The Bulging Box

First, note that the square inscribed in an unit circle has side length $√2$ . Thus, the cube inscribed in the solid as defined by the three inequality equations has volume $2√2$ . Next, the six caps making up the rest of the volume have a total volume as found by integration: $6\int _{ \dfrac { 1 }{ \sqrt { 2 } } }^{ 1 }{ { (2\sqrt { 1-{ z }^{ 2 } } ) }^{ 2 }dz } =16-10\sqrt { 2 }$

Thus, the total volume is exactly $16-8\sqrt { 2 }$

and so $a + b + c = 16 - 8 + 2 = 10$

We can even use the prismoid formula, provided that we dissected this solid correctly: one central cube and six "lids". Due to symmetry, if we draw the cross section, we will discover that the cube has side length $r \sqrt { 2 }$ . The volume of each lid is $\frac { h }{ 6 } \left( 0+{ r }^{ 2 }\left( \frac { 5 }{ 2 } -\sqrt { 2 } \right) +2{ r }^{ 2 } \right) ,$ where $h = r\left(1-\frac{\sqrt 2}{2}\right)$ .

Adding the volumes of the cube with six times the lid yields $(16-8\sqrt{2}){r}^{3}$ .

I skipped the geometric steps due to the lack of a diagram, but it takes about one page to show.