Chapter 3.3, Problem 50E

To determine

To find: The explanation for the instantaneous rate of change of the volume with respect to the radius should equal the surface area.

Answer to Problem 50E

When the volume of the sphere is divided into small layers such that the surface area is $4\pi r^2$ and infinitesimally small width of $\Delta r$ . Then, the change in the volume is $4\pi r^2\Delta r$ and the volume is divide by the change in the radius $\Delta r$ that is $4\pi r^2$ .

Explanation of Solution

Given Data:

The volume $V$ of the sphere of radius $r$ such that $\left(\frac{4}{3}\right)\pi r^3$ and then the surface area $A$ is $4\pi r^2$ and $\frac{dV}{dr}=A$ .

Calculation:

Consider the volume of the radius is $V=\frac{4\pi r^3}{3}$ . Then, the volume of the radius with respect to r is,

  $\frac{dV}{dr}=\frac{4\pi }{3}\left(\frac{d\left(r^3\right)}{dr}\right)$

Consider that when the volume of the sphere is divided into small layers such that the surface area is $4\pi r^2$ and infinitesimally small width of $\Delta r$ . Then, the change in the volume is $4\pi r^2\Delta r$ and the volume is divide by the change in the radius $\Delta r$ that is $4\pi r^2$ .

Thus, the change in the volume with respect to radius is,

  $\frac{dv}{dr}=4\pi r^2$