Line integrals and Stokes' theorem

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Homework Help Overview

The discussion revolves around the application of Stokes' theorem in the context of line integrals. Participants are examining a specific problem involving parametrization of a curve and the calculation of integrals related to a vector field.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessity of using Stokes' theorem and whether the original poster's calculations align with its application. There are inquiries about the correctness of the parametrization used for the curve and suggestions to verify its accuracy against known equations.

Discussion Status

Some participants have provided guidance on the importance of verifying the parametrization and have raised questions about its correctness. There is an ongoing exploration of the implications of Stokes' theorem and the calculations involved, with no explicit consensus reached on the correct approach yet.

Contextual Notes

There are indications that the original poster's parametrization may not accurately describe the intended curve, which is a point of contention in the discussion. Participants emphasize the need for clarity in mathematical notation and the importance of proper parametrization in solving the problem.

WMDhamnekar
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Homework Statement
https://mathhelpforum.com/attachments/1667128375169-png.45062/
Relevant Equations
##F=\langle z,x,y \rangle, z = 2x + 2y -1## and ##z = x^2 + y^2 ##
1667404967015.png

My answer:
1667404950652.png
 
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Please type out your attempts rather than post an image. Posting an image makes it more difficult to read and makes it impossible to quote particular parts of it.

To answer your question: You would just compute the integrals as usual.
 
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You haven't used Stokes' theorem. You just calculated the first integral. (I assume. I didn't check your work.) Stokes' theorem tells you that if you were to calculate the other integrals, you will get the same result. So in this problem, you're really just verifying that the theorem works in this particular case, not actually using it.

So forget about using the theorem and just calculate the other integrals.
 
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Just curious, which book is this problem from?
 
vela said:
You haven't used Stokes' theorem. You just calculated the first integral. (I assume. I didn't check your work.) Stokes' theorem tells you that if you were to calculate the other integrals, you will get the same result. So in this problem, you're really just verifying that the theorem works in this particular case, not actually using it.

So forget about using the theorem and just calculate the other integrals.
I computed line integral as follows:
Curve ##D = (x-1)^2 + (y-1)^2 = 1## Now I parametrize it ##x = \cos(t) , y= \sin(t) , z = 2\cos(t) + 2\sin(t) ##

##F (x,y,z) = zdx + xdy +y dz, r(t) = (\cos(t))i + (\sin(t))j +(\cos(t) +\sin(t))k , x'(t) = - \sin(t), y'(t)= \cos(t), z'(t)= -2\sin(t) +2\cos(t) ##
##\begin{align*}\displaystyle\int_ D F\cdot dr &= \displaystyle\int_0^{2\pi} (2\cos(t)+ 2\sin(t))(-\sin(t)) +(\cos(t))(\cos(t)+(\sin(t))(-2\sin(t) + 2\cos(t))dt\\
&= \displaystyle\int_0^{2\pi} \cos^2(t)-4\sin^2(t)-4\sin(t)\cos(t) dt\\
&= -3\pi \end{align*}##
 
Last edited:
Your parametrization does not describe the correct curve.
 
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Orodruin said:
Your parametrization does not describe the correct curve.
If my parametrization of the curve is wrong, what is your suggestion for correct parametrization of the curve?
 
WMDhamnekar said:
If my parametrization of the curve is wrong, what is your suggestion for correct parametrization of the curve?
I believe it would be more instructive if you first verified that it is indeed wrong and then think about what you could do to correct it. What do you get if you insert the parametrization in, for example, the equation for the paraboloid ##z = x^2 + y^2## or the plane ##z = 2x + 2y - 1##?

Edit: Or indeed into ##(x-1)^2 + (y-1)^2 = 1## ...

Edit 2: On a TeXnical note: Please use \cos and \sin for the trigonometric functions in LaTeX. Just writing sin and cos makes LaTeX typest it as if it was a product of ##s##, ##i##, and ##n## and ##c##, ##o##, and ##s##, respectively. Note the difference:
$$
sin x \leftrightarrow \sin x \qquad cos x \leftrightarrow \cos x
$$
 

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