Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A curve $y(x)$ of length $2a$ is drawn between the points (0,0) and (a,0) in such a way that the solid obtained by rotating the curve about the $x$-axis has the largest possible volume. Find $y\left(\frac{a}{2}\right)$.

share|improve this question
5  
What have you tried? –  Tara B Mar 29 '13 at 22:58
    
I get an ugly differential equation. –  copper.hat Mar 30 '13 at 0:20
2  
I am curious to know if there is a nice answer to this that doesn't involve solving the ODE from Euler Lagrange... –  copper.hat Mar 30 '13 at 16:01
3  
Where is this question from? is this homework? a home-exam? –  nbubis Jul 2 '13 at 20:04

3 Answers 3

Maximize $\displaystyle\int_0^a\pi y^2\,\mathrm{d}x$ while maintaining $\displaystyle\int_0^a\sqrt{1+y'^2}\,\mathrm{d}x=2a$. In other words:

find the $y$ so that $$ \int_0^a y\,\delta y\,\,\mathrm{d}x=0\tag{1} $$ for every $\delta y$ so that $$ \int_0^a\frac{y'\,\delta y'}{\sqrt{1+y'^2}}\,\mathrm{d}x=0\tag{2} $$ Integrating $(2)$ by parts gives $$ \int_0^a\frac{y''}{\left(1+y'^2\right)^{3/2}}\,\delta y\,\mathrm{d}x=0\tag{3} $$ $(1)$ and $(3)$ imply that there is a constant $b$ so that $$ -\frac{y''}{\left(1+y'^2\right)^{3/2}}=2b^2y\tag{4} $$ $(4)$ describes the critical $y$:

$\hspace{3.5cm}$enter image description here

Multiply $(4)$ by $y'$, integrate, and solve for $y'$ $$ (1+y'^2)^{-1/2}=b^2y^2+c^2\\[6pt] y'=\pm\sqrt{\frac1{(b^2y^2+c^2)^2}-1}\tag{5} $$ Since $y''$ is always one sign, it must be negative. Thus, $y'$ is positive until $y$ reaches maximum, then it is negative until $y=0$.

The length of the curve is $$ \begin{align} &2\int_0^{\frac{\sqrt{1-c^2}}{b}}\sqrt{1+y'^2}\frac{\mathrm{dy}}{y'}\\ &=2\int_0^{\frac{\sqrt{1-c^2}}{b}}\frac{\mathrm{d}y}{\sqrt{1-(b^2y^2+c^2)^2}}\\[6pt] &=\frac{2\sqrt{1-c^2}}b\int_0^1\frac{\mathrm{d}y}{\sqrt{1-((1-c^2)y^2+c^2)^2}}\\ &=\frac2{b\sqrt{c^2+1}}\mathrm{EllipticK}\left(\frac{c^2-1}{c^2+1}\right)\tag{6} \end{align} $$ The width of the curve (distance between $x$-intercepts) is $$ \begin{align} &2\int_0^{\frac{\sqrt{1-c^2}}{b}}\frac{\mathrm{dy}}{y'}\\ &=2\int_0^{\frac{\sqrt{1-c^2}}{b}}\frac{(b^2y^2+c^2)\,\mathrm{d}y}{\sqrt{1-(b^2y^2+c^2)^2}}\\[6pt] &=\frac{2\sqrt{1-c^2}}b\int_0^1\frac{((1-c^2)y^2+c^2)\,\mathrm{d}y}{\sqrt{1-((1-c^2)y^2+c^2)^2}}\\ &=\frac2{b\sqrt{c^2+1}}\left((c^2+1)\mathrm{EllipticE}\left(\frac{c^2-1}{c^2+1}\right)-\mathrm{EllipticK}\left(\frac{c^2-1}{c^2+1}\right)\right) \tag{7} \end{align} $$ Solving for the $c$ that makes $(6)$ the double of $(7)$ yields $$ c=0.26829153853907013879\tag{8} $$ For that $c$, $(7)$ gives the width of the curve to be $$ a=\frac1b1.29027749675509851046\tag{9} $$ Equations $(4)$ and $(5)$ say that $y$ increases from $0$ to its maximum and returns to $0$ symmetrically. Thus, $y(a/2)$ is the maximum of $y$. Equation $(5)$ says that $y'=0$ when $$ y=\frac{\sqrt{1-c^2}}{b}\tag{10} $$ Answer: Therefore, combining $(8)$, $(9)$, and $(10)$ yields $$ y(a/2)=0.74661285489041830537\,a\tag{11} $$


