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Math Help - Surface Integrals help

  1. #1
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    Surface Integrals help

    Hi, I'm having a little trouble to solve these problems. May anyone help me please?

    What is the area of the spherical surface X^2+Y^2+Z^2=9 inside the cylinder
    X^2+Y^2=3x ?

    What is the area of the spherical surface X^2+Y^2+Z^2=9 inside the cylindrical X^2+Y^2=4 ?

    What is the area of the spherical surface X^2+Y^2+Z^2=16 inside the cylindrical Y^2+Z^2=4Z ?

    Thanks, and sorry for my English because it is not my native language
    Last edited by nathanfsu; November 29th 2008 at 09:06 PM.
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  2. #2
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    anyone?
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  3. #3
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    When you draw it, the approach becomes immediately clear right? Just integrate under that section of intersection:

    S=2\mathop\int\int\limits_{\hspace{-15pt}D} \sqrt{(f_x)^2+(f_y)^2+1}\;dA

    where:

    f(x,y)=\sqrt{9-y^2-x^2}

    D=\left\{(x,y): (x-3/2)^2+y^2=(3/2)^2\right\}

    Just complete the square in  x^2+y^2=3x to get D.

    I haven't actually tried to do the integration though. May need to use spherical or other means to evaluate it. Also, it's easy to draw. Here's the Mathematica code:

    Code:
    pic1 = ContourPlot3D[{x^2 + y^2 + z^2 == 9, 
        x^2 + y^2 == 3*x}, {x, -3, 3}, 
       {y, -3, 3}, {z, -3, 3}, ContourStyle -> 
        {Opacity[0.4], LightPurple}]
    Attached Thumbnails Attached Thumbnails Surface Integrals help-ball-cylinder.jpg  
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  4. #4
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    hm my problem is that I can't solve the integration, I tried polar and spherical coordinates, but I think I'm not using the right integration limit

    Any Help
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  5. #5
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    So what's the integral you get? We can just integrate over the first quadrant right since it's symmetric and just multiply by 2 and that's just the top so multiply by 2 again but for now, just the top over the first quadrant I get:

    S_{uq1}=3\int_0^3\int_{0}^{\sqrt{3x-x^2}}\frac{1}{\sqrt{9-(x^2+y^2)}}dydx

    where we'll set S_{uq1} be the surface area over the top part just over the first quadrant.

    The x^2+y^2 part looks amendable to polar coordinates. Mathematica reports for that integral S_{uq1}=9/2(\pi-2). Don't see immediately how to do the integration though.
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  6. #6
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    Yeah, I got the same integration. The problem is that after I tried to put in polar coordinates, the integration that came is very difficult to solve without cas
    X=1,5 + R.Cos(A)
    Y=R.Sen(A)
    0<=R<=1.5
    0<=A<=pi

    it would be:

    2*intg*intg R/(Sqrt(-Rcos(A)-R^2+6.75))drda
    I stop here; don't know how to solve, or if this is right
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  7. #7
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    Hey, this ain't no hill:

    S_{uq1}=3\int_0^3\int_{0}^{\sqrt{3x-x^2}}\frac{1}{\sqrt{9-(x^2+y^2)}}dydx

    You can do the first part right:

    \int_0^{\sqrt{3x-x^2}} \frac{dy}{\sqrt{(9-x^2)-y^2}}

    That's just an inverse sine but I ran into problems with that so I converted it to an inverse tan. When I simplify it and apply the limits, I get for the outer integral::

    \int_0^3 \arctan\left(\sqrt{\frac{3}{x}}\right) dx

    Now, I think we can integrate that by parts. You'll need to check all this cus' I went through it quick but I think the principle is sound.



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  8. #8
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    Thanks man
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Shouldn't the outer integral be  \int_0^3 \arctan\left(\sqrt{\frac{x}{3}}\right) dx ?
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  10. #10
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    Quote Originally Posted by chiph588@ View Post
    Shouldn't the outer integral be  \int_0^3 \arctan\left(\sqrt{\frac{x}{3}}\right) dx ?
    I'm not sure; I didn't go through it rigorously . . . Perhaps Nathan could confirm this.
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  11. #11
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    I did on my HP that integral, and the result was
    -ATAN(SQRT(3X)/3)

    How do you do to post math symbols ?
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