# Without Stokes's Theorem – Calculate $\iint_S \operatorname{curl} \mathbf{F} \cdot\; d\mathbf{S}$ for $\mathbf{F} = yz^2\mathbf{i}$ – 2013 10C

2013 10C. Consider the bounded surface S that is the union of $x^2 + y^2 = 4$ for $−2 \le z \le 2$
and $(4 − z)^2 = x^2 + y^2$ for $2 \le z \le 4.$ Sketch the surface.
Use suitable parametrisations for the two parts of S to verify Stokes’s Theorem for
for $\mathbf{F} = (yz^2,0,0)$. picture

Herein, I enquire only about directly computing $\iint_S (\nabla × F )· d\mathbf{S}$.
Denote the $2 \le z \le 4$ cone P, and the $-2 \le z \le 2$ cylinder C. I use only the first paragraph of this.

Then $\mathbf{\nabla × F} = (0, 2yz, -z^2)$,
$\iint_P (\nabla × F ) · d\mathbf{S} = \iint_{x^2 + y^2 \le z^2, z = 2} (\nabla × F ) \cdot \color{darkred}{\mathbf{n}} \, dA \\ = \iint_{x^2 + y^2 \le z^2, z = 2} (♦, ♦, \underbrace{-z^2}_{=-4} ) \cdot \color{darkred}{(0, 0,-1)} \, dA = \iint_{x^2 + y^2 \le z^2, z = 2} 4 dA = 4\pi(2)^2$.

♦ denote objects that don’t need to be computed because they’re dot-producted with 0.

$\large{2.}$ To ellya especially, is it necessary to parameterise the P piece? Comparing my work to yours, I see that we differ only by a negative sign? Yet my work has far fewer steps. Does it not function?
How would one determine that the correct normal vector is $\color{green}{\mathbf{n} = (0, 0, 1)}$?

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I would parameterize from the start, so on $C$ $-2\le z\le 2$ and $x^2+y^2=4$. This is a cylinder of radius $2$ centred at the origin with height $4$.

So let $x=2\cos\phi,y=2\sin\phi$ where $0\le\phi\le 2\pi$ now we parametrise our surface $C$ as $\sigma (\phi,z)=(2\cos\phi,2\sin\phi,z)$, and now $F=(2z^2\sin\phi,0,0)$.

Here we let $z$ increase from -2 to 2 so with this orientation, the normal $n=\sigma_\phi\times\sigma_z = \left| \begin{array}{ccc} i & j & k \\ -2\sin\phi & 2\cos\phi & 0 \\ 0 & 0 & 1 \end{array} \right|=(2\cos\phi,2\sin\phi,0)$

So we have $\int\int_C(\nabla\times F)\cdot dS=\int_0^{2\pi}\int_{-2}^2(\nabla\times F)\cdot(2\cos\phi,2\sin\phi,0)dzd\phi$

Now $\nabla\times F=\left| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz^2 & 0 & 0 \end{array} \right|=(0,\frac{\partial}{\partial z}(yz^2),-z^2)=(0,2yz,-z^2)=(0,4z\sin\phi,-z^2)$

So $\iint_C(\nabla\times F) \cdot d\mathbf{S}=\int_0^{2\pi}\int_{-2}^2(0,4z\sin\phi,-z^2)\cdot(2\cos\phi,2\sin\phi,0)dz~d\phi \\ =\int_0^{2\pi}\int_{-2}^2 8z\sin^2 \phi \, dz \, d\phi =4 \int_0^{2\pi} (1-\cos2\phi) \, d\theta \; \int_{-2}^2 z \, dz =4 \color{#009900 }{[z^2]^2_{-2}} \; [\phi-\frac{1}{2}\sin 2\phi]^{2\pi}_0=\color{#009900 }{0}$

I took a different route because the “$\frac{\partial z}{\partial y}\times\frac{\partial z}{\partial x}$” didn’t make sense to me. They are not vectors.

To do the $P$ integral I would also parametrize, but this time you are integrating over a cone.

Parametrising $P$, here we let $\sigma(\phi,z)=((4-z)\cos\phi,(4-z)\sin\phi,z)$, here

$n=\sigma_\phi\times\sigma_z =\left| \begin{array}{ccc} i & j & k \\ (z-4)\sin\phi & (4-z)\cos\phi & 0 \\ -\cos\phi & -\sin\phi & 1 \end{array} \right|=((4-z)\cos\phi,(4-z)\sin\phi,4-z)$

so $\int\int_P (\nabla\times F)\cdot dS=\int_0^{2\pi}\int_2^4(0,2z(4-z)\sin\phi,-z^2)\cdot((4-z)\cos\phi,(4-z)\sin\phi,4-z)dzd\phi$

$=\int_0^{2\pi}\int_2^42z(4-z)^2\sin^2\phi-z^2(4-z)dz d\phi$

$=\int_0^{2\pi}\int_2^4z(4-z)^2(1-\cos 2\phi)-(4z^2-z^3)dzd\phi$

$=\frac{20}{3}(\phi-\frac{1}{2}\cos 2\phi|_0^{2\pi})-2\pi(\frac{44}{3})$

$=2\pi(\frac{20}{3}-\frac{44}{3})=-2\pi(\frac{24}{3})=-8(2)\pi=-16\pi$

1. Cross producting two vectors produces a vector normal to both of them. Since $z$ is oriented positively, $\sigma_\phi\times\sigma_z$ produces the outward normal which is what we want, taking the other order gives us an inward normal.

2. In my belief you got lucky with your computation, since the integral you performed was on the base of the cone which is not actually part of the surface.

I believe your main issue was with the reasoning for taking the normal to be $\sigma_\phi\times\sigma_z$ over $\sigma_z\times\sigma_\phi$. This may seem strange Please refer here.

Actually had a go at this a few weeks ago but kept getting the wrong answer. I’ll just do the cone, since I think Ellya’s calculation is correct for the cylinder. However for the cone, if we do a line integral around the base of cone we get an answer of $-16\pi$, which disagrees with his result. Here’s my proof for the cone:

Let $I$ be the surface integral over the cone with normal $\mathbf n$ taken in an upwards direction.

$$I= \iint_P (\nabla \times \mathbf F)\cdot d\mathbf S = \iint_P (0,2yz,-z^2)\cdot \mathbf n d\mathbf S$$

Now we will use the formula for a flux integral $\iint_S \mathbf F\cdot \mathbf n \, d\mathbf S = \iint_D (F_1, F_2, F_3) \cdot \color{green}{(-\partial_x f, \partial_y f, 1)} \, dA = \iint_D (-F_1 \partial_x f – F_2 \partial_y f +F_3 ) \, dxdy$,
where $D$ is a projection of $S$ onto the x-y plane, and $f(x,y)$ is the equation of the surface. For us, $f_x’ = \frac{-x}{\sqrt{x^2 + y^2}}$ and $f_y’ = \frac{-y}{\sqrt{x^2 + y^2}}$. So,

$$I = \iint_D \frac{2y^2 z}{\sqrt{x^2 + y^2}} – z^2 dxdy$$

Noting that $z = 4-\sqrt{x^2 + y^2}$ and then letting $x=r\cos\theta$, $y=r\sin \theta$,

$$I= \int_0^{2\pi} \int_0^2 8r^2 \sin^2 \theta -2r^3 \sin^2 \theta -16 r +8r^2 -r^3 drd\theta = -16\pi$$