Volume: Since we are maximizing the volume, why not compute the volume: $$ \begin{align} &2\int_0^{\frac{\sqrt{1-c^2}}{b}}\pi y^2\frac{\mathrm{dy}}{y'}\\ &=2\pi\int_0^{\frac{\sqrt{1-c^2}}{b}}\frac{y^2(b^2y^2+c^2)\,\mathrm{d}y}{\sqrt{1-(b^2y^2+c^2)^2}}\\[6pt] &=2\pi\left(\frac{\sqrt{1-c^2}}b\right)^3\int_0^1\frac{y^2((1-c^2)y^2+c^2)\,\mathrm{d}y}{\sqrt{1-((1-c^2)y^2+c^2)^2}}\\ &=\frac{2\pi\sqrt{c^2+1}}{3b^3}\left(\mathrm{EllipticK}\left(\frac{c^2-1}{c^2+1}\right)-c^2\mathrm{EllipticE}\left(\frac{c^2-1}{c^2+1}\right)\right) \tag{12} \end{align} $$ Combining $(8)$, $(9)$, and $(12)$ gives $$ \text{Volume}=1.21275710583444588043\,a^3 $$

share|improve this answer
1  
Thanks for this nice answer confirming my brute force approach. +1 –  Shuhao Cao Jul 3 '13 at 16:16
    
Must the extremal $y$ be differentiable on the whole interval? –  zyx Jul 8 '13 at 2:07
    
@zyx: $y'$ is assumed to be in $L^1$ since the curve is assumed to be rectifiable. However, to apply the machinery of the calculus of variations, a certain amount of smoothness is assumed. –  robjohn Jul 8 '13 at 13:29
    
I meant, in (4) is the conclusion that $b$ is constant or piecewise constant, and in the second case whether there can be a piecewise-smooth $y$ that does better. There are examples in Gelfand-Shilov's book on calculus of variations where this kind of issue can make a difference to the answer. I think the smooth extremum you found is the correct answer for this problem but do not see how to prove it without a separate argument about smoothing corners in ways that reduce surface area more so (in a suitable sense) than volume. –  zyx Jul 8 '13 at 16:01
    
@zyx: $b$ is constant since we have assumed that for any $\delta y$ so that $(3)$ holds, $(1)$ also holds. Note that there is a $k$ so that $$ \int_0^a\left(y-k\frac{y''}{\left(1+y'^2\right)^{3/2}}\right)\frac{y''}{\left(1+‌​y'^2\right)^{3/2}}\,\mathrm{d}x=0 $$ if $\frac{y''}{\left(1+y'^2\right)^{3/2}}$ is identically $0$, any $k$ will do. Otherwise, $$ k=\left.\int_0^ay\frac{y''}{\left(1+y'^2\right)^{3/2}}\,\mathrm{d}x \middle/\int_0^a\left(\frac{y''}{\left(1+y'^2\right)^{3/2}}\right)^2\,\mathrm{d}‌​x\right. $$ –  robjohn Jul 8 '13 at 16:35

Summary of my answer:

  1. Some references for the classical isoperimetric problem of maximizing the area under a curve given a fixed arc length.

  2. Reformulating the revoluted volume problem into a non-constraint optimization problem.

  3. Example of why maximizing the revolution volume is not equivalent to maximizing the area under a curve given a fixed arc length.

  4. Numerical approach and the approximation to $y(a/2)$ is $0.74a$ with the approximated maximized revoluted volume $1.212108 a^3$.


Let's say that the curve $y$ that maximizes the revolution volume subject to length $l$:

$$\text{Maximize }V[y] = \pi\int^a_0 y(x)^2\,dx\quad \text{subject to}\quad \int^a_0 \sqrt{1+\big(y'(x)\big)^2}\,dx = l.\tag{$\dagger$}$$

This is a modified version of the famous Dido's problem.


Isoperimetric approach for maximizing the AREA: If we allow the curve being defined as a parametric curve rather than a well-defined function's graph $(x,y(x))$, then the Isoperimetric theorem says:

The area surrounded by the straight line and a curve is maximum when the curve is a part of the circle with arclength fixed.

