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Math Help - Flux 2

  1. #1
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    Flux 2

    I need to find the flux of this vector field (in the pic) that goes through this plan (in the pic) and z goes from 0 to 1.
    How am I suppose to do that?
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  2. #2
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    Quote Originally Posted by asi123 View Post
    I need to find the flux of this vector field (in the pic) that goes through this plan (in the pic) and z goes from 0 to 1.
    How am I suppose to do that?
    Do you know that when the surface is defined in the form z = f(x, y) then:


    \int \int_S F \cdot dS = \int \int_S F \cdot n dS = \int \int_{R_{xy}} F \cdot \left( \frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k \right) \, dx \, dy


    since n = \frac{\frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k}{\sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1}} and dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy.


    After making the required substitutions your flux integral becomes \int \int_{R_{xy}} 2y - 2x + 1 \, dx \, dy, where R_{xy} is the region of the xy-plane defined by x^2 + y^2 = 1. I'd suggest switching to polar coordinates to get this double integral.

    Note that the region R_{xy} is found by substituting z = 1 into z = \sqrt{x^2 + y^2}.


    Of course, you could always switch to cylindrical coordinates at the very start of the problem .......
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Do you know that when the surface is defined in the form z = f(x, y) then:


    \int \int_S F \cdot dS = \int \int_S F \cdot n dS = \int \int_{R_{xy}} F \cdot \left( \frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k \right) \, dx \, dy


    since n = \frac{\frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k}{\sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1}} and dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy.


    After making the required substitutions your flux integral becomes \int \int_{R_{xy}} 2y - 2x + 1 \, dx \, dy, where R_{xy} is the region of the xy-plane defined by x^2 + y^2 = 1. I'd suggest switching to polar coordinates to get this double integral.

    Note that the region R_{xy} is found by substituting z = 1 into z = \sqrt{x^2 + y^2}.


    Of course, you could always switch to cylindrical coordinates at the very start of the problem .......
    Check this out, is that what you meant?
    Attached Thumbnails Attached Thumbnails Flux 2-scan0002.jpg  
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Do you know that when the surface is defined in the form z = f(x, y) then:


    \int \int_S F \cdot dS = \int \int_S F \cdot n dS = \int \int_{R_{xy}} F \cdot \left( \frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k \right) \, dx \, dy


    since n = \frac{\frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k}{\sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1}} and dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy.

    After making the required substitutions your flux integral becomes \int \int_{R_{xy}} 2y - 2x + 1 \, dx \, dy, where R_{xy} is the region of the xy-plane defined by x^2 + y^2 = 1. I'd suggest switching to polar coordinates to get this double integral.

    Note that the region R_{xy} is found by substituting z = 1 into z = \sqrt{x^2 + y^2}.


    Of course, you could always switch to cylindrical coordinates at the very start of the problem .......
    mr. fantastic, I'd like to suggest something to improve your notation.

    Instead of typing

    Code:
    [tex]\int \int_S F \cdot dS[/tex]
    And getting \int \int_S F \cdot dS, you can type

    Code:
    [tex]\iint\limits_S F \cdot dS[/tex]
    and get \iint\limits_S F \cdot dS, which looks neater.

    --Chris
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  5. #5
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    Quote Originally Posted by asi123 View Post
    Check this out, is that what you meant?
    Looks fine. Do you know what your integral terminals will be?
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