Line Integral over Curve

**veritaserum2002** · April 27th 2010, 10:00 PM

Calculate $\int_{C} z dx+x dy+y dz$

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

**Failure** · April 28th 2010, 02:37 AM

Originally Posted by veritaserum2002

Calculate $\int_{C} z dx+x dy+y dz$

where C is the curve obtained by intersecting surfaces $z=x^2$ and $x^2+y^2=4$ , 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.

**HallsofIvy** · April 28th 2010, 02:49 AM

$x^2+ y^2= 4$ is a circle with center at the origin and radius 2. A standard parmeterization is $x= 2cos(\theta)$ $y= 2sin(\theta)$ . $z= x^2$ , then, becomes $z= 4cos^2(\theta)$ .

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

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 $\theta$ going from 0 to $2\pi$

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