Similar argument could be made by a physical argument using membrane between ideal gas and vacuum, please refer to Mark's answer here. By the theorem, $$ y\left(\frac{a}{2}\right) = r(1-\cos\theta),\tag{A} $$ where $r$ and $\theta \in (0,\pi)$ satisfies: $$2r\theta = l,\quad \text{and}\quad 2r\sin\theta = a. $$ When $\theta = \pi/2$, it is a half circle and $l = \pi a/2$. When $l = 2a$, there exists a $\pi/2<\theta<\pi$, and $$ y\left(\frac{a}{2}\right) = \frac{a}{\theta}\left(1 + \sqrt{1-\frac{\theta^2}{4}}\right), \quad \text{where } \sin \theta = \frac{\theta}{2}. $$


Calculus of variation: If we are not allowed to have something like this. So we parametrize the curve $t\mapsto (x(s),y(s))$ for $s \in (0,l)$, so that $$ \int^a_0\sqrt{1+\big(y'(x)\big)^2}\,dx = \int^l_0 1 ds= l , \quad \text{and } (x')^2 + (y')^2 = 1. $$ Now $$ V[y] = \pi\int^a_0 y(x)^2\,dx = \pi\int^l_0 y^2 x'\,ds = 2\pi\int^l_0 y^2 \sqrt{1-(y')^2}\,ds. \tag{$\ddagger$} $$ Now the problem is transformed to an unconstraint problem, unfortunately the E-L equation is as follows: $$ 2y\sqrt{1-(y')^2} + \frac{d}{ds}\left(\frac{y^2 y'}{\sqrt{1-(y')^2}}\right) = 0.\tag{1} $$ Using another perturbation approach, letting $$\lim_{\epsilon\to 0}\frac{d}{d\epsilon} V[y+\epsilon v] =0$$ the variational problem is as follows: for any smooth $v$ $$ \int^l_0 \frac{2yv - 2y (y')^2 - y^2 y' v'}{\sqrt{1-(y')^2}} \,ds = 0.\tag{2} $$ Both (1) and (2) are subject to the boundary condition $y(0) = y(l) = 0$.


Equivalence?! No!

Question: Is problem $(\dagger)$ equivalent to $$\text{Maximize }A[y] = \pi\int^a_0 y(x)\,dx\quad \text{subject to}\quad \int^a_0 \sqrt{1+\big(y'(x)\big)^2}\,dx = l\,?$$

The answer is no. Let $l = \pi a/2$, then the largest area for a fixed arc length by formula (A) is achieved when $y$ is a half circle by the isoperimetric theorem: $$ y = \sqrt{\left(\frac{a}{2}\right)^2-\left(x-\frac{a}{2}\right)^2} \implies \frac{d}{dx}\frac{y'}{\sqrt{1+(y')^2}} = -\frac{d}{dx}\frac{x-a/2}{a/2} = -\frac{2}{a}.$$ So that the E-L equation of the Lagrange multiplier for maximizing the area under $y$ is satisfied. $$ 1-\frac{d}{dx} \frac{\lambda y'}{\sqrt{1+(y')^2}} =0, \quad \text{ where }\lambda = -\frac{a}{2}. $$ However, there does not exists any single $\lambda \in \mathbb{R}$ such that the E-L equation for the volume maximization problem holds if $y$ is a half-circle: $$ y-\frac{d}{dx} \frac{\lambda y'}{\sqrt{1+(y')^2}} =0. $$ According to WolframAlpha, equation of this kind should somehow involve the elliptic integral.


Numerical approach: After some furious trial and error for different functions satisfying the E-L equations for problem $(\dagger)$ and $(\ddagger)$, I am suggesting (as of now), a closed-form solution is impossible. So we turn to some numerical approach setting $a = 1$, $l=2$. Equi-distant sample points are: $(x_n,y_n)$, with $0\leq n\leq N+1$, step size $h$, $(x_0,y_0) = (0,0)$, and $(x_N,y_N) = (1,0)$.

