Question

The volume inside a sphere of radius ris$\mathbf{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\tau \tau r}}^{\mathbf{3}}$. Then$\mathbf{dV}\mathbf{=}\mathbf{4}{\mathbf{\tau \tau r}}^{\mathbf{2}}\mathbf{dr}\mathbf{=}\mathbf{Adr}$whereis the area of the sphere. What is the geometrical meaning of the fact that the derivative of the volume is the area? Could you use this fact to find the volume formula given the area formula?

Short answer

Think of the ball as being built up from a series of shells of thickness, whose volume is$\mathrm{dV}=4{\mathrm{\tau \tau r}}^2\mathrm{dr}$

The formula for the volume is$V\left(R\right)={\int }_{r=0}^{R}A\left(r\right)dr$ .

Step-by-step solution

Given Condition

The volume inside a sphere of radius r is $\mathrm{V}=\frac{4}{3}{\mathrm{\tau \tau r}}^3$androle="math" localid="1659147146563" $\mathrm{dV}=4{\mathrm{\tau \tau r}}^2\mathrm{dr}=\mathrm{Adr}$ where A is the area of the sphere.

Concept of derivative of the volume is the area

The derivative is the difference between the volume of a slightly larger sphere and a slightly smaller sphere. That difference is the surface area.

Calculate the volume of the shell

The geometrical meaning of the fact that the derivative of the volume is the area is that the rate of change of the volume of the shell is equal to the surface area of the shell. Volume formula can be found using the given area formula.

Think of the ball as being built up from a series of shells of thickness. The volume of the shell is calculated as below.

$$\begin{gathered} \mathbf{V}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\mathbf{sin\theta d\theta }{\mathbf{\int }}_{\mathbf{r}}^{\mathbf{2}\mathbf{x}}\mathbf{d\varphi }{\mathbf{\int }}_{\mathbf{r}}^{\mathbf{r}\mathbf{+}\mathbf{dr}}{\mathbf{r}}^{\mathbf{2}}\mathbf{dr} \\ \mathbf{}\mathbf{}=4\mathrm{\pi }\frac{1}{3}{\overline{){\mathrm{r}}^3}}_{\mathrm{r}}^{\mathrm{r}+\mathrm{rd}} \\ =\frac{4\mathrm{\pi }}{3}\left({\left(\mathrm{r}+\mathrm{dr}\right)}^3-{\mathrm{r}}^3\right) \\ =\frac{4\mathrm{\pi }}{3}\left[{\mathrm{r}}^3{\left(1+\frac{\mathrm{dr}}{\mathrm{r}}\right)}^3-{\mathrm{r}}^3\right] \\ \approx \frac{4\mathrm{\pi }}{3}\left[{\mathrm{r}}^3\left(1+3\frac{\mathrm{dr}}{\mathrm{r}}\right)-{\mathrm{r}}^3\right] \\ =4{\mathrm{\pi r}}^2\mathrm{dr} \end{gathered}$$

The formula for the volume is$V\left(R\right)={\int }_{r=0}^{R}A\left(r\right)dr$ .