Stokes' theorem for the vector field $\vec{F}(x,y,z)=(y,-x,xz)$ on the surface $z=9-x^2-y^2$

Question

Verify Stokes' theorem for the vector field $\vec{F}(x,y,z)=(y,-x,xz)$ on the surface $z=9-x^2-y^2$ with $z\geq 0$.

So I've tried finding the parametrisation which is $$\vec{r}(\theta,r)=(r\cos\theta,r\sin\theta,9-r^2), \qquad r\in [0,3], \ \theta\in [0,2\pi].$$Its normal vector is $$\vec{N}=\frac{\partial \vec{r}}{\partial \theta}\times \frac{\partial \vec{r}}{\partial r}=\dots=((9r-2r^2)\cos\theta,(9r-2r^2)\sin\theta,-r)$$ But then I checked a random normal vector on Geogebra and it just doesn't seem perpendicular. Is it actually wrong?

I've been trying to look for mistakes but I can't find any. Is it actually like that?

And then for the rest of the problem, I found that $\nabla\times\vec{F}=(0,-r\sin\theta,-2)$, so \begin{align} \iint_S (\vec{\nabla}\times\vec{F})\cdot\vec{dS} &=\iint_S (\vec{\nabla}\times\vec{F})\cdot\vec{N}d\theta dr \\ &=\int_0^{2\pi}\int_0^3 (-(9-r^2)^2 r\sin\theta+2r)drd\theta \\ & =\dots \\ &=18\pi, \end{align} while the solutions says the answer is $-18\pi$. Have I done anything wrong? Is the normal vector incorrect too?

Answer 1

Given the vector field $\mathbf{F} : \mathbb{R}^3 \to \mathbb{R}^3$ of law:

$$ \mathbf{F}(x,y,z) := (y,-x,xz) $$

and given the support of a surface:

$$ \Sigma := \left\{(x,y,z) \in \mathbb{R}^3 : z=9-x^2-y^2,\,z \ge 0\right\} $$

whose natural parameterization turns out to be:

$$ \mathbf{r}(u,v) = \left(u\cos v,u\sin v,9-u^2\right), \quad \quad \text{with} \; (u,v) \in D := [0,3] \times [0,2\pi) $$

by definition, the flux of $\nabla \land \mathbf{F}$ through $\Sigma$ is equal to:

$$ \begin{aligned} \Phi & := \iint\limits_{\Sigma} (\nabla \land \mathbf{F})\cdot\mathbf{n}\,\text{d}S \\ & = \iint\limits_D (\nabla \land \mathbf{F}) \cdot \left(\mathbf{r}_u(u,v) \land \mathbf{r}_v(u,v)\right)\text{d}u\,\text{d}v \\ & = \iint\limits_D \left(0,\,u^2-9,\,-2\right) \cdot \left(2u^2\cos v,2u^2\sin v,\,u\right)\text{d}u\,\text{d}v \\ & = \int_0^3 \text{d}u \int_0^{2\pi} \left(-2u-18u^2\sin v+2u^4\sin v\right)\text{d}v \\ & = \int_0^3 -4\pi\,u\,\text{d}u \\ & = \boxed{-18\pi}. \end{aligned} $$

Then, as required by the exercise, we verify by applying the rotor theorem:

$$ \Phi = \int\limits_{\partial\Sigma^+} \mathbf{F}\cdot \mathbf{t}\,\text{d}s = \int_0^{2\pi} \left(3\sin v,-3\cos v,0\right)\cdot\left(-3\sin v,3\cos v,0\right)\text{d}v = \int_0^{2\pi} -9\,\text{d}v = \boxed{-18\pi} $$

and with this the exercise can be considered concluded.

Answer 2

Without any use of polar coordinates:

The position vector of the given surface $S$ is $\vec r=\langle x,y,9-x^2-y^2\rangle$, $D: x^2+y^2\leq 9,$ and its area element vector is $\vec{dS}=(\vec r_x\times\vec r_y)dxdy=\langle2x,2y,1\rangle dxdy$. We find the curl of the given vector field $\vec F=\langle y,-x,xz\rangle$ as $\vec\nabla\times\vec F=\langle 0,-z,-2\rangle$. Hence, the flux of the curl is equal to $$\iint_S\vec\nabla\times\vec F\cdot\vec{dS}=\iint_D(-18y+2x^2y+2y^3-2)dxdy=\iint_D(-2)dxdy=(-2)(9\pi)=-18\pi$$ where the integral of the odd degree monomials is zero due to symmetry.

The position vector of the boundary contour $\partial S$, $x^2+y^2=9, z=0$ is $\vec r=\langle x,y,0\rangle$. Hence the contour integral can be computed as $$\int_{\partial S}\vec F\cdot d\vec r=\int_{\partial S}ydx-xdy=\iint_D((-x)_x-y_y)dxdy=\iint_D(-2)dxdy=-18\pi$$ by use of the Green's theorem. Note that $\iint_Ddxdy=Area(D)=9\pi.$