Results 1 to 3 of 3

Thread: Evaluate line integral with 3 dimensions

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    52

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    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 $\displaystyle G(x,y,z)=x^2y^2z$, then $\displaystyle \nabla G(x,y,z)=\left<2xy^2z,2x^2yz,x^2y^2\right>=\mathbf {F}$, Therefore, this line integral is conservative!!! Define the contour as the vector valued function $\displaystyle \mathbf{r}=\left<\cos t,\sin t, 2\right>$ with $\displaystyle 0\leq t\leq 2\pi$.

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

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

    Does this make sense?
    Last edited by Chris L T521; May 15th 2010 at 06:00 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,793
    Thanks
    3035
    Once you have determined that the field is conservative (that the differential is exact), for example, by observing that $\displaystyle \frac{\partial(2xy^2z)}{\partial y}= \frac{(\partial2x^2yz)}{\partial x}$, $\displaystyle \frac{\partial(2xy^2z)}{\partial z}= \frac{\partial(x^2y^2)}{\partial x}$, and that $\displaystyle \frac{\partial(2x^2yz)}{\partial z}= \frac{\partial(x^2y^2)}{\partial y}$, 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, $\displaystyle x= cos(\theta)$, $\displaystyle y= sin(\theta)$ with $\displaystyle z= 1$. Then $\displaystyle dx= -sin(\theta)d\theta$, $\displaystyle dy= cos(\theta)d\theta$ and $\displaystyle dz= 0$. $\displaystyle d\vec{s}= (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta$. Integrating around the path in the "positive orientation" would be integrating from 0 to $\displaystyle 2\pi$ and in the "negative orientation", from $\displaystyle 2\pi$ to 0 (or, equivalently, from 0 to $\displaystyle -2\pi$).

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

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

Similar Math Help Forum Discussions

  1. Evaluate the line integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 18th 2011, 08:38 AM
  2. [SOLVED] Evaluate line integral (parabola)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 4th 2010, 08:19 PM
  3. Evaluate Line Integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Nov 1st 2010, 11:56 AM
  4. Evaluate the line integral...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 17th 2008, 01:04 PM
  5. Evaluate the line integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jun 8th 2007, 08:15 AM

Search Tags


/mathhelpforum @mathhelpforum