Question

The volume inside a sphere of radius \(r\) is \(V=\frac{4}{3} \pi r^{3} .\) Then \(d V=4 \pi r^{2} d r=A d r\) where \(A\) is 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

The rate of change of the volume of a sphere with respect to its radius is equal to its surface area. By integrating the area formula with respect to the radius, the volume formula can be derived.

Step-by-step solution

Understand the Given Formulas

The problem provides two important formulas: the volume of a sphere, given by \[ V = \frac{4}{3} \pi r^3 \]and the relationship between the differential changes in volume and radius,\[ dV = 4\pi r^2 \, dr \]which states that the differential change in volume (dV) is equal to the surface area of the sphere (A) times the differential change in radius (dr).

Interpret the Geometrical Meaning

To interpret the geometrical meaning, note that \[ dV = A \, dr \]indicates that the rate of change of volume with respect to the radius equals the surface area of the sphere. This means that as you infinitely slightly increase the radius of the sphere, the increase in volume is approximated by the area of the sphere times that tiny increment in radius.

Use the Relationship to Derive Volume Formula

To derive the volume formula from the area formula, recognize that the given relationship can be written in terms of derivatives:\[ \frac{dV}{dr} = 4\pi r^2 \]. Integrate both sides with respect to r to find V:\[ V = \int 4\pi r^2 \, dr \].

Compute the Integral

Perform the integration on the right-hand side to get:\[ V = 4\pi \int r^2 \, dr \].The integral of \( r^2 \) is \( \frac{r^3}{3} \), so:\[ V = 4\pi \left( \frac{r^3}{3} \right) + C \]where C is the constant of integration. Assuming the volume V is zero when r is zero, C must be zero.Thus, we obtain:\[ V = \frac{4}{3} \pi r^3 \]

Key concepts

Derivatives and Their Geometric Significance

A derivative measures how a function changes as its input changes. In this problem, the derivative of the volume of a sphere with respect to its radius is given by \( \frac{dV}{dr} = 4\pi r^2 \).
The derivative here tells us the rate at which the volume increases as the radius increases. This rate is actually the surface area of the sphere!
This means that if we increase the radius of the sphere slightly, the increase in volume can be approximated by the surface area times the small increment in radius. Understanding this helps us see how closely related volume and surface area are in three-dimensional shapes.

Integration to Derive Volume from Surface Area

Integration is the reverse process of differentiation. When we integrate the derivative of a function, we recover the original function. Given that \( \frac{dV}{dr} = 4\pi r^2 \),
we can integrate this to find the volume formula of a sphere.
By integrating both sides of \( \frac{dV}{dr} = 4\pi r^2 \) with respect to \(r\), we get:
\[ V = \int 4\pi r^2 \, dr \] Performing this integral gives us: \[ V = 4\pi \left( \frac{r^3}{3} \right) = \frac{4}{3} \pi r^3 \] This confirms that the volume inside a sphere of radius \(r\) is \( \frac{4}{3} \pi r^{3} \).
Using integration helps us connect the rate of change (surface area) to find the total volume of the sphere.

The Relationship between Volume and Surface Area

In a sphere, the surface area is closely linked to the volume. Their relationship can be derived and understood through calculus. The surface area of a sphere is given by \(A = 4\pi r^2\).
When we look at the differential change in volume with respect to the radius: \(dV = 4\pi r^2 \, dr\), it shows how every small increase in radius results in an increase in volume proportional to the surface area times the small radius change. Translating this into practical terms, if we imagine blowing up a balloon (a sphere) by a very tiny amount, the small increase in air inside (volume) is related to the skin of the balloon's outer surface area (surface area) times the tiny increase in radius. This fascinating relationship bridges the concepts of surface area and volume in a geometrically meaningful way, enhancing our understanding of 3D shapes.