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Math Help - Flux help

  1. #1
    n22
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    Flux help

    Hi ,I would appreciate some guidance on this .thanks.
    Evaluate the flux of the vector field F=18zi-12j+3yk across that part of the plane 2x+3y+6z=12 which is located in the first octant of the plane 2x+3y+6z=12 .Use the unit normal to the plane that points in the positive z-direction .


    So far I have found the normal to the plane ... as 2i+3j+6k/7

    I was intending on using the eqn Iintergral F  n ds =
    not sure what to do with ds=sqrt(...)
    I am not sure from where to where I am integrating ...flux
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  2. #2
    MHF Contributor

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    Re: Flux help

    You mean "that part of the plane 2x+3y+6z=12 which is located in the first octant". The additional "of the plane 2x+ 3y+ 6z= 12" confused me!

    When x and y are 0, that becomes 6z= 12 or z= 2 so one corner of the triangle is (0, 0, 2). When y and z are 0, that becomes 2x= 12 or x= 6 so (6, 0, 0) is another corner of the triangle.
    And when x and z are 0, that becomes 3y= 12 or y= 4 so (0, 4, 0) is the third corner. Yes, the unit normal to the plane is (2i+ 3j+ 6k)/7 (note the parentheses1)

    The "sqrt" in "sqrt(...)dx" is that "7", the length of the vector.

    Another way of looking at surface integrals, which I prefer, is this: We can write any curve, in an xyz coordinate system, as x= f(s,t), y= g(s,t), z= h(s, t) where s and t are the parameters.
    And we can write that as the vector equation \vec{r}(s, t)= f(s,t)\vec{i}+ g(s,t)\vec{j}+ h(s,t)\vec{k}. The two derivatives, \vec{r}_s= f_s\vec{i}+ g_s\vec{j}+ h_s\vec{k} and \vec{r}_t=f_t\vec{i}+ g_t\vec{j}+ h_t\vec{k} are vectors lying in the tangent plane to the surface so that \vec{r}_s\times \vec{r}_t is normal to the surface. Further, we have \vec{n}dS= \vec{r}_s\times\vec{r}_t dt ds.

    Here, our surface is the plane 2x+ 3y+ 6z= 12 which we can write as z= 2- (1/3)x- (1/2)y and write x= 3s, y= 2t, z= 2- s- t. The "position vector" or a point on the plane would be \vec{r}= 3s\vec{i}+ 2t\vec{j}+ (2- s- t)\vec{k}. The derivatives are \vec{r}_s= 3\vec{i}- vec{k} and \vec{r}_t= 2\vec{j}- \vec{k}. The cross product of those two vectors (positive z component) is 2\vec{i}+ 3\vec{j}+ 2\vec{k} so that \vec{n}dS= (2\vec{i}+ 3\vec{j}+ 2\vec{k})dsdt.

    The integrand vector function is \vec{F}= 18z\vec{i}- 12\vec{j}+ 3y\vec{k} so \vec{F}\cdot\vec{n}dS= (36z-36+ 6y)dsdt= (2- s-t- 36+6t)dsdt= (5t-s- 34)dsdt

    Projecting 2x+3y+ 6z= 12 into the xy-plane (z= 0) gives 2x+ 3y= 12 or y= 4- (2/3)x while x runs from 0 to 6. Since s= x and t= y, The integral will be
    \int_0^6\int_0^{4- (2/3)x} (5t- s- 3)dt ds
    Thanks from n22
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