Computing line integral using Stokes' theorem

  • #1
WMDhamnekar
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Homework Statement:
Compute ## \oint_C x^2z dx + 3xdy -y^3 dz ## where C is the unit circle ##x^2 +y^2 =1## oriented counter-clockwise
Relevant Equations:
##\int_C f\cdot dr = \iint\limits_\Sigma curl(f)\cdot n d\sigma##
##curl([x^2z, 3x , -y^3],[x,y,z]) =[-3y^2 ,x^2,3]##
The unit normal vector to the surface ##z(x,y)=x^2+y^2## is ##n= \frac{-2xi -2yj +k}{\sqrt{1+4x^2 +4y^2}}##
##[-3y^2,x^2,3]\cdot n= \frac{-6x^2y +6xy^2}{\sqrt{1+4x^2 + 4y^2}}##
Since ##\Sigma## can be parametrized as ##r(x,y) = xi + yj +(x^2 +y^2)k## for (x,y) in the region ##D = \{(x,y): x^2 + y^2 \leq 1\}## then,

##\begin{align*}\iint\limits_{\Sigma} (curl f) \cdot n d\sigma &= \iint\limits_{D} curl f \cdot n \parallel \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y}\parallel dA\\
&=\iint\limits_{D} 3 - 2x^2y + 6xy^2 dA\\
&= \int_0^{2*\pi}\int_0^1 (3 - 2r^3 \cos^2{(\theta)}\sin{(\theta)} + 6r^3\cos{(\theta)}\sin^2{(\theta)})r dr d\theta \\
&= 3\pi
\end {align*}##

But the answer provided by author is ##3\pi## How is that? Where I am wrong?
 
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Answers and Replies

  • #2
pasmith
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You say that the integral is taken over the curve [itex]x^2 + y^2 = 1[/itex], but you don't specify a value for [itex]z[/itex]; is [itex]z = 0[/itex] intended? The answer seems to be consistent with that, but your choice of surface has [itex]z = 1[/itex] on [itex]x^2 + y^2 = 1[/itex]. Is there a reason why you did not use the surface [itex]\{ x^2 + y^2 \leq 1, z = 1 \}[/itex] with unit normal [itex]\mathbf{k}[/itex]?

Looking at the integral you attempted, you should have [tex]
\iint_D 3 - 2x^2 y + 6xy^2\,dA = \int_0^{2\pi} \int_0^1 (3 - 2r^3 \cos^2 \theta \sin \theta + 6r^3 \cos \theta \sin^2 \theta)r\,dr\,d\theta.[/tex] Have another go at it.
 
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  • #3
WMDhamnekar
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I got the answer ##3\pi## using z=1. If I take z=0, that answer also would be ## 3\pi##.

z is the function of (x,y). ## x^2 + y^2 =1##

So, I took z = 1
 
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  • #4
pasmith
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It is not actually necesary to use Stokes' Theorem here; assuming [itex]z[/itex] is constant on [itex]C[/itex] we have [tex]
\begin{split}
\int_C x^2z\,dx + 3x\,dy - y^3\,dz &= \int_C x^2z\,dx + 3x\,dy \\
&= \int_0^{2\pi} -z \cos^2 \theta \sin \theta + 3\cos^2 \theta\,d\theta \\
&= \int_0^{2\pi} -z\tfrac{1}{2}(1 + \cos 2\theta) \sin \theta + \tfrac{3}{2}(1 + \cos 2 \theta)\,d\theta \\
&= \int_0^{2\pi} \tfrac 32\,d\theta \\
&= 3\pi\end{split}[/tex] independent of [itex]z[/itex].
 
  • #5
WMDhamnekar
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It is not actually necesary to use Stokes' Theorem here; assuming [itex]z[/itex] is constant on [itex]C[/itex] we have [tex]
\begin{split}
\int_C x^2z\,dx + 3x\,dy - y^3\,dz &= \int_C x^2z\,dx + 3x\,dy \\
&= \int_0^{2\pi} -z \cos^2 \theta \sin \theta + 3\cos^2 \theta\,d\theta \\
&= \int_0^{2\pi} -z\tfrac{1}{2}(1 + \cos 2\theta) \sin \theta + \tfrac{3}{2}(1 + \cos 2 \theta)\,d\theta \\
&= \int_0^{2\pi} \tfrac 32\,d\theta \\
&= 3\pi\end{split}[/tex] independent of [itex]z[/itex].
You can also compute the aforesaid integral without ##x^2zdx## because z =0. Isn't it?

Why did you ignore ##1+\cos{2\theta}## in the last integral?
 
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  • #6
pasmith
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You can also compute the aforesaid integral without ##x^2zdx## because z =0. Isn't it?

The above shows that the result is the same for any [itex]z[/itex].

Why did you ignore ##1+\cos{2\theta}## in the last integral?

The integrals of cosine and sine over a whole number of periods is zero. [itex]\cos 2\theta \sin \theta[/itex] can be expressed as a linear combination of [itex]\sin 3\theta[/itex] and [itex]\sin \theta[/itex] and its integral therefore vanishes.
 
  • #7
WMDhamnekar
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The above shows that the result is the same for any [itex]z[/itex].



The integrals of cosine and sine over a whole number of periods is zero. [itex]\cos 2\theta \sin \theta[/itex] can be expressed as a linear combination of [itex]\sin 3\theta[/itex] and [itex]\sin \theta[/itex] and its integral therefore vanishes.
Would you tell me how would you express ##\cos2\theta \sin\theta## as a linear combination of [itex] \sin3\theta [/itex] and [itex] \sin \theta ?[/itex]

1670684221234.png
 
  • #8
renormalize
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Would you tell me how would you express ##\cos2\theta \sin\theta## as a linear combination of [itex] \sin3\theta [/itex] and [itex] \sin \theta ?[/itex]
$$\cos2\theta\sin\theta=\left(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\right)\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)=\frac{e^{3i\theta}-e^{-3i\theta}+e^{-i\theta}-e^{i\theta}}{4i}=\frac{\sin3\theta-\sin\theta}{2}$$
 
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  • #9
pasmith
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Would you tell me how would you express ##\cos2\theta \sin\theta## as a linear combination of [itex] \sin3\theta [/itex] and [itex] \sin \theta ?[/itex]

Use [tex]\sin(2\theta \pm \theta) = \sin(2\theta) \cos \theta \pm \sin(\theta) \cos (2 \theta)[/tex]
 
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