## anonymous 3 years ago Hi there! Please help me get started. Evaluate the line integral: $\int_c xyz^2ds$ C is the line segment from (-1, 5, 0) to (1, 6, 4) So I know that $\int_c F\bullet dr=F\bullet T ds$ and that $ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$

1. anonymous

hold on for just a second I think I got this.

2. anonymous

Nope, never mind. I don't quite know what to do with (-1,5,0) to (1, 6, 4). I'm used to seeing and equation for C.

3. anonymous

4. anonymous

I can try

5. anonymous

thank you. Am I supposed to find an equation with those points? (-1,5,0) to (1, 6, 4)

6. anonymous

You need to paramaterize the line so that you can set up your limits properly.

7. anonymous

So I'm thinking $x(t) = 2t-1$$y(t)=t+5$$z(t)=4t$ And C then simply becomes the closed interval [0,1]

8. anonymous

Why?

9. anonymous

how did you come up those equations?

10. anonymous

I made them up! I can parametrize however I want to, so long as I am careful about the limits of t. If we were doing a surface integral, we would have to parametrize with two variables.

11. anonymous

So with that we can just plug into the original integral:$\int\limits_{t=0}^{t=1}x(t)y(t)z(t)^2ds=\int\limits_{t=0}^{t=1}x(t)y(t)z(t)^2\sqrt{dx^2+dy^2+dz^2}=$$\int\limits_{t=0}^{t=1}x(t)y(t)z(t)^2\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt$ Now all you have to do is replace with the functions I gave you for x, y, and z and integrate. If it's not integrable then we can use Green's Theorem.

12. anonymous

Actually we can't even use Green's theorem rite now since we're in 3D...

13. anonymous

Can you please explain how you came up with the parametrization?

14. anonymous

Okay - Do you have a visual understanding for what $\int\limits_{C}^{}F(x, y, z)ds$ means?

15. anonymous

I imagine a line

16. anonymous

erm in this case it works but it doesnt have to be a line.

17. anonymous

18. anonymous

Well what you SHOULD imagine when you see the contour integral above is that you're adding up all the differentially small "pieces" of the contour (in this case, a line) - ds - multiplied by a weight function which has a value for every point in space (in this case, xyz^2) - F(x, y, z).

19. anonymous

So I like to think of it as linear density and we're adding up all of the differential weights of the wire $\lambda(x, y, z) ds$ The result in this scenario is the mass of the wire which is in the shape of C, our contour.

20. anonymous

Aaaah I see. Ok lets say for example that we have this line |dw:1345003327080:dw| would xyz^2 be the function of this line, and we're integration from it from (-1, 5, 0) to (1, 6, 4)

21. anonymous

Well, its actually a line in your problem, whereas you drew a curve.

22. anonymous

oh, I'm sorry

23. anonymous

And xyz^2 is just some scalar function that gives a magical value to each point in 3D-space. My physical interpretation of this is linear density, if that helps.

24. anonymous

I think I'm starting to get a picture now. So back to the parametrization. I'm sure you used some sort of formula.

25. anonymous

Well, not really. You just got to be clever. But to parameterize a line, you don't really have to be that clever.

26. anonymous

Parametrization is a whole thing to learn on its own. When you get into surface integrals, You'll have to be doing even more complex things with it.

27. anonymous

Do you need help getting how to parameterize a line?

28. anonymous

29. anonymous

Ok. So you need something which makes a movement from (-1, 5, 0) to (1, 6, 4). That path is C (note: direction matters)

30. anonymous

So I decided, because I can set up my variable t, which I invented, however I want, that I want it to go from 0 to 1. In that time, I decided that my path would be covered.

31. anonymous

x needs to change from -1 to 1 in 1 second, so I assume you can figure out what line satisfies that.

32. anonymous

similar approach for y and z.

33. anonymous

As you can see the decision to make t go from 0 to 1 is completely arbitrary. I could have made it go from 0 to pi if I wanted. But if you actually go ahead an try to do that and parametrize, you'll notice that there will be a factor of (1/pi) attached to all the t's in my x(t) y(t) and z(t) equations. The point is that this arbitrary decision is not giving any more information into the problem.

34. anonymous

That makes sense. Thank you! Gtg for now but I'll be back to complete this problem.

35. anonymous

well im not gonna wait on you I cant promise I'll be back to help.

36. TuringTest

a more methodical way to parameterize a line segment from $$P_0=(x_0,y_0,z_0)$$ to $$P_1=(x_1,y_1,z_1)$$ is with the formula$\vec r(t)=P_0+t(P_1-P_0)~;~~~~0\le t\le1$which leads to$x(t)=(1-t)x_0+tx_1$$y(t)=(1-t)y_0+ty_1~;~~~~0\le t\le1$$z(t)=(1-t)z_0+tz_1$

37. TuringTest

after simplification you still get what @vf321 has, but that is a fool-proof approach to the idea

38. anonymous

oh thank you @TuringTest. These were the formulas I was looking for!

39. TuringTest

welcome guess where I got them from mostly ? http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx

40. TuringTest

as usual :P

41. anonymous

Dang it! I forgot again!

42. TuringTest

once you remember you'll be as quick as me on these things!

43. anonymous

This is what I came up with: $\sqrt{21}(16)\left[\frac{2}{5}-\frac{9}{4}-\frac{5}{3}\right]$

44. TuringTest

ok, let me check...

45. TuringTest

I got the same :)

46. anonymous

Good, thanks =D

47. anonymous

But the parameterization's not the hard part. You have to plug into the integral and solve that still.