Calculating the Line Integral of F over C: Stokes' Theorem and Symmetry

[greg_rack]

Homework Statement

Use Stokes' theorem to evaluate $\iint_{\textbf{S}}^{}curl \textbf{F}\cdot d\textbf{S}$, with $\textbf{F}=<xyz, xy, x^2yz>$ and S consisting of the top and the four sides(no bottom) of the cube with vertices $(\pm 1,\pm 1,\pm 1)$

Relevant Equations

Computable form of a line integral and Stokes' theorem

From Stokes we know that $\iint_{\textbf{S}}^{}curl \textbf{F}\cdot d\textbf{S}=\int_{C}^{}\textbf{F}\cdot d\textbf{r}$.
Now, we can calculate the surface integral of the curl of F by calculating the line integral of F over the curve C.
The latter ends up being 0(I calculated it parametrizing all the segments of the cube, that was more than tedious).
I was wondering whether this result could've been determined/deduced even prior to performing all the calculations, maybe due to symmetry of the surface and/or to its "interaction" with this particular vector field; with my very poor abstraction skills, I couldn't find anything... I thought probably rewriting $\int_{C}^{}\textbf{F}\cdot d\textbf{r}$ in the scalar form $\int_{C}^{}xyz\ dx + xy \ dy+ x^2yz \ dz$ could have helped, but I still don't see the trick, if any.

 

[Orodruin]

The latter ends up being 0(I calculated it parametrizing all the segments of the cube, that was more than tedious).

What do you mean by this? All of the edges of the cube? Those are not the boundary of your surface. Only the bottom square is.

You could also use Stokes' theorem to change the integral surface to the bottom of the cube and only do that integral (which is trivially zero because of the anti-symmetry in $y$ of the $z$-component of the field).

 

[greg_rack]

What do you mean by this? All of the edges of the cube? Those are not the boundary of your surface. Only the bottom square is.

Yes sorry, I was looking at the wrong graph whilst typing. Using the bottom square is what I actually did.

You could also use Stokes' theorem to change the integral surface to the bottom of the cube and only do that integral (which is trivially zero because of the anti-symmetry in $y$ of the $z$-component of the field).

I don't get why it should be zero by considering the surface as the bottom one though. Could you elaborate on this?

 

[Orodruin]

Yes sorry, I was looking at the wrong graph whilst typing. Using the bottom square is what I actually did.

I don't get why it should be zero by considering the surface as the bottom one though. Could you elaborate on this?

The surface normal is given by $\vec e_z$ and the z-component of the curl is given by
$$
(\nabla \times \vec F)_z = \partial_x F_y - \partial_y F_x = y - xz.
$$
Therefore, the integral is given by
$$
\int_S (y - xz) dx\, dy = \int_S (y + x) dx\, dy
$$
where $S$ is the bottom square and we have used that $z = -1$ on this surface. The surface is symmetric around $y = 0$ while $y$ is anti-symmetric and so the integral of the first term is zero. The same argument holds for the second term but for the $x$ coordinate.

Alternatively, you can just consider that the four sides of the square are along the x/y directions. For the sides in the x-direction, the relevant component of $\vec F$ is the x-component $xyz$, which is antisymmetric about $x = 0$ while the intervals of integration are symmetric. The same argument goes for the integrals in the y-direction.

 

[greg_rack]

where $S$ is the bottom square and we have used that $z = -1$ on this surface. The surface is symmetric around $y = 0$ while $y$ is anti-symmetric and so the integral of the first term is zero. The same argument holds for the second term but for the $x$ coordinate.

That is very clear now, thank you so much!

Alternatively, you can just consider that the four sides of the square are along the x/y directions. For the sides in the x-direction, the relevant component of $\vec F$ is the x-component $xyz$

I got lost again here: for the sides of which the normal is oriented in the x direction, aren't of importance the y and z components of $\vec F$, instead of the x? Aren't those causing the "curl" we're interested in dotting with the normal?

 

[Orodruin]

I got lost again here: for the sides of which the normal is oriented in the x direction, aren't of importance the y and z components of F→, instead of the x? Aren't those causing the "curl" we're interested in dotting with the normal?

What you quoted here regards the circulation integral, not the surface integral.

 

[greg_rack]

What you quoted here regards the circulation integral, not the surface integral.

Ohhh, crap, I'm sorry for making so much confusion!
To try and clarify things up then, Stokes' th. relates the circulation around a boundary curve to the surface integral of the circulation of the vector field(true?) of interest; what we did here was arrive to "standard" scalar surface integral(s), starting from solving a circulation one over the same surface.

But indeed, isn't the integral we have started from, $\iint_{S}^{}curl\textbf{F}\cdot d\textbf{S}$, giving us somehow a measure of the "curl living inside the surface"? And isn't this curl caused, in the case of a side with it's normal pointing in the +x, by the y and z components of $\vec F$?