3D line integral question Let $C$ be the curve of intersection of the cylinder $x^2 + y^2 = 1$ and the surface $z = xy$, oriented counterclockwise around the cylinder. Compute the integral $\int_C y\,dx + z\,dy + x\,dz$.
 A: Parameterize as follows:
$$x=\cos\theta;\space y=\sin\theta;\space z=\cos\theta\sin\theta=\frac{1}{2}\sin2\theta$$
$$\vec{r}(\theta)=\langle{\cos\theta,\sin\theta,\frac{1}{2}\sin2\theta}\rangle,0\le\theta<2\pi$$
$$\oint_Cy\space{dx}+z\space{dy}+x\space{dz}=\int_{0}^{2\pi}(-\sin^2{\theta}+\frac{1}{2}\sin{2\theta}\cos{\theta}+\cos{\theta}\cos{2\theta})\space d\theta$$
From there you can make use of trig identities to evaluate the integral.
A: To do line integrals over curves in any dimension, we must first parametrize the curve with some parameter and then express the integrand and differentials in that parameter. Basically perform a variable substitution. This reduces the integral to a one dimensional integral. 
The curve C
  lies on the cylinder with unit radius described by the equation, $x^{2}+y^{2}=1$
 . This cylindrical geometry calls for use of the polar coordinates $\left(r,\theta,z\right)$
 . 
$$
x = r\cos\theta,
y = r\sin\theta,
z = z
$$
 The cylinder equation tells us how x
  and y
  relate to each other and the surface equation tells us how z
  is related to x,y
 . In terms of the polar coordinates we can now write the curve C
  as the vector parametrized by the parameter $\theta$
  [which is conveniently the polar coordinate of azimuthal angle].
$$
\vec{C}=\left(\cos\theta,\sin\theta,\cos\theta\sin\theta\right)
$$
,r=1
  on the cylinder.
Now, the differentials will be changed to:
$$
dx = d\left(\cos\theta\right)=-\sin\theta d\theta,
dy = d\left(\sin\theta\right)=\cos\theta d\theta,
dz = d\left(\cos\theta\sin\theta\right)=-\sin^{2}\theta d\theta+\cos^{2}\theta d\theta
$$ 
The integral now becomes:
$$
\int_{0}^{2\pi}-\sin^{2}\theta d\theta+\cos^{2}\theta\sin\theta d\theta+\cos\theta\left(\cos^{2}\theta-\sin^{2}\theta\right)d\theta =
$$
$$ 
-\frac{1}{2}\int_{0}^{2\pi}\left(1-\cos2\theta\right)d\theta-\int_{0}^{2\pi}\cos^{2}\theta d\left(\cos\theta\right)+\int_{0}^{2\pi}\left(1-2\sin^{2}\theta\right)d\left(\sin\theta\right)  
$$
Where we have used the trig identities:
$$ \sin^{2}\theta = \frac{1-\cos2\theta}{2},
\cos^{2}\theta+\sin^{2}\theta = 1
$$ 
and the parameter $\theta$
  runs through a full $2\pi$
  rotation.
