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Let $$\vec{F} (x, y, z) = xy\hat{i} +(4x - yz)\hat{j} + (xy - z^{1/2}) \hat{k},$$ and let $C$ be a circle of radius $R$ lying in the plane $x + y + z = 5$. If $$\int_C \vec{F} \cdot d\vec{r} = \pi \sqrt{3},$$ where $C$ is oriented in the counterclockwise direction when viewed from above the plane, what is the value of $r$?

This question has some context and the version of Stokes Theorem to use.

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  • $\begingroup$ What are your thoughts on the problem? Please explain what you tried and we will be happy to provide hints $\endgroup$ – gt6989b Dec 9 '14 at 20:36
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    $\begingroup$ See my comment on the related question. math.stackexchange.com/questions/1059792/… Ironically, I'm the one who usually asks a student to show me their thinking first. See my many answers to other questions. I'll keep working, may be able to type further thoughts here. $\endgroup$ – nickalh Dec 9 '14 at 22:34
  • $\begingroup$ @gtg989b Thanks for the offer. Keep the suggestions coming. $\endgroup$ – nickalh Dec 9 '14 at 23:31
  • $\begingroup$ @gtg989b What's the equation of a circle on a slant in 3 dimensions? $\endgroup$ – nickalh Dec 10 '14 at 1:28
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We know that $$\oint_C \vec{F} \cdot d\vec{r}=\pi\sqrt{3}$$

By the Stokes theorem, we can write that

$$ \oint_C \vec{F} \cdot d\vec{r}=\iint_S \nabla \times \vec{F}\cdot d\vec{S} = \iint_S \nabla \times \vec{F}\cdot \vec{n}, \; dS$$

where $S$ is the surface inside the circle $C$ (the disc with radius $r$), lying in the plane $x+y+z=5$, and $\vec{n}$ is a unitary normal vector to this surface adequately orientated, i.e. $\vec{n}=\frac{1}{\sqrt{3}}(1,1,1)$. $\nabla \times \vec{F}$ represents the curl of $\vec{F}$, and is equal to $(x+y,-y,4-x)$.

It follows that

$$ \iint_S \nabla \times \vec{F}\cdot \vec{n}, \; dS = \frac{1}{\sqrt{3}} \iint_S x+y-y+4-x \; dS = \frac{4}{\sqrt{3}} \;Area(S)=\frac{4}{\sqrt{3}} \pi r^2. $$

Solving for $r$ yields:

$$ r=\frac{\sqrt{3}}{2}. $$

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  • $\begingroup$ (+1) Good Job. :) But you were lucky! What would you do if $\vec{n}=\frac{1}{\sqrt{6}}(1,1,2)$? :) $\endgroup$ – H. R. Dec 8 '15 at 15:01
  • $\begingroup$ Yes, I agree with you, if terms don't simplify we would have to go with your option…but this looks like a textbook exercise, and if the textbook is "well thought", computations often simplify the way we expect them to! $\endgroup$ – Kuifje Dec 8 '15 at 15:03
  • $\begingroup$ Yeah! There should be some trick in "well thought" exercises! :) $\endgroup$ – H. R. Dec 8 '15 at 15:07
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According to your last comment, I think the main challenge of this question is to write the parametric equation of the circle in 3D. To do this, we first note that the parametric equation of the plane can be written as

$$ {\bf{r}}(x,y) = x {\bf{i}} + y {\bf{j}} + (5-x-y) {\bf{k}} \tag{1}$$

where $x$ and $y$ are the parameters. Next, suppose that the center of the circle is at

$$ {\bf{c}} = 5 {\bf{k}} \tag{2}$$

and hence the vector ${\bf{r-c}}$ always lie in our plane. In order to find a vector such that it lies on a circle with radius $R$ in the plane, we require that

$$\begin{array}{} {\bf{r-c}} = x {\bf{i}} + y {\bf{j}} - (x+y) {\bf{k}}\\ ({\bf{r-c}}) \cdot ({\bf{r-c}}) = R^2 \\ x^2+y^2+(x+y)^2=R^2 \\ 2x^2+2y^2+2xy=R^2 \\ x^2+y^2+xy=\frac{R^2}{2} \end{array} \tag{3}$$

Now, this last equation represent an ellipse rotated with angle $\alpha=-\frac{\pi}{4}$ around the origin.

$\qquad \qquad \qquad \qquad$enter image description here

This suggest us to do the following change of coordinates

$$\begin{array}{} x = \cos\alpha \, u + \sin\alpha \, v \\ y = -\sin\alpha \, u + \cos\alpha \, v \end{array} \tag{4}$$

which represent a rotation. This will turn the last equation in $(3)$ into

$$(\sqrt{3}u)^2+v^2=R^2 \tag{5}$$

Also, we can go further and do another change of coordinates

$$\begin{array}{} u = \frac{1}{\sqrt{3}} r \, \cos\theta \\ v = r \, \sin\theta \end{array} \tag{6}$$

and this will turn $(5)$ into the simple equation

$$r=R \tag{7}$$

To conclude, we introduce the new curvelinear coordinates

$$\begin{array}{} x = \frac{\cos\alpha}{\sqrt{3}} \, r \, \cos\theta + \sin\alpha \, r \, \sin\theta \\ y = - \frac{\sin\alpha}{\sqrt{3}} r \, \cos\theta + \cos\alpha \, r \, \sin\theta \\ z=z \end{array} \tag{8}$$

and do the further computations in this new coordinates which simplifies the whole arithmetic.

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