Volume's approximation $V_h$ can be found by $$ V_h = \sum^{N}_{n=0}\pi h\frac{(y_n +y_{n+1})^2}{4} = \mathbf{y}^T A\mathbf{y}, $$ with $\mathbf{y} = (y_1,\ldots,y_n)$, and $A$ is a tri-diagonal matrix $$ A =\frac{\pi h}{4}\begin{pmatrix} 2 & 1 & & 0\\ 1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & 1 & 2 \end{pmatrix}, $$ The arclength constraint is: $$ S_h = \sum^{N}_{n=0} \sqrt{(y_{n+1}-y_n)^2 + h^2} = 2. $$ This is a very hard nonlinear quadratic programming problem. Using $101\times 101$ grid points on $[0,1]^2$ (step size $h=10^{-2}$), $y(1/2)\approx 0.74$, blue curve is the curve maximizing the revoluted volume, while the red curve is a scaled circle that has the same arclength ($\mathrm{error}<O(10^{-2})$)

curve

In computing the approximation, I found that the amount of work looking for all curves that satisfies the constraint is insanely large which is $O(2^{100})$! So here I admit I cheated the big way by assuming the following:

  • The curve achives maximum at $x=1/2$, non-descreasing from $x=0$ to $x=1/2$, and is symmetric about $x=1/2$.

Hence now the problem turns into a pure integer programming, we want to find the increment step vectors for the first half $(v_1,\ldots,v_{N/2})$, with each entry non-negative. The curve's $y$-coordinates can be then written as the following with a lower-triangular matrix $B$:

$$ \begin{pmatrix}y_1\\ \vdots \\y_{N/2} \end{pmatrix} =h\begin{pmatrix} 1 & 0 & 0& 0\\ 1 & 1 & \ddots & 0\\ 1& \ddots & 1 & 0 \\ 1 &1 & 1 & 1 \end{pmatrix}\begin{pmatrix}v_1\\ \vdots \\v_{N/2} \end{pmatrix}. $$ Let $v = (v_1,\ldots,v_{N/2},v_{N/2},\ldots,v_1)^T$, then $$ y = \begin{pmatrix}B & 0\\ E & -B \end{pmatrix}v = Cv, $$ where $E$ is a matrix with all its entries being $1$. Notice the boundary condition is already incorporated for we consider only the increment (we can set $v_0 = 0$), now the problem changes to:

$$\text{Maximize } V_h[v]= v^T C^T ACv \\ \text{subject to } v_n \geq 0, \sum^{N/2}_{n=1} v_n \leq N,\;\text{and } S_h[v] = \sum^{N/2}_{n=0} 2h\sqrt{ v_n^2 + 1} = 2. \tag{$\star$}$$

In real simulation, I used Monte Carlo for $(\star)$ (I am not an expert in nonlinear optimization, there should be a more state-of-art approach rather than this brute force approach). The numerical procedures roughly are:

  1. Generating an enormous amount of random numbers in scaled Dirichlet distribution for integer (non-negative numbers with sum fixed) with a fixed sum from $N-1,N-2,\ldots, N/2$. This fixed sum is the total number of steps $y$ goes up in the grids from $x=0$ to $x=1/2$. Notice $N/2$ is the highest number of increment steps in a perfect circle, for the arclength is $2>\pi/2$ so we wanna at least choose increment step more than that heuristically, unless there exists some zig-zag curves that maximize the volume (zig-zag won't pass the arclength constraint though).

  2. Checking which increment step vectors are satisfying the arclength constraint within certain relative error $\epsilon$ (I chose $0.05$ in simulation).

  3. Compute the volume then choose the maximum one.

  4. Decreasing the $\epsilon$ for the arclength, perturb the increment step vector, repeat.

The eventual $y(1/2)= h\sum_{n=1}^{N/2} v_n$ for that maximizer increment step vector. What I got was $$ y(1/2)\approx 0.74, \text{ with } V[y] \approx 1.212108. $$


Lastly the problem reminded me of a youtube video long time ago: Using soap bubble to find the Steiner points for four fixed nodes.

share|improve this answer
2  
Your plot looks like the plot of the differential equation in my answer. Did you get a value for $y(a/2)$? –  robjohn Jul 3 '13 at 5:55
    
I agree that there may be no closed form solution for $y$, but given the constants of integration involved in getting the differential equations for $y$, there is a closed form in terms of elliptic functions for the length and the width of the curve. –  robjohn Jul 3 '13 at 6:02
    
