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Math Help - Evaluate line integral with 3 dimensions

  1. #1
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    Evaluate line integral with 3 dimensions


    Where C is the loop with z=2

    with a clockwise orientation when viewed from above.


    As usual, I have trouble with the third dimension. I have a problem like this without the third dimension, and I have problems with the third dimension but other shapes (parabola, etc). I'm confused with this one.

    Edited: can you show me both clockwise and counterclockwise orientations? I'm not really following how to differentiate that part.
    Last edited by ban26ana; May 14th 2010 at 09:25 AM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ban26ana View Post

    Where C is the loop with z=2

    with a clockwise orientation when viewed from above.


    As usual, I have trouble with the third dimension. I have a problem like this without the third dimension, and I have problems with the third dimension but other shapes (parabola, etc). I'm confused with this one.

    Edited: can you show me both clockwise and counterclockwise orientations? I'm not really following how to differentiate that part.
    Observe that if G(x,y,z)=x^2y^2z, then \nabla G(x,y,z)=\left<2xy^2z,2x^2yz,x^2y^2\right>=\mathbf  {F}, Therefore, this line integral is conservative!!! Define the contour as the vector valued function \mathbf{r}=\left<\cos t,\sin t, 2\right> with 0\leq t\leq 2\pi.

    Then by the fundamental theorem of line integrals, \int_C\mathbf{F}\cdot d\mathbf{r}=\int_C\nabla G\cdot d\mathbf{r}=G(\mathbf{r}(b))-G(\mathbf{r}(a)).

    Thus, \int_0^{2\pi}\nabla G\cdot d\mathbf{r}=\left.\left[x^2y^2z\right]\right|_{\left<1,0,2\right>}^{\left<1,0,2\right>}=  0.

    Does this make sense?
    Last edited by Chris L T521; May 15th 2010 at 06:00 AM.
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  3. #3
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    Once you have determined that the field is conservative (that the differential is exact), for example, by observing that \frac{\partial(2xy^2z)}{\partial y}= \frac{(\partial2x^2yz)}{\partial x}, \frac{\partial(2xy^2z)}{\partial z}= \frac{\partial(x^2y^2)}{\partial x}, and that \frac{\partial(2x^2yz)}{\partial z}= \frac{\partial(x^2y^2)}{\partial y}, you don't really have to find an "anti-derivative"- the integral around any closed path is 0.

    ban26ana, if the differential were not exact, any one dimensional path can be written in terms of a single parameter. Here, since the path is the circle with center at (0, 0), in the z= 1 plane, with radius 1, so we can use the "standard" parameterization for the unit circle, x= cos(\theta), y= sin(\theta) with z= 1. Then dx= -sin(\theta)d\theta, dy= cos(\theta)d\theta and dz= 0. d\vec{s}= (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta. Integrating around the path in the "positive orientation" would be integrating from 0 to 2\pi and in the "negative orientation", from 2\pi to 0 (or, equivalently, from 0 to -2\pi).

    Here, \vec{F}= 2xy^2z\vec{i}+ 2x^2yz\vec{j}+ x^2y^2\vec{k} = 2cos(\theta)sin^2(\theta)\vec{i}+ 2cos^2(\theta)sin(\theta)\vec{j}+ cos^2(\theta)sin^2(\theta)\vec{k} so \vec{F}\cdot d\vec{s}= (-2cos^2(\theta)sin^2(\theta)+ 2cos^2(\theta)sin^2(\theta))d\theta= 0.

    Since it is true, for any function G(x,y,z) with continuous second derivatives, that \nabla\times \nabla G= \vec{0}, Chris L T521's observation that this \vec{F}(x,y,z)= \nabla G for a specific G, is the same as saying \nabla\times\vec{F}(x,y,z)= 0 and so, by Stoke's theorem, \oint\vec{F}\cdot d\vec{s}= \int\int \nabla\times\vec{F}\cdot d\vec{S}= \int\int \vec{0}\cdot d\vec{S}= 0.
    Last edited by HallsofIvy; May 15th 2010 at 02:43 AM.
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