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 $$ \Omega : 2-\sqrt{1-z^2}<r<2+\sqrt{1-z^2}. $$ 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 $$\begin{eqnarray} 2\int_\Omega dxdydz &=&2\int_{-1}^1 \int_0^{2\pi}\int_{2-\sqrt{1-z^2}}^{2+\sqrt{1-z^2}}rdrd\theta dz\\ &=&2\pi\int_{-1}^1 r^2\big|^{2+\sqrt{1-z^2}}_{2-\sqrt{1-z^2}} \;dz\\ &=&16\pi \int_{-1}^1 \sqrt{1-z^2} \;dz\\ &=&16\pi \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \cos^2 u \;du=8\pi^2. \end{eqnarray}$$


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$$\int_{\delta\Omega}\langle f,v\rangle dS=8\pi^2$$


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.