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Math Help - different approach on calculating flux threw area..

  1. #1
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    different approach on calculating flux threw area..

    calculate
    <br />
\iint_{M}^{}\vec{F}\vec{dS}<br />
    where
    <br />
\vec{F}=(e^y,ye^x,x^2y) <br />
    M is a part of hyperboloid x^2+y^2
    which is located at 0<=x<=1 and 0<=y<=1 ,and its normal vector points outside
    like this :
    http://i28.tinypic.com/f9p63r.gif

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

    x^2+y^2=r
    <br />
=\int_{0}^{2\pi}\int_{0}^{1}\frac{(e^y,ye^x,x^2y) \cdot (2x,2y,-1)}{1}rdrd\theta<br />
    how to what are the intervals for r
    i just guessed its from 0 to 1
    i dont know how to know the upper interval here

    except that
    is this method ok?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    first of all z = x^2 +y^2 is a parabaloid

    secondly 0 < x < 1 and 0 < y < 1

    is a square not a circle so you want to stay with rectangular coordinates

    with limits 0 to 1 for both x and y
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  3. #3
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    thanks
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  4. #4
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    what if i have only this area
    z^2=16-x^2 -y^2
    with out any limitations
    here i can use polar coordinated
    i put z=0 -> r=4
    and it generates from a point so in the start r=0
    correct?
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  5. #5
    MHF Contributor Calculus26's Avatar
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    If z > 0 is a stated condition then In this case you're region in the x-y plane is indeed a circle of radius 4 so convert to polar and r varies from 0 to 4 and theta from 0 to 2pi.
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