Computing line integral using Stokes' theorem

In summary: to rewrite \sin(2\theta) \sin \theta = \tfrac 12 (\cos \theta \sin 3\theta - \cos 3\theta \sin \theta) and then use \cos (2\theta) = 1 - 2 \sin^2 \theta to write \cos (2\theta) \sin \theta = \tfrac 12 (\sin 3\theta - \sin \theta + \sin \theta - \sin 3\theta), and then cancel the \sin 3\theta terms.
  • #1
WMDhamnekar
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Homework Statement
Compute where C is the unit circle oriented counter-clockwise
Relevant Equations

The unit normal vector to the surface is

Since can be parametrized as for (x,y) in the region then,



But the answer provided by author is How is that? Where I am wrong?
 
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  • #2
You say that the integral is taken over the curve , but you don't specify a value for ; is intended? The answer seems to be consistent with that, but your choice of surface has on . Is there a reason why you did not use the surface with unit normal ?

Looking at the integral you attempted, you should have Have another go at it.
 
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  • #3
I got the answer using z=1. If I take z=0, that answer also would be .

z is the function of (x,y).

So, I took z = 1
 
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  • #4
It is not actually necesary to use Stokes' Theorem here; assuming is constant on we have independent of .
 
  • #5
pasmith said:
It is not actually necesary to use Stokes' Theorem here; assuming is constant on we have independent of .
You can also compute the aforesaid integral without because z =0. Isn't it?

Why did you ignore in the last integral?
 
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  • #6
WMDhamnekar said:
You can also compute the aforesaid integral without because z =0. Isn't it?

The above shows that the result is the same for any .

Why did you ignore in the last integral?

The integrals of cosine and sine over a whole number of periods is zero. can be expressed as a linear combination of and and its integral therefore vanishes.
 
  • #7
pasmith said:
The above shows that the result is the same for any .
The integrals of cosine and sine over a whole number of periods is zero. can be expressed as a linear combination of and and its integral therefore vanishes.
Would you tell me how would you express as a linear combination of and

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  • #8
WMDhamnekar said:
Would you tell me how would you express as a linear combination of and
 
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  • #9
WMDhamnekar said:
Would you tell me how would you express as a linear combination of and

Use
 
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