Line integral over vector field

[leehufford]

Homework Statement

Let F be a vector field F = [-y³, x³+e^-y,0]

The path in space is x² + y² = 25, z = 2.

My parametrization is r(t) = [5cos(t),5sin(t),2]

Homework Equations

Line integral is the integral of F(r(t)) * r'(t) dt, where here the asterisk * is for the dot product, not normal multiplication.

The Attempt at a Solution

The way I have always done line integrals is to express the field F in terms of r(t), or F(r(t)), then dot that with r'(t), then integrate. But this method leads to terms like cos(t)e^-5sin(t)... so I am thinking my old way of doing these line integrals isn't going to work here. If you would like me to post my attempt at a solution I will, but hopefully someone can help me do this line integral in a different way or point out an error. It's been a little over a year since I've done line integrals and I'm a little rusty. Thanks in advance for the help,

Lee

 

[jedishrfu]

I think you need to post the start of your solution namely the actual line integral and then show your parameterization of it. That would be a good start before any mentor will help. We really need to see your work.

Did you figure in the length of dr ie divide by the length of dr that was the one thing I forgot on my line Integrals but I don't see that it's the trick you're looking for?

Can you use greens theorem in a plane to solve it ie integrate the equivalent surface integral? This is why greens theorem is so valuable.

 

[leehufford]

Thanks for the reply.

The field F is [-y3, x3+e^-y,0]. The parametrization is r(t) = [5cos(t),5sin(t),2].

So dr/dt = [-5sin(t), 5cos(t), 0].

Now F(r(t)) = [-5sin³(t), 125cos³(t) + e^-5sin(t), 0].

And F(r(t)) (dot) dr/dt = something I'm not sure how to integrate. This is the reason why I am assuming there is another way. You might be on to something there with Green's theorem, but I'm not sure how to use that in this context (to parametrize or not) etc. Like I said before, when I was in multivariate calculus (vector calculus if you like) I always used the F(r(t)) (dot) dr/dt method, and it worked well for me. Thanks again,

Lee

 

[jedishrfu]

http://en.m.wikipedia.org/wiki/Green's_theorem

 

[SteamKing]

Thanks for the reply.

The field F is [-y3, x3+e-y,0]. The parametrization is r(t) = [5cos(t),5sin(t),2].

So dr/dt = [-5sin(t), 5cos(t), 0].

Now F(r(t)) = [-5sin³(t), 125cos³(t) + e^-5sin(t), 0].

And F(r(t)) (dot) dr/dt = something I'm not sure how to integrate. This is the reason why I am assuming there is another way. You might be on to something there with Green's theorem, but I'm not sure how to use that in this context (to parametrize or not) etc. Like I said before, when I was in multivariate calculus (vector calculus if you like) I always used the F(r(t)) (dot) dr/dt method, and it worked well for me. Thanks again,

Lee

I think you need to check your calculation of F(r(t)) again. I don't think you have calculated the x-component of F correctly.

In any event, once you have correctly determined the parametric form of F * dr/dt, why can't you evaluate the integral w.r.t. the parameter t? After all, F * dr/dt becomes a scalar quantity, not a vector.

 

[leehufford]

I see that I forgot to cube the -5 in the x term. So:

Integral(F(r(t)) dot dr/dt)dt = Integral(625sin⁴(t) + 625cos⁴(t) + 5e^-5sin(t))dt from 0 to 2pi. I have no idea how to integrate 5e^-5sin(t) with respect to the parameter t. I'm also a little rusty on the powers of sine/cosine, but I'm sure I can find some trick for that. Thanks for the reply!

Lee

Edit: I just noticed I didn't have -y as an exponent in the vector field (forgot superscript) in my earlier posts, sorry about that typo!

 

[vela]

Shouldn't $e^{-5\sin t}$ be multiplied by $\cos t$? It's pretty straightforward to evaluate
$$\int e^{-5\sin t}\cos t\,dt$$ with the appropriate substitution.

 

[leehufford]

Ah, yes. Thank you for noticing that. It's very late at night where I am.

