Surface Integral of a Vector Field Over a Torus

Question

Let $S$ be the surface obtained after rotating $(x-2)^2+z^2=1$ around the $z$-axis. What is the value of $$\int_{S}\mathbf{F\cdot n } dA$$ where $$\mathbf{F}=(x+\sin(yz), y+e^{x+z}, z-x^2\cos(y))$$

Answer 1

Observe that

$\partial \mathbf F_x / \partial x = \partial \mathbf F_y / \partial y = \partial \mathbf F_z / \partial z = 1, \tag{1}$

which readily implies that

$\nabla \cdot \mathbf F = 3, \tag{2}$

so if we use Gauss's divergence theorem we obtain

$\int_{S}\mathbf{F\cdot n } dA = \int_\Omega \nabla \cdot \mathbf F dV = \int_\Omega 3dV = 3\text{Vol}(\Omega), \tag{3}$

where $\Omega$ is the region bounded by $S$ and $\text{Vol}(\Omega)$ is the volume of $\Omega$.

We have

$\text{Vol}(\Omega) = 4\pi^2, \tag{4}$

as is explained in this wikipedia page on the torus. Such being the case,

$\int_{S}\mathbf{F\cdot n } dA = 12\pi^2. \tag{5}$

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

Answer 2

Let S be the torus and D be the inner of the torus. Then by Stokes formula, we have $\int_S F.n dA=\int_D div(F)dVol$ Then the following is easy to calculate.