Question

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 \[ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]

Answer 1

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

Answer 2

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.

Answer 3

Please help @satellite73

Answer 4

I can try

Answer 5

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

Answer 6

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

Answer 7

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]

Answer 8

Why?

Answer 9

how did you come up those equations?

Answer 10

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.

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

I imagine a line

Answer 16

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

Answer 17

Can you please elaborate?

Answer 18

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).

Answer 19

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.

Answer 20

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

Answer 21

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

Answer 22

oh, I'm sorry

Answer 23

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.

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

Do you need help getting how to parameterize a line?

Answer 28

Yes please

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

similar approach for y and z.

Answer 33

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.

Answer 34

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

Answer 35

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

Answer 36

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\[x(t)=(1-t)x_0+tx_1\]\[y(t)=(1-t)y_0+ty_1~;~~~~0\le t\le1\]\[z(t)=(1-t)z_0+tz_1\]

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

as usual :P

Answer 41

Dang it! I forgot again!

Answer 42

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

Answer 43

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

Answer 44

ok, let me check...

Answer 45

I got the same :)

Answer 46

Good, thanks =D

Answer 47

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