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

This Question is Closed

MathSofiyaBest ResponseYou've already chosen the best response.0
hold on for just a second I think I got this.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Please help @satellite73
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
thank you. Am I supposed to find an equation with those points? (1,5,0) to (1, 6, 4)
 one year ago

vf321Best ResponseYou've already chosen the best response.2
You need to paramaterize the line so that you can set up your limits properly.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
So I'm thinking \[x(t) = 2t1\]\[y(t)=t+5\]\[z(t)=4t\] And C then simply becomes the closed interval [0,1]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
how did you come up those equations?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
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.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
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.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Actually we can't even use Green's theorem rite now since we're in 3D...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Can you please explain how you came up with the parametrization?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Okay  Do you have a visual understanding for what \[\int\limits_{C}^{}F(x, y, z)ds\] means?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
erm in this case it works but it doesnt have to be a line.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Can you please elaborate?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
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).
 one year ago

vf321Best ResponseYou've already chosen the best response.2
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
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)
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Well, its actually a line in your problem, whereas you drew a curve.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
And xyz^2 is just some scalar function that gives a magical value to each point in 3Dspace. My physical interpretation of this is linear density, if that helps.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I think I'm starting to get a picture now. So back to the parametrization. I'm sure you used some sort of formula.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Well, not really. You just got to be clever. But to parameterize a line, you don't really have to be that clever.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
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.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Do you need help getting how to parameterize a line?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Ok. So you need something which makes a movement from (1, 5, 0) to (1, 6, 4). That path is C (note: direction matters)
 one year ago

vf321Best ResponseYou've already chosen the best response.2
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.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
x needs to change from 1 to 1 in 1 second, so I assume you can figure out what line satisfies that.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
similar approach for y and z.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
That makes sense. Thank you! Gtg for now but I'll be back to complete this problem.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
well im not gonna wait on you I cant promise I'll be back to help.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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\[\vec r(t)=P_0+t(P_1P_0)~;~~~~0\le t\le1\]which leads to\[x(t)=(1t)x_0+tx_1\]\[y(t)=(1t)y_0+ty_1~;~~~~0\le t\le1\]\[z(t)=(1t)z_0+tz_1\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
after simplification you still get what @vf321 has, but that is a foolproof approach to the idea
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
oh thank you @TuringTest. These were the formulas I was looking for!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
welcome guess where I got them from mostly ? http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Dang it! I forgot again!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
once you remember you'll be as quick as me on these things!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
This is what I came up with: \[\sqrt{21}(16)\left[\frac{2}{5}\frac{9}{4}\frac{5}{3}\right]\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
ok, let me check...
 one year ago

vf321Best ResponseYou've already chosen the best response.2
But the parameterization's not the hard part. You have to plug into the integral and solve that still.
 one year ago
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