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Math Help - 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 \alpha : z = \frac{x^2}{2}, y = 0 with 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  <br />
SA = 2 \pi \int_a^b y \sqrt{1+y'^2}\,dx <br />
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    Quote Originally Posted by Danny View Post
    It's a standard formula in Calc II  <br />
SA = 2 \pi \int_a^b y \sqrt{1+y'^2}\,dx <br />
    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 z=\frac{x^2+y^2}{2} which can be parametrized as \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.

    \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= 2\pi\int r\sqrt{r^2+1}dr
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    Quote Originally Posted by Spec View Post
    Or you can recognize that the rotation will produce an elliptical paraboloid z=\frac{x^2+y^2}{2} which can be parametrized as \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.

    \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= 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 z=\frac{x^2+y^2}{2} which can be parametrized as \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.

    \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= 2\pi\int r\sqrt{r^2+1}dr
    as you know that rotation will produce an elliptical paraboloid z=\frac{x^2+y^2}{2}
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