To my original post, though... is there an alternative form of this line integral that would've made the calculation simpler? Green's theorem was mentioned, but I wasn't sure exactly how to apply that to this case. Thanks again,

Lee

 

[SteamKing]

Ah, yes. Thank you for noticing that. It's very late at night where I am.

To my original post, though... is there an alternative form of this line integral that would've made the calculation simpler? Green's theorem was mentioned, but I wasn't sure exactly how to apply that to this case. Thanks again,

Lee

Green's Theorem can be used to convert a double integral over a 2-D region into a line integral on the path enclosing that 2-D region. However, in this problem, you are asked to evaluate a line integral on a certain path. Sometimes, you just have to slog thru the calculations with (hopefully) no errors.

 

[vela]

You're trying to calculate
$$\oint \vec{F}(\vec{r})\cdot d\vec{r} = \oint F_x\,dx + F_y\,dy + F_z\,dz.$$ Does seeing it written that way make it easier for you to see how to apply Green's theorem?

 

[leehufford]

It's been a while since I used Green's theorem. I am assuming that F_x, F_y and F_z are just the pieces of the original field vector equation and dx, dy and dz would be the derivative of the function representing the curve and not the parametrized curve. Is that right? Thanks to everyone for all of your help.

Lee

 

[leehufford]

So I started with ∮Fxdx+Fydy+Fzdz and got the same integrand I did before using F(r(t)). I also tried using the double integral of dF₂/dx - dF₁/dy dA, where here the d's are partial derivatives (I cannot for the life of me find the symbols). The integrand is simple... 3x² + 3y². For dA, do I have to convert to polar coordinates? I my book, the integrand is actually a constant, so they have a constant times the double integral of dA, which is just the area, so they put the area of the circle in and they're done. Sorry to ask so many questions on one problem, I'm just rusty with this stuff and really want to understand it well. Thanks again for everyone who has helped so far (especially vela).

-Lee

 

[vela]

So I started with ∮Fxdx+Fydy+Fzdz and got the same integrand I did before using F(r(t)). I also tried using the double integral of dF₂/dx - dF₁/dy dA, where here the d's are partial derivatives (I cannot for the life of me find the symbols). The integrand is simple... 3x² + 3y². For dA, do I have to convert to polar coordinates?

You don't have to. You should be able to write down the integral in terms of cartesian coordinates, where dA = dx dy, and evaluate it, but it would be a lot easier to evaluate the integral after converting to polar coordinates.

 

[leehufford]

So I converted to polar coordinates to take advantage of the Pythagorean identity. My final answer was 937.5 pi. If you feel like confirming this, that would be awesome. If not, no worries, you've already helped me a ton. Thanks again for everything.

Lee

 

[vela]

That's not correct. You can kind of check your answer by calculating the line integral both ways, as a path integral and using Green's theorem. You should get the same answer either way. If the answers don't match, you made a mistake somewhere.

If you want help in tracking down where you made mistakes, you'll need to show us your work. (Oftentimes, as you type in your work, you will spot errors you overlooked before.)

 

[leehufford]

Certainly. I started with the double integral of 3x² + 3y² dA, Using x = rcos(theta), y = rsin(theta)

Now: double integral of 3r²cos²(theta) + 3r²sin²(theta) r dr d(theta). (extra r from the Jacobian)
My limits of integration are 0 to 5 for r, and 0 to 2pi for theta. (The circle is radius 5).

Now factoring out 3r², cos²(theta) + sin²(theta) =1, so, combining the 3r² with the r from the Jacobian, I have:

single integral of (3/4)r⁴ evaluated from 0 to 5 d(theta). 5⁴(3/4) is 1875/4. So

Integral of 1875/4 d(theta) from 0 to 2pi gives (1875/4)(theta) evaluated from 0 to 2pi gives (1875/2)pi or 937.5 pi.

Unfortunately, writing out my work didn't reveal to me where the mistake was made.

 

[vela]

Oh, that answer is correct. You had a different answer ($24\pi$) in your post before. I guess you fixed it while I was replying to it.

 

[leehufford]

Yea I used 2 as the radius at first for some reason, but corrected it within minutes so I didn't bother to announce an edit. You must've replied in that short window. Thank you so much for your help. I will pay it forward by trying to help some other people with their homework questions.

Lee