# Thread: Evaluate line integral with 3 dimensions

1. ## 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.

2. Originally Posted by ban26ana  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?

3. 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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