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Math Help - Surface Area Question

  1. #1
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    Surface Area Question

    Find the surface area of the cone z=2sqrt(x^2+y^2) and above a region in the xy-plane with area 5.

    If anyone could help me with this problem, I would really appreciate it.
    Thanks.
    John Marcum
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  2. #2
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    Quote Originally Posted by OntarioStud
    Find the surface area of the cone z=2sqrt(x^2+y^2) and above a region in the xy-plane with area 5.
    If anyone could help me with this problem, I would really appreciate it.
    Thanks.
    John Marcum
    Hello, John,

    as you may know the term \sqrt{x^2+y^2} describes the distance between the origin and a point in the xy-plane.

    If you draw a plane cut through the cone along the z-axis it'll look like the diagram which I attached.

    With your cone you get:  r = \sqrt{x^2+y^2}.
    The slope s is s=\sqrt{r^2+4r^2}=r\cdot \sqrt{5}.

    The surface area of a cone contains the base circle and the area, which is produced by the sloping line:

    A_S=\pi r^2+\pi r s

    A_{surface}=\pi r^2+ \pi \cdot r \cdot s = \pi(x^2+y^2)+\pi\sqrt{x^2+y^2}\cdot \sqrt{5(x^2+y^2}
    Notice: I've corrected my mistake. I'm sorry if the previous result caused any irritation.
    Simplify to: (1+\sqrt{5}) \cdot \pi \cdot (x^2+y^2)=(1+\sqrt{5}) \cdot \pi \cdot r^2

    This surface has the value 5. So you get: 5=(1+\sqrt{5}) \cdot \pi \cdot r^2. Solve for r and re-substitute. You'll get the equation:
    \sqrt{x^2+y^2}=\sqrt{\frac{5(\sqrt{5}-1)}{4\cdot \pi}

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails Surface Area Question-kegel2310706.gif   Surface Area Question-kegel310706.gif  
    Last edited by earboth; July 31st 2006 at 09:47 PM.
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  3. #3
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    earboth which software you use for functions of two variables?
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    earboth which software you use for functions of two variables?
    Hello, TPH,

    Derive 6

    Greetings

    EB
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