Computing line integral using Stokes' theorem

[WMDhamnekar]

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?

 

[pasmith]

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 <br /> \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.

 

[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

 

[pasmith]

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

 

[WMDhamnekar]

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

 

[pasmith]

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.

 

[WMDhamnekar]

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 ?

 

[renormalize]

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

$$\cos2\theta\sin\theta=\left(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\right)\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)=\frac{e^{3i\theta}-e^{-3i\theta}+e^{-i\theta}-e^{i\theta}}{4i}=\frac{\sin3\theta-\sin\theta}{2}$$

 

[pasmith]

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)