How Do You Calculate a Line Integral for a Helical Path?

[harvellt]

Homework Statement

\inty dx +x dy + z dz

c= helix x = 3 cos t
y = 3 sin t
z = 4t
0\leqt\leq2\Pi

Homework Equations

\int F(x,y,z) ds
ds=\sqrt{[Fx(x,y,z)]+[Fy(x,y,z)]+[Fz(x,y,z)]}dt (still learning latex the partial derivatives are suposed to be squared.)
(also can't figure out how to put in limits of intagration)

The Attempt at a Solution

ds = 5

5\int(3 sint + 3 cos t + 4t) =
5[(-3 cos t + 3 sin t + 2t^{2}) evaluated from 0 to 2\Pi

=40\Pi^{2}

My real question is whenever you evaluate sin or cos around all the way around from 0 to 2\Pi is it supposed to be zero? So both the first terms drop out and your left with just 2t^{2}?

 

[Billy Bob]

Homework Equations

\int F(x,y,z) ds
ds=\sqrt{[Fx(x,y,z)]+[Fy(x,y,z)]+[Fz(x,y,z)]} (still learning latex the partial derivatives are suposed to be squared.)
(also can't figure out how to put in limits of intagration)
[/tex]

Really that "ds" integral is not relevant. The actual problem uses dx, dy, and dz, and the evaluation is pretty straightforward. Find dx, dy, and dz in terms of t, for example dx=-3\sin t\,dt, etc. Then substitute for x, y, z, dx, dy, and dz.

The integral will have \sin^2 t, etc. in it, so unfortunately you won't get an immediate 0.

P.S. Example tex: \int_0^{2\pi} \sin^2 t \, dt (click it to see the code)

 

[harvellt]

so I end up with substitution with:

\\INT_0^{2\\pi}\\ 3sin t + 3cos t 4t \\sqrt\\{-3 sin^2 t + 3 cos^2 + 16}

humm missing something in there to make the intageral and sqrt show up.

 

[Billy Bob]

so I end up with substitution with:

\\INT_0^{2\\pi}\\ 3sin t + 3cos t 4t \\sqrt\\{-3 sin^2 t + 3 cos^2 + 16}

If you mean \int_0^{2\pi}\ 3sin t + 3cos t 4t \sqrt{-3 sin^2 t + 3 cos^2 + 16}, then are you still trying to use ds? You shouldn't, because there is no ds in the given problem. This is a different kind of problem. There is no square root involved.

humm missing something in there to make the intageral and sqrt show up.

Too many slashes. Actually "click" to see the code. If you "hover" it will show too many slashes.

 

[harvellt]

\int_0^{2\pi}\ ( 3sin t + 3cos t + 4t ) \sqrt{-3 sin^2 t + 3 cos^2 + 16}dt
I guess I am more confused than I thaught. I thaught that was the definitation of a line integral involved the ds?

 

[Billy Bob]

The integral \int_C F(x,y,z)\, ds is not the same as the integral \int_C P(x,y,z)\,dx + Q(x,y,z)\,dy + R(x,y,z)\, dz. Your original problem is of the second form, but it appears you are trying to do it using a definition of the first form.

 

[whitay]

Is it conservative?

If so, then you could use fundamental theorem of line integrals.

 

[HallsofIvy]

No, it's not conservative. harvellt, as Billy Bob said, this line integral does NOT involve "ds".

If x= 3 cos(t) then dx= -3 sin(t) dt.
If y= 3 sin(t) then dy= 3 cos(t) dt.
If z= 4t then dz= 4 dt.

Now just replace every x, y, z and dx, dy, dz with its formula in terms of t:
\int_{t= 0}^{2\pi} ydx+ xdy+ zdz= \int_0^{2\pi} (3 sin(t))(-3 sin^2(t)dt)+ (3cos t)(3cos(t)dt)+ (4t)(4dt)

That's a relatively easy integral.