Volumes and surface areas of 3d shapes.

Question

For some $3$D shapes if the formula for Volume is known then the S.A formula can be found by differentiating $\frac{dV}{dR}$ : e.g in case of sphere $V = \frac{4}{3} \pi R^3$ and S.A is $4 \pi R^2$ (differential). However for a cube with $V = R^3$ the S.A is $ 6 R^2$ (NOT $3 R^2$)

Is there a general rule when $\frac{dV}{dR}$ applies and when it does not ?

See this answer for the sphere. – Taroccoesbrocco Jan 17 '18 at 18:10 — Taroccoesbrocco, Jan 17 '18 at 18:10

score 0 · Answer 1 · answered Jan 17 '18 at 22:09

The reason why this happens is because the circle is the outlier among all 3D objects: the centre of the circle is equidistant to any point on the surface. This mean that rate of change is constant along the surface, so it can be differentiated to get the surface area.

Volumes and surface areas of 3d shapes.

1 Answers1