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Math Help - Parameterization of a volume of sphere.

  1. #1
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    Unhappy Parameterization of a volume of sphere.

    Hi all,


    I would like some help with a parameterization. The parameterization is as follows:


    S is the surface of a sphere with center at (6, −2, 4) and radius 6, V is the volume bounded by S.

    parameterize the volume V, using the parameters (r, θ, φ).

    How do I do this problem?

    I don't even know where to begin.

    Any help much appreciated.
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  2. #2
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    Re: Parameterization of a volume of sphere.

    Hey mcleja.

    Basically you need to go from cartesian to polar co-ordinates in three dimensions. Take a look at this wiki entry:

    List of common coordinate transformations - Wikipedia, the free encyclopedia
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  3. #3
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    Re: Parameterization of a volume of sphere.

    Woud I use the formula for the sphere x^2+y^2+z^2=sqrt(3). Then use the conversion formulas to get (sqrt(3)sin(φ)cos(θ),sqrt(3)sin(θ)sin(φ),sqrt(3)co s(φ))?
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  4. #4
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    Re: Parameterization of a volume of sphere.

    Here are the points of the sphere: \{(x,y,z) \in \mathbb{R}^3 \mid (x-6)^2 + (y+2)^2 + (z-4)^2 \le 36\}.

    x = r\sin \theta \cos \varphi
    y = r\sin \theta \sin \varphi
    z = r\cos \theta

    So, the points of the set are \{(r,\theta,\varphi) \mid (r\sin\theta \cos \varphi - 6)^2 + (r\sin\theta \sin \varphi +2)^2 + (r\cos\theta -4)^2 \le 36\}. Is that how you want it parametrized? I don't understand what you are looking for.
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  5. #5
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    Re: Parameterization of a volume of sphere.

    Now you need to get your limits in terms of the polar co-ordinates.

    The limits for a sphere are easy in polar space since r is from 0 to 6 and the angles are just exhaustive (from 0 to 2pi and 0 to pi if I recall correctly).

    Since all the limits are separate (i.e. they all are independent and don't depend on other variables) the integration is a lot easier which is why you do this in polar space and not cartesian space.
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