# Flux integral using Stokes' Theorem

Calculate (via two methods) the flux integral $$\int_S (\nabla \times \vec{F}) \cdot \vec{n} dS$$ where $\vec{F} = (y,z,x^2y^2)$ and $S$ is the surface given by $z = x^2 +y^2$ and $0 \leq z \leq 4$, oriented so that $\vec{n}$ points downwards.

I have applied Stokes' Theorem and resulted with $$\oint_{\partial S} \vec{F} \cdot \vec{n} d\vec{r}.$$ Further, I calculated the normal vector $$(rsin(\theta), rcos(\theta), -r)$$ such that it is pointing downwards. I am stuck now because I cannot interpret the meaning of $\vec{r}$.

My second method would be to direcly compute curl, $$\text{curl} \vec{F} = (2x^2y - 1,-2xy^2,-1).$$ Compute the normal via gradient, $$\nabla z = (2x,2y,-1)$$ Substitute everything to get, $$\int_S \dots dS.$$

I'm stuck now and having difficulty finding the boundary in terms of $x$ and $y$. Can anyone please lend me a hand? Thanks.

• Stokes' theorem gives you a relationship between the surface integral over surface $S$ and the line integral on the boundary of $S$. Notice the boundary of your surface is just a circle of radius $2$ at $z=4$ so take your parameterization of the curve to be $r=(2\cos t,2\sin t, 4)$, and integrate $F\cdot n$ from $t=0$ to $t=2\pi$. The surface is a paraboloid so use the parameterization $r=(u\cos v,u\sin v, u^{2})$, express $\triangledown \cdot F$ in this paramterization and integrate over $S$. – Matematleta Jun 4 '15 at 12:00
• Chilango, are these two methods? I just computed the integral of $F \cdot n$ and it gave out $2 \pi$ which is close to the answer, $4 \pi$, so I probably made an error during integration. – Jester Tran Jun 4 '15 at 12:08
• In all these caes you need to choose a parameterization. In the case of the line integral, it will be a space curve. In the case of the surface, it will involve two variables $u$ and $v$. The parameterization you choose is determined by what the surface (and its boundary look like). After you get the parameterization, you just substitute into the formula, and do the line integral and/or surface integral, as required. – Matematleta Jun 4 '15 at 12:12
• Thanks, Chilango. To clarify, why does integrating the boundary give the answer? Intuitively, I don't see why it gives the surface integral. Also, with your parameterization for the surface, what is wrgon ith $r = (u,v,u^2 +v^2)$? – Jester Tran Jun 4 '15 at 12:17
• The proof of Stokes' theorem is not trivial but it's really just a computation, following your nose to verify the formula. It says that, under certain conditions, you can recover all the "information" about a surface just by looking at the boundary. Notice that Stokes Thm says also that you can use ANY surface that has the same boundary as the one you are looking at as the answer shows. It's much easier than integrating over the paraboloid as I suggested, but you will get the same result. Your paramterizaton is ok but switiching to polar coordiantes is probably easier. – Matematleta Jun 4 '15 at 12:23

One consequence of the stoke's theorem is that you can change the surface of the integration as long as you don't touch its edges (or the 'lid'). So Let's pick the most convenient surface that can be formed by the lid. That area would be the circle $x^2+y^2<4=z$, a two dimensional surface parallel to $z=0$. So the normal vector is $-\hat z$

$$\iint_{x^2+y^2<4}(\nabla \times F).(-\hat z)dS$$

$$\int _{-2}^2\int _{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}+1dydx=4\pi$$

Method 2: It is difficult to explain why this should be so, but, you should use

$$\int_{x^2+y^2=4}F.\bar {dl}$$

We switch to cylindrical. $\bar {dl}=-\hat \phi rd\phi=-(-\sin\phi \hat x+cos\phi \hat y)2d\phi$ We get:

$$\int_0^{2 \pi } 2 \left(2 \sin ^2(\phi)-4 \cos (\phi)\right) \, d\phi=4\pi$$

If you do not understand these, you need to work on simpler examples (this is not a difficult example, however it is messy and would divert your attention from understanding the point of vector calculus)

• Thanks, grdgfgr. My problem sheet, sadly, does not have any simpler examples so I'll have a look online. – Jester Tran Jun 4 '15 at 12:20
• How did you get $(\nabla \times F) \cdot (-\hat z)$ to be $+1$? – Jester Tran Jun 4 '15 at 12:23
• You found that the z component of $\nabla \times F$ was -1. Unit vector is $-\hat z$ – grdgfgr Jun 4 '15 at 12:24
• It wasn't a mistake. – grdgfgr Jun 4 '15 at 12:35
• Do you mind elaborating? Why are we looking at the $z$ component and why is the normal vector unit aswell? – Jester Tran Jun 4 '15 at 12:40