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Math Help - Surface area of a sphere

  1. #1
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    Surface area of a sphere

    Find the surface ara of the sphere x^2 + y^2 + z^2=4z that lies inside the paraboloid z=x^2+y^2, using cylindrical co-ords.

    What I've done:
    equations of the sphere and paraboloid in cylindrical co-ordinates are r^2+z^2=4z and z=r^2.
    substituting and rearranging these gives
    z+z^2=4z\; \; \Rightarrow z^2-3z=0\; \; \Rightarrow z(z-3)=0\; \; \Rightarrow z=(0,3)

    r^2+r^4=4r^2\; \; \Rightarrow r^4-3r^2=0\; \; \Rightarrow r^2(r^2-3)=0\; \; \Rightarrow r=(0,\sqrt{3})

    this gives the points of intersections (r,z)=(0,0) \; and\; (\sqrt{3},3)


    And that's as far as I can get. I'm having a problem with the partial derivatives.

    To get them I've done this:
    x^2 + y^2 + z^2=4z\; \; \Rightarrow z^2=4z-x^2 - y^2\; \; \Rightarrow z=\sqrt{4z-x^2-y^2}

    When I differentiate, am I right in assuming the 'z' on the RHS disappears as well? Does it go to 1 or 0?
    ie.
    \frac{\partial z}{\partial x}=\frac{-x}{\sqrt{4-x^2-y^2}}
    or:
    \frac{\partial z}{\partial x}=\frac{-x}{\sqrt{-x^2-y^2}}
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  2. #2
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    Nov 2009
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    Hint: Use implicit differentiation with respect to x.

    What you did is not correct.
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