Not Sphere

The solid with maximum volume is tricylinder, the intersection of three cylinders:

$\begin{cases} x^2+y^2\le1\\ x^2+z^2\le1\\ y^2+z^2\le1.\\ \end{cases}$

The cross-section of the solid in a plane perpendicular to the $z$ -axis and passing through the point $(0,0,z)$ , whose area is denoted by $A(z)$ , is given by

$\left\{(x,y)\,|\,x^2+y^2\le1,|x| \le \sqrt{1-z^2},|y| \le \sqrt{1-z^2}\right\},$

where $-1\le z\le 1$ . See the animation with Desmos.

Therefore, we have

$A(z)=\begin{cases} 4(1-z^2), & z^2\ge \dfrac 12\\ \pi-4\arcsin z+4z\sqrt{1-z^2}, & z^2\le \dfrac 12 \end{cases}$

Using the definition of volume, we get volume of the solid

$\begin{aligned} V&=\int_{-1}^1A(z)\,dz \\&=2\left(\int_0^{1/\sqrt 2}\left(\pi-4\arcsin z+4z\sqrt{1-z^2}\right)\,dz+\int_{1/\sqrt 2}^14\left(1-z^2\right)\,dz\right) \\&=\boxed{16-8\sqrt 2}. \end{aligned}$

I reasoned out the shape as the intersection of three cylinders This shape can also be thought of as an interior cube of side $\sqrt{2}$ with a curved cap on each side One-quarter of each cap is part of a sphere. so viewing this down the middle shows half a segment of a circle travelling across the bottom, the height is $\sqrt{1-x^{2}}-\frac{\sqrt{2}}{2}$ and width is $2x$

so the volume of this quarter-cap is to integrate the product of these from $0$ to $\frac{\sqrt{2}}{2}$ which is about $.07741$

Multiply this by $24$ and add the cube $\sqrt{2}^{3}=2.8284$ and get $\boxed{4.686}$

(Incidentally, I used Tinkercad for these drawings. It's not the fanciest, but it gets the job done. Plus I'm now ready to 3-d print this cool shape)

We can dissect this solid into a cube of side $\sqrt{2}$ and six caps with a base of $\sqrt{2}$ . The volume is

$(\sqrt{2})^3 + 6 \displaystyle \int _{ \frac { 1 }{ \sqrt { 2 } } }^{ 1 }{ { \left( 2\sqrt { 1-{ x }^{ 2 } } \right) }^{ 2 } } dx = 16-8\sqrt{2}$

where $x$ is the distance from the center of the solid. The cross sections of the caps are squares, with side length of $2\sqrt { 1-x^2 }$

Hint: The equations of this solid are $x^2 + y^2 < 1 \land x^2 + z^2 < 1 \land y^2 + z^2 < 1$ The solid is called tricylinder. The exact answer is $16 - 8 \sqrt{2}$ .

Whilst not as elegant as the other solutions, this problem is reasonably easy to solve via Monte-Carlo simulation. One (admittedly rough) approach to this is to take a random sample of points from the uniform distribution over a cube of side length two centred on the origin and, for each point, test whether it is within the unit circle as seen from the three axes. We then simply multiply the proportion of points passing the test by the volume of the cube and reach an estimate.

The following code tested ten million points in around 125 seconds and consistently gave an answer within 0.002 of the actual solution.

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from numpy import random

n = 10**7
successes = 0

for _ in range(n):
    point = map(lambda x: 2*x-1, random.rand(3))
    inside = True
    for i,j in ((0,1),(1,2),(2,0)):
        if point[i]**2 + point[j]**2 > 1:
            inside = False
            break
    if inside:
        successes += 1

print successes*8./n