b y \sqrt{1+y'^2}\,dx
    $
    I do not know this formula.
    Could you explain better?
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  5. #5
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    Quote Originally Posted by Apprentice123 View Post
    I do not know this formula.





    Could you explain better?
    Are you familar with the formula's for volumes of revolution?

    I need a starting place for our discussion.
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  6. #6
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    Or you can recognize that the rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$ which can be parametrized as $\displaystyle \sigma(u,v)=\left(u,v, \frac{u^2+v^2}{2}\right)$

    A switch to polar coordinates will then basically produce the same integral as Danny showed you.

    $\displaystyle \iint_D \left| \vec {\sigma'_u}\times\vec {\sigma'_v}\right|dudv=\{\text{Polar coordinates}\}=\int \int_0^{2\pi}\sqrt{r^2+1} \cdot rd\varphi dr=$ $\displaystyle 2\pi\int r\sqrt{r^2+1}dr$
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  7. #7
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    Quote Originally Posted by Spec View Post
    Or you can recognize that the rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$ which can be parametrized as $\displaystyle \sigma(u,v)=\left(u,v, \frac{u^2+v^2}{2}\right)$

    A switch to polar coordinates will then basically produce the same integral as Danny showed you.

    $\displaystyle \iint_D \left| \vec {\sigma'_u}\times\vec {\sigma'_v}\right|dudv=\{\text{Polar coordinates}\}=\int \int_0^{2\pi}\sqrt{r^2+1} \cdot rd\varphi dr=$ $\displaystyle 2\pi\int r\sqrt{r^2+1}dr$
    Ok. Thanks
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  8. #8
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    Quote Originally Posted by Spec View Post
    Or you can recognize that the rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$ which can be parametrized as $\displaystyle \sigma(u,v)=\left(u,v, \frac{u^2+v^2}{2}\right)$

    A switch to polar coordinates will then basically produce the same integral as Danny showed you.

    $\displaystyle \iint_D \left| \vec {\sigma'_u}\times\vec {\sigma'_v}\right|dudv=\{\text{Polar coordinates}\}=\int \int_0^{2\pi}\sqrt{r^2+1} \cdot rd\varphi dr=$ $\displaystyle 2\pi\int r\sqrt{r^2+1}dr$
    as you know that rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$
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