Sphere with radius 3 cm, an algebraic coincidence
- Input
- r=3
- Expected output
- V ≈ 113.10 cm³, S ≈ 113.10 cm²
r=3 is the only radius where (4/3)πr³ and 4πr² give the same number, 36π; at any other radius the two values pull apart.
sphere volume calculator
Sphere volume is V=(4/3)πr³, growing with the cube of the radius; surface area is S=4πr², growing with the square. The two formulas land on the exact same number, 36π, precisely when r=3, a purely algebraic coincidence, not a physical one.
r=3 is the only radius where (4/3)πr³ and 4πr² give the same number, 36π; at any other radius the two values pull apart.
The ratio S/V ≈ 0.4286 matches 3/7 exactly, confirming the area-to-volume relationship is always 3/r.
Real Earth is a slightly flattened ellipsoid, not a perfect sphere; the calculation treats the mean radius as constant in every direction.
Use any unit you prefer (cm, m, inches…). The area/volume will be in the corresponding squared/cubed unit. For example, if you enter cm, area will be in cm².
Because setting (4/3)πr³=4πr² and simplifying leads to r=3; at that specific radius, the two values coincide at 36π≈113.10, but they stay in different units (cm³ and cm²), so it is an algebraic coincidence, not a physical equivalence.
Yes: dividing S by V cancels π and r², leaving S/V=3/r. For a sphere with radius 7 cm, that ratio equals 3/7≈0.4286, matching the computed values V≈1436.76 cm³ and S≈615.75 cm².
It is a close approximation, not the exact figure: using the 6,371 km mean radius as if Earth were a perfect sphere gives V≈1.0832×10¹² km³, close to geophysical estimates, but real Earth is slightly flattened at the poles.
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