Line Integral - Stokes theorem

[AwesomeTrains]

Homework Statement

Hello
I was given the vector field: $\vec A (\vec r) =(−y(x^2+y^2),x(x^2+y^2),xyz)$ and had to calculate the line integral $\oint \vec A \cdot d \vec r$ over a circle centered at the origin in the $xy$-plane, with radius $R$, by using the theorem of Stokes.

Another thing, when calculating $∇× \vec A$ I used Cartesian coordinates but when doing the integration I changed to polar coordinates, is that okay? Or do I have to do it all in Cartesian coordinates, because I did curl in Cartesian coordinates?

Homework Equations

Stokes equation:

The Attempt at a Solution

My result is: $2R^4π$ Is that correct ? I evaluated both sides and got the same.

Kind regards Alex

 

[vela]

Another thing, when calculating $∇× \vec A$ I used Cartesian coordinates but when doing the integration I changed to polar coordinates, is that okay? Or do I have to do it all in Cartesian coordinates, because I did curl in Cartesian coordinates?

It's fine to change coordinates to make evaluating the integral easier whenever you want.

My result is: $2R^4π$ Is that correct ? I evaluated both sides and got the same.

I didn't actually do the integral on paper so I might have missed something, but the answer looks right to me.

 

[AwesomeTrains]

Okay, thanks for the reply :)

 

[rude man]

I got 4πR⁴.

del x A = 4R² k
dA = R²dθ/2 k
∫2R²R²dθ from 0 to 2π = 4πR⁴.

 

[AwesomeTrains]

Thanks for doing the calculation but why is $dA=\frac{R^2}{2} d \theta$. My curl vector is the same.
I was using $dxdy=rdrd \theta$

 

[rude man]

Thanks for doing the calculation but why is $dA=\frac{R^2}{2} d \theta$. My curl vector is the same.
I was using $dxdy=rdrd \theta$

I took the differential area as that of a triangle of area 1/2 bh = 1/2 R Rrdθ = 1/2 R²dθ
b = base
h = height
Proof: A = ∫dA = 1/2 R²∫dθ from 0 to 2π = πR².

You can do it your way too! Just requires integrating twice instead of once:
dA = r dr dθ
A = ∫∫r dr dθ= R²/2 (2θ) = πR².
So how about showing how you got 2πR⁴? One of us must have stumbled a bit in the dA department.

 

[vela]

I got 4πR⁴.

del x A = 4R² k
dA = R²dθ/2 k
∫2R²R²dθ from 0 to 2π = 4πR⁴.

The curl is $\nabla\times\vec{A} = 4r^2 \hat{k}$. It depends on variable $r$, not constant $R$, so you can't use that area element.

 

[AwesomeTrains]

Thanks for the replies guys. I'll write out my calculation then you can see where the mistake lies.
$\nabla \times \vec A = (xz - 0)i + (0 - yz)j + (3x^2+y^2 - (-x^2 - 3y^2))k$
Chose a vektor normal to the circle:
$d \vec f = 0i+0j+k$
The dot product is then:
$\nabla x \vec A \cdot d \vec f = 4x^2+4y^2 = 4r^2$
Then I integrated over the circle, and changed to spherical coordinates where I used the area element: $dA =rdrd \phi$:
$\int_{0}^{2\pi} \int_{0}^R 4r^2rdrd \phi = \int_{0}^{2Pi} \int_{0}^R 4r^3drd \phi = 2R^4\pi$

 

[rude man]

I agree, I should not have made r constant in the curl expression.

 

[AwesomeTrains]

Is $2R^4\pi$ then the correct result?

 

[vela]

Yes.

 

[rude man]

Is $2R^4\pi$ then the correct result?

Yes, I say it is, and you were right to use r dr dθ as the elemental area.

 

[AwesomeTrains]

Okay, thanks for the help, will be back with some more problems :)

 

[rude man]

Okay, thanks for the help, will be back with some more problems :)

Please do! I'll try not to mess up next time ...