# Evaluate the line integral

1. Jul 13, 2009

### harvellt

1. The problem statement, all variables and given/known data

$$\int$$y dx +x dy + z dz

c= helix x = 3 cos t
y = 3 sin t
z = 4t
0$$\leq$$t$$\leq$$2$$\Pi$$

2. Relevant 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 cant figure out how to put in limits of intagration)

3. 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$$^{2}$$?

Last edited: Jul 13, 2009
2. Jul 13, 2009

### Billy Bob

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)

3. Jul 13, 2009

### 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.

Last edited: Jul 13, 2009
4. Jul 13, 2009

### Billy Bob

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.

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

5. Jul 13, 2009

### 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?

6. Jul 13, 2009

### 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.

7. Jul 14, 2009

### whitay

Is it conservative?

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

8. Jul 14, 2009

### HallsofIvy

Staff Emeritus
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.