anonymous
  • anonymous
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 \[ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hold on for just a second I think I got this.
anonymous
  • 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.
anonymous
  • anonymous
Please help @satellite73

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anonymous
  • anonymous
I can try
anonymous
  • anonymous
thank you. Am I supposed to find an equation with those points? (-1,5,0) to (1, 6, 4)
anonymous
  • anonymous
You need to paramaterize the line so that you can set up your limits properly.
anonymous
  • 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]
anonymous
  • anonymous
Why?
anonymous
  • anonymous
how did you come up those equations?
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
Actually we can't even use Green's theorem rite now since we're in 3D...
anonymous
  • anonymous
Can you please explain how you came up with the parametrization?
anonymous
  • anonymous
Okay - Do you have a visual understanding for what \[\int\limits_{C}^{}F(x, y, z)ds\] means?
anonymous
  • anonymous
I imagine a line
anonymous
  • anonymous
erm in this case it works but it doesnt have to be a line.
anonymous
  • anonymous
Can you please elaborate?
anonymous
  • 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).
anonymous
  • 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.
anonymous
  • 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)
anonymous
  • anonymous
Well, its actually a line in your problem, whereas you drew a curve.
anonymous
  • anonymous
oh, I'm sorry
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
Well, not really. You just got to be clever. But to parameterize a line, you don't really have to be that clever.
anonymous
  • 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.
anonymous
  • anonymous
Do you need help getting how to parameterize a line?
anonymous
  • anonymous
Yes please
anonymous
  • 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)
anonymous
  • 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.
anonymous
  • anonymous
x needs to change from -1 to 1 in 1 second, so I assume you can figure out what line satisfies that.
anonymous
  • anonymous
similar approach for y and z.
anonymous
  • 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.
anonymous
  • anonymous
That makes sense. Thank you! Gtg for now but I'll be back to complete this problem.
anonymous
  • anonymous
well im not gonna wait on you I cant promise I'll be back to help.
TuringTest
  • 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\[x(t)=(1-t)x_0+tx_1\]\[y(t)=(1-t)y_0+ty_1~;~~~~0\le t\le1\]\[z(t)=(1-t)z_0+tz_1\]
TuringTest
  • TuringTest
after simplification you still get what @vf321 has, but that is a fool-proof approach to the idea
anonymous
  • anonymous
oh thank you @TuringTest. These were the formulas I was looking for!
TuringTest
  • TuringTest
welcome guess where I got them from mostly ? http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx
TuringTest
  • TuringTest
as usual :P
anonymous
  • anonymous
Dang it! I forgot again!
TuringTest
  • TuringTest
once you remember you'll be as quick as me on these things!
anonymous
  • anonymous
This is what I came up with: \[\sqrt{21}(16)\left[\frac{2}{5}-\frac{9}{4}-\frac{5}{3}\right]\]
TuringTest
  • TuringTest
ok, let me check...
TuringTest
  • TuringTest
I got the same :)
anonymous
  • anonymous
Good, thanks =D
anonymous
  • anonymous
But the parameterization's not the hard part. You have to plug into the integral and solve that still.

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