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 $\int_C y\,dx + z\,dy + x\,dz$.

• use polar parametrization $x=\cos \theta, y=\sin \theta, z=\cos \theta \sin \theta$ Commented May 26, 2013 at 18:03

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.

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.