Evaluate line integral with 3 dimensions

Aug 2008
52
1

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:

Chris L T521

MHF Hall of Fame
May 2008
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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?
 
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HallsofIvy

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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, \(\displaystyle \oint\vec{F}\cdot d\vec{s}= \int\int \nabla\times\vec{F}\cdot d\vec{S}= \int\int \vec{0}\cdot d\vec{S}= 0\).
 
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