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Math Help - Surface area of a disc

  1. #1
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    Surface area of a disc

    I have a surface defined as  z(x^2+y^2)=4. I have calculated the surface integral but now I just want to calculate the contribution from the disc z=4. How do i do this?
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  2. #2
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    Re: Surface area of a disc

    Pretty straight forward, isn't it? If z= 4, your equation becomes 4(x^2+ y^2)= 4 so x^2+ y^2= 1. That's a circle with radius 1.
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  3. #3
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    Re: Surface area of a disc

    I thought so but the answer is 15pi, not pi
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  4. #4
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    Re: Surface area of a disc

    The answer to what question?
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  5. #5
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    Re: Surface area of a disc

    The surface integral was 33pi. In part (b), I had to calculate volume integral where V is the region defined by  0<x^2+y^2<4/z and 1<z<4. This was 48pi.

    Now I have to verify divegence theorem. Well, I need to show 33pi + contribution from z=4 + contribution from z=1 = 48pi.

    Sorry if I was not clear enough.
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  6. #6
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    Re: Surface area of a disc

    In cylindrical coordinates, x^2+ y^2= \frac{4}{z} becomes z= \frac{4}{r^2}. Of course, in order to have a closed figure, we must have the "top" and "bottom", z= 4, and z= 1.

    The divergence theorem says that \int\oint\vec{F}d\vec{S}= \int\int\int \nabla\cdot\vec{F} dV.

    If you want to find the volume of the figure, \int\int\int dV, using that, you must have a vector function, \vec{F}, such that \nabla\cdot\vec{F}= 1. What vector function did you use?

    Of course, the top of the figure, z= 4, is the disk x^2+y^2\le 1 while the bottom of the figure, z= 1, is the disk x^2+ y^2= 4 which means that, to find the volume directly, you must integrate the constant 1 from z= 0 to z= 4 over the disk x^2+ y^2= 1 which is simply 1 times 4 times the area of that disk, 4\pi. Then, integrate the constant 1 from z= 0 to z= \frac{4}{x^2+ y^2}= \frac{4}{r^2} (in cylindrical coordinates) for r= 1 to 2 (and \theta from 0 to 2\pi, of course).
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