# Line Integral over Curve

• Apr 27th 2010, 10:00 PM
veritaserum2002
Line Integral over Curve
Calculate $\displaystyle \int_{C} z dx+x dy+y dz$

where C is the curve obtained by intersecting surfaces $\displaystyle z=x^2$ and $\displaystyle x^2+y^2=4$, oriented so that the z-coordinate increases along C.
• Apr 28th 2010, 02:37 AM
Failure
Quote:

Originally Posted by veritaserum2002
Calculate $\displaystyle \int_{C} z dx+x dy+y dz$

where C is the curve obtained by intersecting surfaces $\displaystyle z=x^2$ and $\displaystyle x^2+y^2=4$, oriented so that the z-coordinate increases along C.

That C has to be "oriented so that the z-coordinate increases along C" is a ridiculous requirement that cannot be satisfied generally, because C is a closed curve. So moving along C we may see the z-coordinates increase, only to see them necessarily decrease later on.
• Apr 28th 2010, 02:49 AM
HallsofIvy
$\displaystyle x^2+ y^2= 4$ is a circle with center at the origin and radius 2. A standard parmeterization is $\displaystyle x= 2cos(\theta)$ $\displaystyle y= 2sin(\theta)$. $\displaystyle z= x^2$, then, becomes $\displaystyle z= 4cos^2(\theta)$.

That is, a parameterization of the path is $\displaystyle x= 2cos(\theta)$, $\displaystyle y= 2sin(\theta)$, $\displaystyle z= 4cos^2(\theta)$.

However, as veritaserum2002 said, in going around that curve, in either direction, the z value, as well as both x and y values, both increase and decrease so "oriented so that the z-coordinate increases along C" makes no sense.

The "positive orientation" for this curve, the orientation so that the "oriented surfaces" (normal axis pointing in the positive z direction) were "on the left" as you move around the curve, is with $\displaystyle \theta$ going from 0 to $\displaystyle 2\pi$