The Paraboidal Minima

The scaling of the paraboloid is defined by a: $z=f(r)= ar^2$ .

The enclosed volume is $πr^2ar^2-\int_0^r {2πx ax^2}dx$ $= π(ar^4- ar^4/2) = \frac{π}{2}ar^4$ . This will be set to 1 later.

The area is a bit more work: The flat 'top' area $A_f=πr^2$ and

the curved area: $A_q=\int_0^{r} 2πx\sqrt{dx^2+dy^2}$ $= 2π\int_0^{r} x\sqrt{1+ 4a^2x^2}dx$ .

Substitute $t= \sqrt{1+ 4a^2x^2}$ , then $dt^2=2tdt=d 1+ 4a^2x^2=8a^2xdx$ so that $xdx=tdt/4a^2$ . Now we have

$A_q = \frac{2π}{4a^2}\int_1^{\sqrt{1+ 4a^2r^2}} t^2dt$ $= \frac{π}{6a^2}((1+ 4a^2r^2)^{\frac{3}{2}} -1)$

$A= A_f+A_q=πr^2+ \frac{π}{6a^2}((1+ 4a^2r^2)^{\frac{3}{2}} -1)$

I used the area and volume formulas in the Python code below, which outputted

... a=3.276471, A=5.6750128905 height=1.44425, diameter=1.32785, difference=0.11641

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import math;

# Fill the 'cup' of paraboloid z=ar^2 with 1 unit volume of water, what is r at water level?
def r_unitvolume(a):
    return (2/(a*math.pi))**(1/4);

# Total surface area (curved & flat) of 1unit volume water in the paraboloid z=ar^2
def Area(a):
    r=r_unitvolume(a);
    return math.pi/(6*a**2)*((1+4*a**2*r**2)**(3/2)-1) + math.pi*r**2;

# Find the value for a that minimizes total surface area of the paraboloid
a=1.0;
A0=Area(a);
delta = 0.4;
while abs(delta)>0.000001:
    print ('a={:.6f}, A={:.10f}'.format(a, A0));
    a=a+delta;
    A1=Area(a);
    if (A1>A0): # area is no longer diminishing, go back slowlier
        delta=-delta/2;
    A0=A1;

r=r_unitvolume(a);
diameter=2*r;
height=a*r**2;
print ('height={:.5f}, diameter={:.5f}, difference={:.5f}'.format( height, diameter, height-diameter));

The volume of a paraboloid of equation $z=\frac{h}{R^2}r^2$ , where h is the height and R is the radius of the top circle, is $V=\frac{\pi}{2}hR^2$ .

The lateral area is more tricky:

$A_L=\int^R_0\int^{2\pi}_0\sqrt{1+4\frac{h^2}{R^4}r^2}\cdot r drd\phi=\frac{\pi}{6}\frac{R}{h^2}\left[\left(R^2+4h^2\right)^{3/2}-R^3\right]$

The total Area is the sum of $A_L$ and the top area $A_T$ : $A=A_L+A_T=\frac{\pi}{6}\frac{R}{h^2}\left[\left(R^2+4h^2\right)^{3/2}-R^3\right]+\pi R^2$

This value is what we need to minimize keeping the volume fixed to $V=1$ : we can substitute $h=\frac{2V}{\pi R^2}=\frac{2}{\pi R^2}$ in the definition of $A$ and take the derivative with respect to $R$ .

The derivative is quite ugly but Mathematica managed to solve the equation $\frac{dA}{dR}=0$ : that gives $R=\left[\frac{2}{3}\left(\frac{3}{\pi^2}-\frac{\sqrt{3}}{\pi^2}\right)\right]^{1/6}\simeq 0.6639$ , that implies $h=\frac{2}{\pi R^2}\simeq 1.4443$ .

The request of the problem is to evaluate the quantity $|h-d|=|h-2R|=1.4443-2\cdot0.6639=0.1165$ ,