Line Integral of F(x,y,z) over Unit Circle

[stanford1463]

Homework Statement

What is the line integral of F(x,y,z) = (xy, x, xyz) over the unit circle c(t) = (cost, sint) t E (0,2pi) ?

Homework Equations

integral= (f(c(t))*c'(t))dt)

The Attempt at a Solution

Ok, so I tried solving this like I would any other line integral using the given equation, but it does not work, since I am taking a 3D function on a 2D function (circle) ? So...f(c(t)) cannot be anything?? As cos t can be x, sint can be y, but what is z?? I know you take the dot product, and solve, taking the integral, but I cannot figure out how to find f(c(t)) as one function has 3 variables and the other has 2? Thanks!

 

[Dick]

z is zero. The circle is (cos(t),sin(t),0). The dimensions have to match or you can't take a dot product. If one is three dimensional the other must be as well, even if they don't spell it out.

 

[stanford1463]

Ok thanks! But what if it's something like F(x,y,z)=(xy, x/z, y/z) ? Then if you plug in 0, you get a denominator of zero...

 

[HallsofIvy]

Ok thanks! But what if it's something like F(x,y,z)=(xy, x/z, y/z) ? Then if you plug in 0, you get a denominator of zero...

Didn't you read Dick's response? The problem is that "c(t)= (cos(t), sin(t))" is two dimensional and you CAN'T integerate a 3 dimensional vector function over a two dimensional path. Are you interpreting that to mean that it must be (cos(t), sin(t), 0)? There is no reason to assume that because z is not mentioned, it must be 0. You MUST have some equation involving z, perhaps "z= 0", perhaps "z= 1", in order that the path be defined in three dimensions.

 

[Dick]

I guess I automatically think of THE Unit Circle, as lying in the x-y plane. But Halls is right, they should have specified a z value.