Evaluate line integral with 3 dimensions

**ban26ana** · May 14th 2010, 09:12 AM

$Evaulate \int_{C}^{}F where F=<2xy^2z, 2x^2 yz, x^2y^2 >$
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.

**Chris L T521** · May 14th 2010, 03:39 PM

Originally Posted by ban26ana

$Evaulate \int_{C}^{}F where F=<2xy^2z, 2x^2 yz, x^2y^2 >$
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?

**HallsofIvy** · May 15th 2010, 02:31 AM

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

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