Having problem with dtermining the path over which a line integral is to be evaluated

Question

Let $C$ be the curve of intersection of $z=xy ; x^2+y^2=1$ traversed once in a direction that appears counterclockwise when viewed from High above the $xy$-plane ; then how to evaluate $\int_Cydx+zdy+xdz$ ? I have found a parametrization of the curve which is $\Big(\cos t ,\sin t , \dfrac {\sin 2t}2\Big)$ , but I cannot make any whereabout of what does "counterclockwise when viewed from High above the $xy$-plane "mean . Please help . Thanks in advance

Answer 1

You are right with that parametrization. Here's what this intersection curve looks like when plotted in Mathematica:
img1 http://puu.sh/lINR8/862057487a.png
What it means to be in the clockwise orientation is that, if you looked at this from above, you would be going around the "circloid" in the counterclockwise direction. See the image below:
img2 http://puu.sh/lINUY/e76e85992f.png
I hope that makes sense. Another way to think about the counterclockwise direction is increasing $\theta$. Think about the unit circle; going counterclockwise yields more positive values of $\theta$. Clockwise gives negative values of $\theta$
In this case, you can imagine $t$ as $\theta$. As $t$ increases, you move counterclockwise on the parametrization.
If you plug $C$ into your line integral, you should be able to calculate it quite easily. I will show the first term, $y\,dx$, and let you go from there.
$y=\sin{t}. \frac{\,dx}{\,dt}=-\sin{t}$, so $dx=-\sin{t}\,dt.$
Therefore, $y\,dx=\sin^2{t}\,dt.$
Do the same for the rest of the terms, factor out the $\,dt$ and integrate.
One final note: this parametrization has $t\in[0, 2\pi]$. That means as $t$ ranges from $0$ to $2\pi$, the curve is traversed just once. This is generally the case for parametric equations like the one in this example.