# Computing line integral using Stokes' theorem

• WMDhamnekar
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.

#### WMDhamnekar

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

Looking at the integral you attempted, you should have $$\iint_D 3 - 2x^2 y + 6xy^2\,dA = \int_0^{2\pi} \int_0^1 (3 - 2r^3 \cos^2 \theta \sin \theta + 6r^3 \cos \theta \sin^2 \theta)r\,dr\,d\theta.$$ Have another go at it.

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WMDhamnekar
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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It is not actually necesary to use Stokes' Theorem here; assuming $z$ is constant on $C$ we have $$\begin{split} \int_C x^2z\,dx + 3x\,dy - y^3\,dz &= \int_C x^2z\,dx + 3x\,dy \\ &= \int_0^{2\pi} -z \cos^2 \theta \sin \theta + 3\cos^2 \theta\,d\theta \\ &= \int_0^{2\pi} -z\tfrac{1}{2}(1 + \cos 2\theta) \sin \theta + \tfrac{3}{2}(1 + \cos 2 \theta)\,d\theta \\ &= \int_0^{2\pi} \tfrac 32\,d\theta \\ &= 3\pi\end{split}$$ independent of $z$.

pasmith said:
It is not actually necesary to use Stokes' Theorem here; assuming $z$ is constant on $C$ we have $$\begin{split} \int_C x^2z\,dx + 3x\,dy - y^3\,dz &= \int_C x^2z\,dx + 3x\,dy \\ &= \int_0^{2\pi} -z \cos^2 \theta \sin \theta + 3\cos^2 \theta\,d\theta \\ &= \int_0^{2\pi} -z\tfrac{1}{2}(1 + \cos 2\theta) \sin \theta + \tfrac{3}{2}(1 + \cos 2 \theta)\,d\theta \\ &= \int_0^{2\pi} \tfrac 32\,d\theta \\ &= 3\pi\end{split}$$ independent of $z$.
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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WMDhamnekar said:
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 $z$.

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. $\cos 2\theta \sin \theta$ can be expressed as a linear combination of $\sin 3\theta$ and $\sin \theta$ and its integral therefore vanishes.

pasmith said:
The above shows that the result is the same for any $z$.
The integrals of cosine and sine over a whole number of periods is zero. $\cos 2\theta \sin \theta$ can be expressed as a linear combination of $\sin 3\theta$ and $\sin \theta$ and its integral therefore vanishes.
Would you tell me how would you express ##\cos2\theta \sin\theta## as a linear combination of $\sin3\theta$ and $\sin \theta ?$

WMDhamnekar said:
Would you tell me how would you express ##\cos2\theta \sin\theta## as a linear combination of $\sin3\theta$ and $\sin \theta ?$
$$\cos2\theta\sin\theta=\left(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\right)\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)=\frac{e^{3i\theta}-e^{-3i\theta}+e^{-i\theta}-e^{i\theta}}{4i}=\frac{\sin3\theta-\sin\theta}{2}$$

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WMDhamnekar
WMDhamnekar said:
Would you tell me how would you express ##\cos2\theta \sin\theta## as a linear combination of $\sin3\theta$ and $\sin \theta ?$

Use $$\sin(2\theta \pm \theta) = \sin(2\theta) \cos \theta \pm \sin(\theta) \cos (2 \theta)$$

WMDhamnekar

## 1. What is Stokes' theorem?

Stokes' theorem is a mathematical theorem that relates the line integral of a vector field over a closed curve to the surface integral of the curl of the vector field over the surface enclosed by the curve.

## 2. How is Stokes' theorem used in computing line integrals?

Stokes' theorem allows us to compute a line integral by converting it into a surface integral. This can be useful when the curve is difficult to integrate over, but the surface is easier to work with.

## 3. What are the steps for computing a line integral using Stokes' theorem?

The steps for computing a line integral using Stokes' theorem are:
1. Identify the curve and the vector field.
2. Determine the orientation of the curve.
3. Find the curl of the vector field.
4. Choose a surface that encloses the curve.
5. Compute the surface integral of the curl over the chosen surface.
6. The result of the surface integral is equal to the line integral over the curve.

## 4. What are the advantages of using Stokes' theorem in computing line integrals?

Using Stokes' theorem can make computing line integrals easier and more efficient, as it allows us to convert the integral into a surface integral. This can be particularly useful when the curve is complex or when the vector field is difficult to integrate.

## 5. Are there any limitations to using Stokes' theorem in computing line integrals?

Stokes' theorem can only be used for closed curves and surfaces, so it is not applicable for open curves or surfaces. Additionally, the vector field must be continuously differentiable over the surface enclosed by the curve for the theorem to hold.