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Math Help - another flux through surface question

  1. #1
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    another flux through surface question

    Compute the flux of the vector field,  \vec{F} , through the surface, S.  \vec{F} = y\vec{i} + 7\vec{j} - xz\vec{k} and S is the surface y = x^2 + z^2 with x^2 + z^2 \leq 36 oriented in the positive y direction.

    My answer:

    \int\limits_R ((x^2+z^2)\vec{i} + 7\vec{j} - xz\vec{k}) \cdot (-2x\vec{i} - 2z\vec{k} + \vec{j}) dA


    I don't know if this is correct...
    Last edited by TheRekz; April 22nd 2009 at 07:59 AM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    it looks good-however the orientatation seems a little strange.

    you have a parabaloid aligned along the positive y axis

    Usually we specify inward or outward--here on the half in the first and 8th octants you have an outward normal but in the 2d and 5th you have an inward normal.

    So is this correct?



    and of course your region of integration is a circle of radius 6 in the x-z plane
    Last edited by Jhevon; April 26th 2009 at 04:14 PM.
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  3. #3
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    Quote Originally Posted by TheRekz View Post
    so is it okay?? I don't know what the limits of integral is..
    reason I ask is because I get;

    \int\limits_R (-2x^3+7) dA

    and I don't know how to represent this in polar
    Last edited by TheRekz; April 25th 2009 at 10:29 PM.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    yes the ingrl is okay


    in polar form just think of z as y

    x= rcos(t)

    z = r sin(t)
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  5. #5
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    hmm.. so then it becomes:

    \int_0^{2pi}\int_0^6 (-432(cos(t)^3)+7) rdrd\theta

    am I missing something here?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    I was using t for theta --just lazy

    integral of cos(theta)^3 should not present a pblm
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  7. #7
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    Quote Originally Posted by Calculus26 View Post
    I was using t for theta --just lazy

    integral of cos(theta)^3 should not present a pblm
    yes but i do need to multiply it by r??
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  8. #8
    MHF Contributor Calculus26's Avatar
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    yes since x = rcos(theta) you'll end up integrating r^4 first

    where does -432 come from?
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    Quote Originally Posted by Calculus26 View Post
    yes since x = rcos(theta) you'll end up integrating r^4 first

    where does -432 come from?
    because r is equal to 6, and therefore :

    x = rcos(theta)
    = (6cos(theta))^3
    = 216cos(theta)^3

    multiplying that by -2, that's where I got the -432cos(theta)^3, am I doing something wrong here? so it should be -432cos(theta)^3r, right?
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  10. #10
    MHF Contributor Calculus26's Avatar
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    No r is a variable you take into account that r is 6 when you integrate r from 0 to 6
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  11. #11
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    shouldn't we multiply it by r first and then take the integration with respect to r?
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  12. #12
    MHF Contributor Calculus26's Avatar
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    again x= rcos(theta) when you integrate over the disk of radius 6

    r varies from 0 to 6 it isn't a constant 6

    so you'd have (-2 r^4 cos^3(theta) +7) drd(theta) to integrate
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  13. #13
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    well after that I got:

    \int_0^{2\pi}\frac{-15552(cos^3\theta)}{5}+42 d\theta

    is this correct??
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  14. #14
    MHF Contributor Calculus26's Avatar
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    Yes
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  15. #15
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    Quote Originally Posted by Calculus26 View Post
    Yes
    moving on forward I have:

    \frac{15552(sin(\theta)^3-3sin(\theta))}{15} +42\theta

    evaluated from 0 to 2pi.. resulting in 84pi and webassign doesn't take this answer
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