Calculate $\int_S v \cdot d\vec{n}$ for subset of a sphere.

Question

Let $0 < d < 1$ and let $S$ be the part of $\mathbb{S}^2 \subseteq \mathbb{R}^3$ for which $z > -d$.

Let furthermore $f(x,y,z) = (-y,x,0)^T$ and $v=\operatorname{rot}(f) \implies v = (0,0,2)$.

Let the "normal-vector" $\vec{n}$ points out of the surface.

Calculate the value of $\int_S v \cdot d\vec{n}$ (looking for the flux through $S$) using:

1. Directly
2. Stokes Theorem
3. Gauss Theorem

The following is an example image if $d = 0.5$:

My approach

Direct calculation No idea.. I would need your help here.

Using Stokes

We know that $\int_\gamma \vec{f} \cdot d\vec{s} = \int_S \operatorname{rot}(f) d\vec{n}$

we, therefore, need to parameterize the lower border of $S$: Using trigonometry and polar coordinates we find:

$$(\sqrt{1-d^2}\cos(t), \sqrt{1-d^2}\sin(t), -d)$$

$$\int_0 ^{2\pi} = \begin{pmatrix} -\sqrt{1-d^2}\sin(t) \\ \sqrt{1-d^2}\cos(t) \\ 0 \end{pmatrix} \cdot \begin{pmatrix} -\sqrt{1-d^2}\sin(t) \\ \sqrt{1-d^2}\cos(t) \\ 0 \end{pmatrix} dt = 2\pi (1-d^2)$$ Is this correct?

Using Gauss

We know that $S$ and a "circle for the floor", let's call this circle $C$ engulf a part of a sphere $Q$ with $\partial Q = S \cup C$

We can now calculate $\int_S \vec v \cdot d\vec{n} +\int_C \vec v \cdot d\vec{n}= \int_{\partial Q}\vec{v} d\vec{n} = \int_Q \operatorname{div}(\vec{v})d\mu$

However $\operatorname{div}(\vec{v}) = 0 \implies \int_Q \operatorname{div}(\vec{v})d\mu = 0$ $\implies \int_S \vec v \cdot d\vec{n} = -\int_C \vec v \cdot d\vec{n}$

I am unsure how to continue here and if what I did is even correct.

Thank you so much for helping.

Answer 1

Direct integration:

Note that $\hat{n}=\frac{x\hat{i}+ y\hat{j}+z\hat{k}}{r^2}$. Since $r=1$ on your surface, $\vec{v}\cdot \hat{n}=\frac{2z}{r}=\sin \theta$ where $\theta$ is the angle of a point on S above the equatorial plane.

Then the integral is $$\int_S 2z dA = 2 \int_{\alpha}^{\frac{\pi}{2}} \int_0^{2\pi}z \cos \theta d \varphi d\theta=\pi \cos (2\theta)|_{\alpha}^{\frac{\pi}{2}}=2\pi(1-d^2) $$ where $\alpha = \sin^{-1}(-d)$.

Gauss:

As you have noted, $\operatorname{div}(\mathbf{v})=0$. Let V represent the volume between S and the circular disk "floor" D. Then by Gauss,

$$\int_V\operatorname{div}(\mathbf{v})=\int_{S+D} \mathbf{v} \cdot \hat{n}$$

Since the LHS is zero, we get $$\int_{S} \mathbf{v} \cdot \hat{n}= -\int_{D} \mathbf{v} \cdot \hat{n}$$ Because $\mathbf{v}$ is constant over D, so that integral is just the product of the area of D and the flux out (-2). Noting that the radius of D is $\sqrt{1-d^2}$, we get $$\int_{S} \mathbf{v} \cdot \hat{n}= -\int_{D} \mathbf{v} \cdot \hat{n}= -\pi (1-d^2)(-2)=2\pi(1-d^2)$$