Line integrals and Stokes' theorem

In summary, the conversation is about calculating integrals using parametrization and verifying the use of Stokes' theorem in a specific case. The expert suggests forgetting about the theorem and just calculating the integrals, and also provides a suggestion for correcting an incorrect parametrization. The conversation also includes a note on using correct formatting in LaTeX.
  • #1
WMDhamnekar
MHB
376
28
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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  • #2
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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  • #3
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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  • #4
Just curious, which book is this problem from?
 
  • #5
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:
  • #6
Your parametrization does not describe the correct curve.
 
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  • #7
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?
 
  • #8
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:
$$
sin x \leftrightarrow \sin x \qquad cos x \leftrightarrow \cos x
$$
 

Related to Line integrals and Stokes' theorem

What is a line integral?

A line integral is a mathematical concept used in vector calculus to calculate the work done by a force along a path. It involves integrating a vector function along a curve or line.

What is the purpose of a line integral?

The purpose of a line integral is to calculate the work done by a force along a specific path. It is also used in many other applications such as calculating the flux of a vector field through a surface.

What is Stokes' theorem?

Stokes' theorem is a fundamental theorem in vector calculus that relates line integrals to surface integrals. It states that the line integral of a vector field over a closed curve is equal to the surface integral of the curl of that vector field over any surface bounded by the curve.

How is Stokes' theorem used in real-world applications?

Stokes' theorem is used in many real-world applications, such as fluid dynamics, electromagnetism, and engineering. It allows for the calculation of surface integrals, which are important in determining quantities like fluid flow and electromagnetic flux.

What are some common misconceptions about line integrals and Stokes' theorem?

One common misconception is that line integrals can only be calculated over straight lines. In reality, they can be calculated over any curve or path. Another misconception is that Stokes' theorem only applies to two-dimensional surfaces, when in fact it can also be applied to three-dimensional surfaces.

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