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Math Help - stocks question..

  1. #1
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    stocks question..

    calculate
    <br />
\iint_{M}^{}rot(\vec{F})\vec{dS}<br />
    where
    <br />
\vec{F}=(y^2z,zx,x^2z^2) <br />
    M is a part of z=x^2+y^2 which is located in 1=x^2+y^2
    and its normal vector points outside

    i am used to solve it like this
    <br />
\iint_{M}^{}rot\vec{F}\vec{dS}=\iint_{D}\frac{rot\  vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy<br />
    <br />
\vec{N}=(2x,2y,-1)<br />
    <br />
\iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy<br />
    now i convert into polar coordinates

    x^2+y^2=r
    is this method ok?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    calculate
    <br />
\iint_{M}^{}rot(\vec{F})\vec{dS}<br />
    where
    <br />
\vec{F}=(y^2z,zx,x^2z^2) <br />
    M is a part of z=x^2+y^2 which is located in 1=x^2+y^2
    and its normal vector points outside

    i am used to solve it like this
    <br />
\iint_{M}^{}rot\vec{F}\vec{dS}=\iint_{D}\frac{rot\ vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy<br />
    <br />
\vec{N}=(2x,2y,-1)<br />
    <br />
\iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy<br />
    now i convert into polar coordinates

    x^2+y^2=r
    is this method ok?
    Stocks question or Stokes question?
    Last edited by mr fantastic; September 18th 2009 at 09:15 AM. Reason: Restored original reply
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  3. #3
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    i dont know if it stocks or stokes
    in general it about integration threw some closed circle etc..

    is my method correct
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  4. #4
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    the formal solution is differs so muuch
    http://i30.tinypic.com/14cede9.gif

    is my method ok?
    i didnt solve it with parameters
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  5. #5
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    why are they switching to parameters?
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    why are they switching to parameters?
    Switching to parameters because it has the following advantages:

    (1) The problem may require you to use the Stoke's Theorem, as you have indicated in the title "stocks question";

    (2) Use the Stoke's Theorem reduces the problem to 1D;

    (3) Solving the problem in 2D is very complicated, sometimes even impossible.
    Last edited by mr fantastic; September 18th 2009 at 09:16 AM. Reason: Restored original reply
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  7. #7
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    look at my final integral
    its not hard at all

    what so hard integrating by x
    and then by y

    ??
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    the formal solution is differs so muuch
    http://i30.tinypic.com/14cede9.gif

    is my method ok?
    i didnt solve it with parameters
    You are trying to solve the problem in 2D. That's ok as long as you get everything right and you can work it out. Did you check if you computed the curl  \vec{F} right?

    curl \vec{F} = \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right) e_1 - \left(\frac{\partial F_3}{\partial x}- \frac{\partial F_1}{\partial z}\right) e_2 + \left(\frac{\partial F_2}{\partial x}- \frac{\partial F_1}{\partial y}\right) e_3.
    Last edited by mr fantastic; September 18th 2009 at 09:17 AM. Reason: Restored original reply
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  9. #9
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    i would like to solve it by parameters to
    but if i will put trigonometric variables in my final integral

    its a double integral how it could become a single integral
    even if i substitute dx and dy and the rest
    its still remains a double integral
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    i would like to solve it by parameters to
    but if i will put trigonometric variables in my final integral

    its a double integral how it could become a single integral
    even if i substitute dx and dy and the rest
    its still remains a double integral
    The Stokes Theorem reduces the problem from an integral over an surface ( x^2+y^2= z, which is a surface) to an integral along a curve ( x^2+y^2= 1, which is a curve).
    Last edited by mr fantastic; September 18th 2009 at 09:18 AM. Reason: Restored original reply
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  11. #11
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    Quote Originally Posted by transgalactic View Post
    calculate
    <br />
\iint_{M}^{}rot(\vec{F})\vec{dS}<br />
    where
    <br />
\vec{F}=(y^2z,zx,x^2z^2) <br />
    M is a part of z=x^2+y^2 which is located in 1=x^2+y^2
    and its normal vector points outside

    i am used to solve it like this
    <br />
\iint_{M}^{}curl\vec{F}\vec{dS}=\iint_{D}\frac{rot \vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy<br />
    <br />
\vec{N}=(2x,2y,-1)<br />
    <br />
\iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy<br />
    now i convert into polar coordinates

    x^2+y^2=r
    is this method ok?
    OK. Let's try direct surface integration step by step, instead of using the Stoke's theorem and see how more complicated it is.

    \vec{F}=(y^2z, xz, x^2y^2)
    (Note: the image you provided indicate the third component is x^2y^2, rather than x^2z^2. Anyway, it does not influence the final answer)

    rot \vec{F} = (2x^2y-x)e_1 + (y^2-2xy^2)e_2 + (z-2yz)e_3=
    (2x^2y-x)e_1+ (y^2-2xy^2)e_2 + ( x^2 + y^2-2x^2y-2y^3)e_3

    \vec{dS} = (2xe_1 + 2ye_2 -1e_3) dxdy

    curl \vec{F} \cdot \vec{dS} = -3x^2 - y^2 +2x^2y+4y^3+4x^3y-4xy^3

    \iint_{M}^{}curl\vec{F} \cdot \vec{dS}=\iint_{x^2+y^2\leq 1} (
    -3x^2 - y^2 +2x^2y+4y^3+4x^3y-4xy^3)\, dxdy

    Taking advantage of the center symmetry,
    \iint_{M}^{}curl\vec{F} \cdot \vec{dS}=\iint_{x^2+y^2\leq 1} ( -3x^2 - y^2)\, dxdy

    Let x=r\cos\theta, y=r\sin\theta
    \iint_{M}^{}curl\vec{F} \cdot \vec{dS}= -\int_{0}^{2\pi}\int_{0}^{1} (3\cos^2\theta+\sin^2\theta)\,r^3 \, dr d\theta =-\frac{1}{4}\int_{0}^{2\pi} (3\cos^2\theta+\sin^2\theta)\, d\theta=-\frac{1}{4}\int_{0}^{2\pi} (2+\cos 2\theta)\, d\theta=-\pi
    Last edited by mr fantastic; September 18th 2009 at 09:19 AM. Reason: Restored original reply
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