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

This Question is Closed

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0hold on for just a second I think I got this.

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Nope, 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.

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Please help @satellite73

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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2You need to paramaterize the line so that you can set up your limits properly.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2So I'm thinking \[x(t) = 2t1\]\[y(t)=t+5\]\[z(t)=4t\] And C then simply becomes the closed interval [0,1]

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0how did you come up those equations?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2I 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.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2So 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.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Actually we can't even use Green's theorem rite now since we're in 3D...

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Can you please explain how you came up with the parametrization?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Okay  Do you have a visual understanding for what \[\int\limits_{C}^{}F(x, y, z)ds\] means?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2erm in this case it works but it doesnt have to be a line.

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Can you please elaborate?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Well 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).

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2So 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.

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Aaaah 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)

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Well, its actually a line in your problem, whereas you drew a curve.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2And 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.

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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Well, not really. You just got to be clever. But to parameterize a line, you don't really have to be that clever.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Parametrization 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.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Do you need help getting how to parameterize a line?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Ok. So you need something which makes a movement from (1, 5, 0) to (1, 6, 4). That path is C (note: direction matters)

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2So 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.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2x needs to change from 1 to 1 in 1 second, so I assume you can figure out what line satisfies that.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2similar approach for y and z.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2As 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.

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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2well im not gonna wait on you I cant promise I'll be back to help.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2a 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\[x(t)=(1t)x_0+tx_1\]\[y(t)=(1t)y_0+ty_1~;~~~~0\le t\le1\]\[z(t)=(1t)z_0+tz_1\]

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

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0oh thank you @TuringTest. These were the formulas I was looking for!

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

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Dang it! I forgot again!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2once you remember you'll be as quick as me on these things!

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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2But the parameterization's not the hard part. You have to plug into the integral and solve that still.
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