@robjohn I got $y(1/2)\approx 0.74$, the grid points were $100\times 100$ on $[0,1]^2$, so the error was at least $O(10^{-2})$, I am glad this brute force solution agrees with your answer. –  Shuhao Cao Jul 3 '13 at 12:27
    
This is quite an impressive numerical approach. I imagine you could simply refine the mesh and get more precision. (+1) –  robjohn Jul 3 '13 at 13:38
    
@robjohn Thanks. About being more precise...Hmmm, I used Monte Carlo, and I am not an expert in numerical nonlinear optimization. $100\times 100$ grid size has already a huge amount of computation...(I felt guilty for borrowing dept's server doing this...) Please see my edit for my approach (Wasn't so confident yesterday for I cheated by assuming I know what the shape is gonna be like at first). –  Shuhao Cao Jul 3 '13 at 16:16

When rotating about $x-axis$ maximum volume is equal to maximum area in 2D. Therefore the question can be given as max area given the length. The area to be maximized can be defined as

$$A[y]=\int_{x_1}^{x_2}y\ dx$$ subject to $$2a=\int_{x_1}^{x_2}\sqrt{1+y'\ ^2} dx$$

We can use Lagrange multiplier such as

$$H=y+\lambda \sqrt{1+y'\ ^2}$$ and $$\frac{\partial H}{\partial y}-\frac{d}{dx}\bigg(\frac{\partial H}{\partial y'}\bigg)=0$$ $$1-\frac{d}{dx}\bigg(\frac{\lambda y'}{\sqrt{1+y'\ ^2}}\bigg)=0$$ $$\Rightarrow \frac{x-\alpha}{\lambda}=\frac{y'}{\sqrt{1+y'\ ^2}}$$ If $\bar x=x-\alpha$ $$\frac{\bar x}\lambda=\frac{y'}{\sqrt{1+y'\ ^2}}\Rightarrow y'=\frac{dy}{dx}=\frac{\bar x}{\sqrt{\lambda^2-\bar x^2}}$$ which can be solved for $y(x)$ such that $$y(x)=-\sqrt{\lambda^2-\bar x^2}+\beta=\beta-\sqrt{\lambda^2-\big(x-\alpha\big)^2}$$ The equation must satisfy following conditions $$(0,0)\Rightarrow 0=\beta-\sqrt{\lambda^2-\big(-\alpha\big)^2}$$ $$(a,0)\Rightarrow 0=\beta-\sqrt{\lambda^2-\big(a-\alpha\big)^2}$$ $$2a=\int_{0}^{a}\sqrt{1+y'\ ^2} dx\qquad y'=\frac{x-\alpha}{\sqrt{\lambda^2-\big(x-\alpha\big)^2}}$$ which you can solve to find $\alpha$, $\beta$ and $\lambda$. From $(0,0)$ condition it follows that $$\beta=\sqrt{\lambda^2-\big(\alpha\big)^2}$$ By replacing it into $(a,0)$ condition it follows that $\alpha=a/2$. We can rewrite the equation as a circle equation $$\bigg(y-\sqrt{\lambda^2-\big(a/2\big)^2}\bigg)^2+\bigg(x-a/2\bigg)^2=\lambda^2$$

---------------EDIT-------------------

Although it produces a nice result; the equation is wrong. It assumes that every $dA$ produces the same amount of $dV$. It would have been right when the volume was produced by extrusion; but in rotational case $dA$ away from axis produces bigger $dV$.

share|improve this answer
    
please find exact values of parameters to see optimal solutions. I have doubts that this $y$ maximizes the volume of ratated figure. –  userNaN Jul 2 '13 at 16:44
    
It is directly derived from calculus of variations and by symmetric conditions around $x=a/2$ you can say that it maximizes the area. To find exact values you have to solve an elliptic integral. Maybe it will be possible with a numerical value of $a$. –  Occupy Gezi Jul 2 '13 at 17:01
    
One has $A[y]=\pi \int_0^{2a} y^2\ dx$. –  Christian Blatter Jul 2 '13 at 17:45
2  
Is the problem of maximizing the revolution volume equivalent to maximizing the area subject to a fixed arc length? –  Shuhao Cao Jul 2 '13 at 20:38
1  
@ShuhaoCao: According to our answers, no. –  robjohn Jul 3 '13 at 6:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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