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Compute the line integral of vector field $F(x,y,z)$ = $⟨x^2,y^2,z^2⟩$ where C is the curve of intersection of $z=x+1$ and $x^2+y^2=1$, from the lowest point on the curve to the highest, traversed counterclockwise when seen from above.

These are a plane tilted 45 degrees and a cylinder of radius 1 centered at the origin, so their intersection is a hoop tilted 45 degrees on the $y$ axis. The hoop's low point is (-1,0,0) and its high point is (1,0,2). I can also do the following conversion to cylindrical coordinates to get the equation of the hoop:

$z=1+x$ $\implies$ $z=1+r\cos\theta$

$x^2+y^2 = 1^2$ $\implies$ $r^2 = 1$ $\implies$ $r=1$

∴ $z=1+1\cos\theta$ $\implies$ $z=1+\cos\theta$

This is all fine. Afterwards, however, I don't know how to convert this into a vector field equation, where everything is supposed to be parameterized entirely into $t$. The formula for the line integral of a vector field is:

$\int^b_aF(x(t),y(t),z(t))\cdot r\prime(t) dt$

And I'm not quite sure how to transform my work so that it fits this formula. I think that I should treat $\theta$ as $t$, and the limits $a$ and $b$ should be $\int^{0}_π$, but I'm stuck beyond that. Would anyone be able to tell me how I should be setting this integral up?

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    It doesn't matter really what you call the parameter, you can call it $\theta$ or $t$, just replace $t$ with $\theta$ in your definition of line integrals – 2017-01-26
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    Yes, I understand that, but I'm still left with $z$ and $r$ and $\theta$ when it seems like I should only have $\theta.$ I'm not sure how to fit everything into the integral, even assuming $\theta=t.$ – 2017-01-26
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    R is a position vector that describes the path and is always of the form $\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j}+z(t)\hat{k}$, so you put your parameterizations in for x,y,z and differentiate to get $\vec{r}'(t)$. So for example in yours the position vector would be $\vec{r}(\theta) = \cos{\theta} \hat{i} + \sin{\theta} \hat{j} + (1 + \cos{\theta})\hat{k}$, the vector field F is simplified similarly. – 2017-01-26
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    I think I understand, the main confusion for me was whether $r(\theta)$ could be simplified in that way. I did a similar conversion for $F$ and got $⟨\cos^2\theta,\sin^2\theta,z^2⟩,$ so it now looks like everything is set up. Thank you! – 2017-01-26
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    Yep and Doug mentions another important point about the conservative nature of the vector field – 2017-01-26

1 Answers 1

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$r(x,y,z) = (\cos t, \sin t, 1+\cos t) \\ r'(t) = (-\sin t, \cos t, -\sin t)$

If you integrate from $(\pi, 0)$ as you suggest above, you will traverse the path in the clockwise direction. In this case, it will prove to be irrelevant.

$\int^b_aF(x(t),y(t),z(t))\cdot r\prime(t) dt\\ \int_{\pi}^{2\pi} (\cos^2 t, \sin^2 t, (1+\cos t)^2)\cdot (-\sin t, \cos t, -\sin t) dt\\ \int_{\pi}^{2\pi} -2\sin t\cos^2 t + \cos t\sin^2 t - \sin t - 2 \sin t \cos t \;dt\\ $

And I will let you take it from here.

It is worth noting that F(x,y,z) is a conservative force.
$F = \nabla \frac 13 (x^3 + y^3 + z^3)$

The integral does not depend on the path, only the endpoints.

$\frac 13 (x^3 + y^3 + z^3) |_{(-1,0,0)}^{(1,0,2)}$

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    I understand what you mean about the direction being irrelevant, but why would traversing from π to 0 be clockwise? Wouldn't that be from π to 2π? – 2017-01-26
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    if $(x,y) = (\cos t,\sin t)$ and we are traversing the contour counter-clockwise, $t$ must be increasing. – 2017-01-26
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    What should the traversal be then? The starting point is in the negative $x$ axis, so it MUST be $π$ as the lower limit...right? And from there, my thinking is that if 0 to $π$ is clockwise, $π$ to 0 must be counterclockwise. – 2017-01-26
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    $0$ to $\pi$ is counter clock-wise. Make your unit circle. Where is $(x(0),y(0))$ ... 3-o'clock. Where is $(x(\frac \pi 6),y(\frac \pi 6)$? 2-o'clock. $(x(\frac \pi 3),y(\frac \pi 3)$ is 1 o'clock. Increasing $t$ is counter-clockwise. – 2017-01-26
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    You're absolutely right, I had the image flipped in my head. Thank you for the help! – 2017-01-26