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

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.

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 $$2$$.

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 !

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 $$\Omega : 2-\sqrt{1-z^2} Hence $$\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}$$
For a) it is best to turn to cylindrical coordinates. Note that for a relation $$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 and $$-1 therefore by the definition of torus we conclude that $$2-\sqrt{1-z^2}and $$5-z^2-4\sqrt{1-z^2}so we can write$$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$$