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Math Help - parametrize surface of a sphere contained within a cylinder

  1. #1
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    parametrize surface of a sphere contained within a cylinder

    I'm looking to parametrize the upper surface of a portion of the sphere x^2 + y^2 + z^2 = 16 contained within the cylinder x^2+y^2 = 4y

    since its a sphere, i've parametrize into spherical coordinates with

    x(u,v) = (4sin(u)cos(v), 4sin(u)cos(v),4cos(u))
    with v={0,2pi}

    however, I'm stuck on finding the bounds for u. specifically with the cylinder being x^2+(y+1)^2 = 1, i'm unsure how to set u. i have r = 4sinv, but that doesn't seem to help. Maybe i'm going about the question wrong.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    An alternative: from the parametric equations of the cylinder,


    C \equiv\begin{Bmatrix}x=2\cos t\\y=2+2\sin t\\z=\lambda\end{matrix} \quad (t\in[0,2\pi],\;\lambda\in\mathbb{R})

    and substituting in the equation of the sphere we obtain

    \lambda^2=8(1-\sin t)


    Fernando Revilla
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    thanks for you help. I forgot to mention that i'd like to find the area of the surface also.
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by vicariage View Post
    thanks for you help. I forgot to mention that i'd like to find the area of the surface also.

    Of the cylinder? . Of the sphere? . Limited by ...


    Fernando Revilla
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    Quote Originally Posted by FernandoRevilla View Post
    Of the cylinder? . Of the sphere? . Limited by ...


    Fernando Revilla
    of the upper portion of the sphere contained within the cylinder
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by vicariage View Post
    of the upper portion of the sphere contained within the cylinder
    Then,

    S=\displaystyle\iint_{D}\sqrt{1+\left(\frac{{\part  ial z}}{{\partial x}}\right)^2+\left(\frac{{\partial z}}{{\partial y}}\right)^2}\;dxdy\;,\quad z=\sqrt{16-x^2-y^2}

    where:

    D \equiv x^2+y^2-4y \leq 0


    Fernando Revilla
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    how would i solve the integral.
    I've simplified the integral to 4(int(1/(16-x^2-y^2)) dxdy

    with a D of x^2+y^2-4y I can do a change of variable and end up with
    2u = x and 2v-2 but i'll end up with a very messy integral but ranges of [0,2pi]x[0,1]

    or i can substitute x=rcos*u and y = rsin*u and end up with ranges of [0,2pi]x[0,4sin*u] and an integrand of 1/(16-r^2). and if i continue on, i'll end up arcsin(r/4) and thus a complicated integral.

    is there a way i can paramatize it into spherical coordinates and find ranges for phi? because that would probably make this question more doable...
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