## MathSofiya Group Title 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 $ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$ 2 years ago 2 years ago

1. MathSofiya Group Title

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

2. MathSofiya Group Title

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 Group Title

4. vf321 Group Title

I can try

5. MathSofiya Group Title

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

6. vf321 Group Title

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

7. vf321 Group Title

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 Group Title

Why?

9. MathSofiya Group Title

how did you come up those equations?

10. vf321 Group Title

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 Group Title

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 Group Title

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

13. MathSofiya Group Title

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

14. vf321 Group Title

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

15. MathSofiya Group Title

I imagine a line

16. vf321 Group Title

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

17. MathSofiya Group Title

Can you please elaborate?

18. vf321 Group Title

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 Group Title

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 Group Title

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 Group Title

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

22. MathSofiya Group Title

oh, I'm sorry

23. vf321 Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

Do you need help getting how to parameterize a line?

28. MathSofiya Group Title

29. vf321 Group Title

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 Group Title

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 Group Title

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

32. vf321 Group Title

similar approach for y and z.

33. vf321 Group Title

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 Group Title

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

35. vf321 Group Title

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

36. TuringTest Group Title

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 Group Title

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

38. MathSofiya Group Title

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

39. TuringTest Group Title

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

40. TuringTest Group Title

as usual :P

41. MathSofiya Group Title

Dang it! I forgot again!

42. TuringTest Group Title

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

43. MathSofiya Group Title

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

44. TuringTest Group Title

ok, let me check...

45. TuringTest Group Title

I got the same :)

46. MathSofiya Group Title

Good, thanks =D

47. vf321 Group Title

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