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Math Help - 2 really advanced Calculus questions

  1. #1
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    2 very difficult Calculus questions

    Please, help somehow if you can!
    Attached Thumbnails Attached Thumbnails 2 really advanced Calculus questions-1.jpg   2 really advanced Calculus questions-2.jpeg  
    Last edited by andreas; December 13th 2008 at 12:39 AM.
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  2. #2
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    I did the following(for the first problem):

    I imagined cylinder  x^2+y^2\leq1 , 0\leq Z\leq1

    First I consider the pipe without top and bottom ( x^2+y^2=1 and 0<Z<1 )


    Since F for i, j has components 1 and 1. The integral for pipe will be simply 2\pi

    Now I consider top and bottom discs: integral for bottom disc will be 0 because it's at z=0. The most interesting is top disc ( x^2+y^2\leq 1 and Z=1). For area of top disc I will have the following integral \int z(x^2+y^2)^2 dA but the top z is 1 so integral is \int (x^2+y^2)^2 dA. dA is a little area at the top disc, and I think it is 2\pi rdr where r=\sqrt{x^2+y^2} So finally my integral is \int (r^2)^2 2\pi rdr=2 \pi\int r^5  dr Taking it from 0 to 1 I get \frac{\pi}{3}. So the finall result is: \frac{\pi}{3}+2\pi=\frac{7\pi}{3}

    I have to submit both problems very soon. Can you check if I did first correctly? How should I prove second one?
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  3. #3
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    Quote Originally Posted by andreas View Post
    I did the following(for the first problem):

    I imagined cylinder  x^2+y^2\leq1 , 0\leq Z\leq1

    First I consider the pipe without top and bottom ( x^2+y^2=1 and 0<Z<1 )


    Since F for i, j has components 1 and 1. The integral for pipe will be simply 2\pi Mr F says: I don't think so. See below.

    Now I consider top and bottom discs: integral for bottom disc will be 0 because it's at z=0.The most interesting is top disc ( x^2+y^2\leq 1 and Z=1). For area of top disc I will have the following integral \int z(x^2+y^2)^2 dA but the top z is 1 so integral is \int (x^2+y^2)^2 dA. dA is a little area at the top disc, and I think it is 2\pi rdr where r=\sqrt{x^2+y^2} So finally my integral is \int (r^2)^2 2\pi rdr=2 \pi\int r^5 dr Taking it from 0 to 1 I get \frac{\pi}{3}.

    So the finall result is: \frac{\pi}{3}+2\pi=\frac{7\pi}{3}

    I have to submit both problems very soon. Can you check if I did first correctly? How should I prove second one?
    I don't think the integral of F over the curved surface of the pipe is equal to 2\pi: If you parametrise the cylindrical surface then an infinitesimal element of the curved surface can be written as

    d\vec{S} = (\cos \theta \, i + \sin \theta \, j) (1) d\theta \, dz.

    Then the surface integral of F over the curved surface becomes

    \int_{z = 0}^1 \int_{\theta = 0}^{2 \pi} (\cos \theta + \sin \theta ) \, d \theta \, dz

    since x^2 + y^2 = 1 on the curved surface. I get zero as the answer.

    So I think the integral over the closed surface will actually be \frac{\pi}{3} + 0 = \frac{\pi}{3}. Doing the calculation using the divergence theorem I get the same answer.
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