MathSofiya 2 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. MathSofiya

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

2. MathSofiya

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. MathSofiya

4. vf321

I can try

5. MathSofiya

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

6. vf321

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

7. vf321

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. MathSofiya

Why?

9. MathSofiya

how did you come up those equations?

10. vf321

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. vf321

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. vf321

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

13. MathSofiya

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

14. vf321

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

15. MathSofiya

I imagine a line

16. vf321

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

17. MathSofiya

18. vf321

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. vf321

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. MathSofiya

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. vf321

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

22. MathSofiya

oh, I'm sorry

23. vf321

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. MathSofiya

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. vf321

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

26. vf321

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. vf321

Do you need help getting how to parameterize a line?

28. MathSofiya

29. vf321

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. vf321

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. vf321

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

32. vf321

similar approach for y and z.

33. vf321

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. MathSofiya

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

35. vf321

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. MathSofiya

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. MathSofiya

Dang it! I forgot again!

42. TuringTest

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

43. MathSofiya

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. MathSofiya

Good, thanks =D

47. vf321

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