Derive the volume of a torus using the divergence theorem and Fubini's theorem.

Question

Given is the vector field $f=(x,y,0)$ and the set $\Omega :=((\sqrt{x^2+y^2}-2)^2+z^2 <1)$ (which is obviously a torus)

I'm asked to :
a) Calculate the outward unit normal field $v: \delta \Omega \to \mathbb{R}^3$
b) Calculate $\int_{\delta\Omega}\langle f,v\rangle dS$ using the Divergence Theorem. Hint : It is expected that you again derive the volume of a torus. For that, calculate first the area enclosed of the set $A_z:=((\sqrt{x^2+y^2}-2)^2\leq1-z^2) $for$ -1\leq z \leq 1$ and then use Fubini's theorem.

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a) So here I think that the outward unit field vector is just the gradient of $f$ normalized. So the gradient is $(1,1,0)$ and when normalized, it is $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$.
b) Now, here I'm not really sure. For the area enclosed, I'm obviously supposed to use polar coordinates. So $(r-2)^2\leq1-z^2$. So $\pm(r-2)=\pm\sqrt{(1-z^2)}$. So $r=2+\sqrt{1-z^2}$ or $r=2-\sqrt{1-z^2}$. Now, the area is $\int_{2-\sqrt{1-z^2}}^{2+\sqrt{1-z^2}}\int_{0}^{2\pi}1rdrd\theta=\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{0}^{2\pi}1rdrd\theta$. And so the volume will be $\int_{-1}^{1}\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{0}^{2\pi}1rdrd\theta$
And our divergence is $2$, so we will just multiply the result by $2$.

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Now, my questions are:
Is my procedure for a) correct ?
Is my procedure for b) correct and if yes, am I now just supposed to evaluate that integral ?

Thanks for you help !

Answer 1

For $a)$, we can calculate $v:\delta\Omega \to \mathbb{R}^3$ using parametrization of $\delta \Omega$. Note that $v$ depends only on $\Omega$, and is completely irrelevant to the given vector field $f$. (So your calculation is not valid.) Let $\sqrt{x^2+y^2} = r$. Then we can write $$ \Omega : 2-\sqrt{1-z^2}<r<2+\sqrt{1-z^2}. $$ Hence $\Omega = \Omega_1 -\Omega_2$ where $\Omega_1:r<2+\sqrt{1-z^2}$ and $\Omega_2: r<2-\sqrt{1-z^2}$. We can calculate $v$ on $\delta\Omega_1$ as follows: $$ \nabla_{x,y,z}(r-2-\sqrt{1-z^2})=(\frac{x}{r},\frac{y}{r},\frac{z}{\sqrt{1-z^2}})', $$and by normalizing $$ v(x,y,z) = \sqrt{1-z^2}(\frac{x}{r},\frac{y}{r},\frac{z}{\sqrt{1-z^2}})'. $$ We can check that $v$ is outward. Similarly for $\delta\Omega_2$, $$ \nabla_{x,y,z}(r-2+\sqrt{1-z^2})=(\frac{x}{r},\frac{y}{r},-\frac{z}{\sqrt{1-z^2}})' $$ and $$ v(x,y,z) = -\sqrt{1-z^2}(\frac{x}{r},\frac{y}{r},-\frac{z}{\sqrt{1-z^2}})'. $$ (The sign is reversed.)

For $b)$, it is true that $\int_{\delta\Omega}\langle f,v\rangle dS=\int_\Omega \nabla\cdot f\;dxdydz=2\text{vol}(\Omega).$ We can calculate this using cylindrical coordinate $$ (x,y,z)=(r\cos\theta, r\sin\theta, z). $$ Then, $dxdydz = rdrd\theta dz$ and the given integral becomes $$\begin{eqnarray} 2\int_\Omega dxdydz &=&2\int_{-1}^1 \int_0^{2\pi}\int_{2-\sqrt{1-z^2}}^{2+\sqrt{1-z^2}}rdrd\theta dz\\ &=&2\pi\int_{-1}^1 r^2\big|^{2+\sqrt{1-z^2}}_{2-\sqrt{1-z^2}} \;dz\\ &=&16\pi \int_{-1}^1 \sqrt{1-z^2} \;dz\\ &=&16\pi \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \cos^2 u \;du=8\pi^2. \end{eqnarray}$$

Answer 2

For a) it is best to turn to cylindrical coordinates. Note that for a relation $f(x,y,z)=0$ the gradient $\nabla f(x,y,z)$ is always normal to it, therefore if the equation of torus is $$f(r,\phi ,z)=(r-2)^2+z^2-1=0$$then we have$$\vec v={\nabla f(r,\phi , z)\over |\nabla f(r,\phi , z)|_2}={(r-2,0,z)\over \sqrt{(r-2)^2+z^2}}$$

b) $$\int_{\delta\Omega}\langle f,v\rangle dS{=\int_V \nabla\cdot f(x,y,z)dV\\=\int_V 2dV\\=2\int_{(r-2)^2+z^2<1}rdrd\phi dz}$$note that $1<r<3$ and $-1<z<1$ therefore by the definition of torus we conclude that $$2-\sqrt{1-z^2}<r<2+\sqrt{1-z^2}$$and $$5-z^2-4\sqrt{1-z^2}<r^2<5-z^2+4\sqrt{1-z^2}$$so we can write$$2\int_{(r-2)^2+z^2<1}rdrd\phi dz{=\int_{(r-2)^2+z^2<1}dr^2d\phi dz\\=2\pi \int_{(r-2)^2+z^2<1}dr^2 dz\\=2\pi \int_{-1}^1\int_{r^2=5-z^2-4\sqrt{1-z^2}}^{r^2=5-z^2+4\sqrt{1-z^2}}dr^2 dz\\=16\pi\int_{-1}^1\sqrt{1-z^2}dz\\=8\pi ^2}$$therefore$$\int_{\delta\Omega}\langle f,v\rangle dS=8\pi^2$$