# Calculating Work Along a Helix Using Line Integrals

• PhysicsMajor
In summary, the problem is that you have turned your helix into a straight line. You need to find the bounds on t (using the endpoints given) and take the integral.
PhysicsMajor
Greetings All Again,

I wanted to thank you for the reply on my other problem, it was indeed very helpful and this is a very strange problem. So here goes :

Compute the work done by the force field F(x,y,z) = <4y,2xz,3y> acting on an object as it moves along the helix defined parametrically by
x=2 cos t, y=2 sin t, z=3t, from the point (2,0,0) to the point (-2,0,3(pi)).

Thanks

this is just

$$\int_C 4y \ dx + 2xy \ dy + 3y \ dz$$

where C is the curve you defined above. You know the parameterization. Just solve for the values of $t$ that give you the endpoints (to get the appropriate bounds), and sub in the parameterized x, y, z, dx, dy, dz.

With this problem, assuming i worked it correctly, i got the values for the parameticize. I am confused about what values to plug into x,y, and z. either what i just solved or the other values for x,y, and z. x = 2cost, y = 2sint, and z = 3t.

r(t)=(1-t)<2,0,0> + t<-2,0,3(pi)>

x(t)=-4t+2
y(t)=0
z(t)=3(pi)t

I hope this makes sense.

i have absolutely no idea what you're doing there. The parameterization is given in the question, x(t) = 2cos t, y(t) = 2sin t, z(t) = 3t. All you need to do is find the bounds on t (using the endpoints given) and take the integral.

physicsmajor, what you've done on this probllem, with yoru parametricization, is turned your path from a helix into a straight line. you should be computing along the helix.

now, if your vector field is conservative, you can do that, the straight line approach, which makes things much easier, but, i'd imagine it's not conservative or you wouldn't be told to integrate along the helix.

thanks trance, i actually figured that out a few hours after i made that post. i guess i am having trouble finding the limits to integrate from. I believe they are from 0 < t < 3(pi), but i am not sure.

0<t is correct, but at the second endpoint you need $z=3t=3\pi$, and does $t=3\pi$ solve that?

no it does not, so its from 0<t<(pi)

sounds good to me.

you the man data!

## 1. What is a line integral problem?

A line integral problem is a mathematical concept in which the integral of a function is calculated along a given curve or path. It is used in various fields of science and engineering, such as physics, engineering, and economics, to solve problems related to work, energy, and other physical quantities.

## 2. How is a line integral problem solved?

To solve a line integral problem, the given curve or path is divided into small segments, and the integral of the function is calculated for each segment. The sum of these individual integrals gives the final value of the line integral. This process is known as the Riemann sum method.

## 3. What is the significance of line integrals in science?

Line integrals play a crucial role in various scientific fields, including physics, engineering, and economics. They are used to calculate physical quantities, such as work, energy, and magnetic flux, along curved paths or fields. They also help in studying the behavior of vector fields and determining the properties of a given curve or path.

## 4. Can line integrals be evaluated using any method other than the Riemann sum method?

Yes, apart from the Riemann sum method, there are other methods for evaluating line integrals, such as the Green's theorem and Stokes' theorem. These theorems provide a more efficient and elegant way of solving line integral problems by reducing them to simpler integrals over a specific region.

## 5. In which real-life applications are line integrals used?

Line integrals have various real-life applications, such as in the study of electric and magnetic fields, fluid mechanics, motion planning, and optimization problems. They are also used in economics to calculate the work done in a production process and in computer graphics to create realistic images by simulating light and shadow effects.

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