APM346-2018S > Quiz-B

Quiz-B P1

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Victor Ivrii:
 If the surface is a surface of revolution $z=u(r)$ with $r^2=x^2+y^2$, then
\begin{equation}
S=2\pi\int_{D} \sqrt{1+u_r^2}\,rdr .
\tag{1}
\end{equation}

Write Euler-Lagrange equation and solve it (find general solution).

Jingxuan Zhang:
Define
$$L(r,u_r)=r\sqrt{1+u_r^2}.$$
Then E.-L. $$(L_{u_r})_r=L_u$$ gives
\begin{equation}\label{1-1}\Bigl(\frac{ru_r}{\sqrt{1+u_r^2}}\Bigr)_r=0,\end{equation}
from which we derive
$$\frac{ru_r}{\sqrt{1+u_r^2}}=A\implies u_r^2(r^2-A^2)=A^2\implies u_r=\frac{A}{\sqrt{r^2-A^2}}\implies u=A\cosh^{-1}(r/A)+B.$$

Victor Ivrii:
Typical blinder: after $L_u=0$ instead of integrating $(L_{u'})'=0\implies L_{u'}=c$ started to really differentiate $L_{u'}$, getting to $u''$...

Error $u'L_{u'}-L=c$ : it would work if $L$ did not depend on the argument $r$, but here it did not depend on function $u$.

Bonus:
The same problem (surface of revolution) but now argument is  $z$ (used to be $u$) and the function is $r=R(z)$.
Then
$$
S=2\pi \int_{z_0}^{z_1} \sqrt{1+R'^2} R\,dz.
$$
Solve it! (the answer should be indeed the same as in the original problem)

Andrew Hardy:
answer to bonus question

My Lagrangian is $$ \sqrt{(1+R'^2)}R $$
so my Euler Lagrangian equations are
 
  $$ \frac{\partial L}{\partial R'} = \frac{ R R'}{\sqrt{1+R'^2}} $$
$$ \frac{\partial L}{\partial R} = \sqrt{1+R'^2} $$
$$  \frac{\partial L}{\partial z}  = \frac{R R''} {\sqrt{1+R'^2}}  + \frac{R'^2}{\sqrt{1+R'^2}} - \frac{R'^2R''R}{(1+R'^2)^{3/2}}$$



I'm not sure how to get to the solution you're looking for though. The argument isn't explicitly in the Lagrangian so I don't know how I'm going to integrate.

Victor Ivrii:
Definitely not. What is E-L equation? Does $L$ depend on $z$ (you denoted it by $x$) explicitly?

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