An example: finding the surface of revolution with minimal area

### 12.8.6 An example: finding the surface of revolution with minimal area

In this section, you will find the function y0∈ D={y∈ C1[0,1]:y(0)=1,y(1)=2/3} for which the area of the corresponding surface of revolution is minimal.

The area of the surface of revolution is measured by the functional

F(y)=2 π | ∫ |
1
0
| y(x) | √ |
1+y′(x)2
| dx.

First, set f(y,y′,x)=y(x) √1+y′(x)2 and compute the associated Euler-Lagrange equation:

eq:=euler_lagrange(y(x)*sqrt(1+diff(y(x),x)^2))

| ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ | − |
y | ⎛ ⎝ | x | ⎞ ⎠
√
⎛ ⎜ ⎜ ⎝ |
d
dx
| y | ⎛ ⎝ | x | ⎞ ⎠ | ⎞ ⎟ ⎟ ⎠ |
2
| +1
| =K0, |
d2
dx2
| y | ⎛ ⎝ | x | ⎞ ⎠ | = |
⎛ ⎜ ⎜ ⎝ |
d
dx
| y | ⎛ ⎝ | x | ⎞ ⎠ | ⎞ ⎟ ⎟ ⎠ |
2
| +1
y | ⎛ ⎝ | x | ⎞ ⎠
| ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

You can obtain the stationary function by finding the general solution of the first equation.

sol:=collect(simplify(dsolve(eq[0],x,y)))

| ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ | −K0, |
K0 | ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ | − | ⎛ ⎜ ⎝ | e |
x−c1
K0
| ⎞ ⎟ ⎠ |
2
| −1 | ⎞ ⎟ ⎟ ⎟ ⎟ ⎠
2 e |
x−c1
K0
| ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

(See Section 10.1.18.) Obviously the constant solution −K0 is not in D, so set y0 to be the second element of the above list. That function, which can be written as y0(x)=−K0 cosh(x−c1/K0), is called a catenary.

y0:=sol[1]:; p:=[K_0,c_1]:;

To find the values of K0 and c1 from the boundary conditions, first plot the curves y0(0)=1 and y0(1)=2/3 for K0∈[−1,1] and c1∈[−1,2] to see where they intersect each other.

eq1:=subs(y0,x=0)=1:; eq2:=subs(y0,x=1)=2/3:; implicitplot([eq1,eq2],K_0=-1..1,c_1=-1..2)

Observe that there are exactly two catenaries satisfying the Euler-Lagrange necessary conditions and the given boundary conditions: the first with K0≈ −0.5 and c1≈ 0.6 and the second with K0≈ −0.3 and c1≈ 0.5. You can obtain the values of these constants more precisely by using fsolve.

p1:=fsolve([eq1,eq2],p,[-0.5,0.6]); p2:=fsolve([eq1,eq2],p,[-0.3,0.5])

[−0.56237423894,0.662588703113],[−0.30613431407,0.567138261119] |  |  |  |  |  |  |  |  |  |

You can check, for each catenary, that fy′ y′(x,yk,yk′)>0 holds for k=1,2.

y1:=subs(y0,p,p1):; y2:=subs(y0,p,p2):; D2f:=diff(f,diff(y(x),x),2):; solve([eval(subs(D2f,y=y1,y(x)=y1))<=0,x>=0,x<=1],x); solve([eval(subs(D2f,y=y2,y(x)=y2))<=0,x>=0,x<=1],x)

[],[] |  |  |  |  |  |  |  |  |  |

You can conclude that the strong Legendre condition is satisfied in both cases, so you can proceed by attempting to find the points conjugate to 0 for each catenary. The function y0 depends on two parameters, so use conjugate_equation to find these points easily.

fsolve(conjugate_equation(y0,p,p1,x,0)=0,x=0..1); fsolve(conjugate_equation(y0,p,p2,x,0)=0,x=0..1)

[0.0],[0.0,0.799514772606] |  |  |  |  |  |  |  |  |  |

You can conclude that there are no points conjugate to 0 in (0,1] for the catenary y1, so it minimizes the functional F. However, for the other catenary there is a conjugate point in the relevant interval, therefore y2 is not a minimizer.

You can verify the above conclusions by computing surface areas for catenaries y1 and y2.

int(y1*sqrt(1+diff(y1,x)^2),x=0..1); int(y2*sqrt(1+diff(y2,x)^2),x=0..1)

0.81396915825,0.826468466845 |  |  |  |  |  |  |  |  |  |

You can see that the surface formed by rotating the curve y1 is indeed smaller than the area of the surface formed by rotating the curve y2. Finally, you can visualize both surfaces for convenience.

plot3d([y1*cos(t),y1*sin(t),x],x=0..1,t=0..2*pi, display=yellow+filled)

plot3d([y2*cos(t),y2*sin(t),x],x=0..1,t=0..2*pi, display=yellow+filled)

(See Section 17.2.4 for information on plot3d.)