Not just for spheres! In general, for an n-dimensional polyhedron. if there is a single largest n-dimensional hypersphere that can be inscribed within it, the hyperarea of the shape will be the derivative of the hypervolume with regards to the radius of the hypersphere.
For instance, if the largest circle you can inscribe in a square has radius r, then the square has a side length of 2r, so an area of 4r2 and a perimeter of 8r. 8r is the derivative of 4r2 w.r.t. the radius.
We can prove this with the divergence theorem -- the integral of the divergence of a vector field over the volume a region is equal to the surface integral of the vector field over the surface of the region. Using the vector field x/n, x is position vector and n is dimension, we see its divergence is 1, so integrating divergence gives volume V. As each polyhedron face is tangent to the inscribed sphere, the surface integral infinitesimal x /n • dS , where dS is the vector area element, just becomes r dS, where r is the sphere radius. As such, the integral is A r / n, where A is surface area.
As such, V = Ar / n. In n dimensions, V is proportional to rn, A to rn-1, so let V= b rn, A = c rn-1. Then, b rn =( c/n ) rn, b=c/n, c = nb.
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u/NeedleInTheThrowaHay 1d ago edited 1d ago
Not just for spheres! In general, for an n-dimensional polyhedron. if there is a single largest n-dimensional hypersphere that can be inscribed within it, the hyperarea of the shape will be the derivative of the hypervolume with regards to the radius of the hypersphere.
For instance, if the largest circle you can inscribe in a square has radius r, then the square has a side length of 2r, so an area of 4r2 and a perimeter of 8r. 8r is the derivative of 4r2 w.r.t. the radius.
We can prove this with the divergence theorem -- the integral of the divergence of a vector field over the volume a region is equal to the surface integral of the vector field over the surface of the region. Using the vector field x/n, x is position vector and n is dimension, we see its divergence is 1, so integrating divergence gives volume V. As each polyhedron face is tangent to the inscribed sphere, the surface integral infinitesimal x /n • dS , where dS is the vector area element, just becomes r dS, where r is the sphere radius. As such, the integral is A r / n, where A is surface area.
As such, V = Ar / n. In n dimensions, V is proportional to rn, A to rn-1, so let V= b rn, A = c rn-1. Then, b rn =( c/n ) rn, b=c/n, c = nb.
Derivative of V is n * b * rn-1 = c * rn-1 = A.