Question

Show that the volume of a spherical shell of radius \(r\) and thickness \(d r\) is \(4 \pi r^{2} d r .\) [Hint: This exercise requires calculus.]

Short answer

The volume of a spherical shell of radius \(r\) and thickness \(d r\) is \(4\pi r^2 dr\)

Step-by-step solution

Write down the volume of a sphere

The volume, \(V\) of a sphere with radius \(r\) is given by the equation: \(V=\frac{4}{3}\pi r^3\)

Differentiate the volume equation with respect to radius

The derivative of \(V\) with respect to \(r\) is given by \(\frac{dV}{dr}= 4\pi r^2\)

Calculate the volume of the shell

The volume of the shell, \(dV\) with thickness \(dr\) is given by: \(dV = \frac{dV}{dr} \cdot dr = 4 \pi r^2 dr\)

Key concepts

Calculus in Geometry

In geometry, calculus is an invaluable tool for understanding complex shapes and their properties. It allows us to calculate dimensions like area, volume, and curves that aren't straightforward. For instance, when dealing with three-dimensional objects such as spheres, calculus helps in calculating quantities that involve changes in dimensions.
By finding the rate at which volume changes with respect to change in radius, we precisely estimate the volume of intricate shapes like spherical shells. Using calculus in geometry leads to deeper insight into how small changes in size can affect the overall properties of the shape.
Calculus helps unravel relationships between dimensions and provides an exact measurement rather than an approximation.

Derivative of Volume Formula

The derivative of a function tells us how that function changes as its input changes. In the context of a sphere, the volume formula gives its volume based on radius, expressed as \(V = \frac{4}{3}\pi r^3\).
When we differentiate this formula with respect to the radius \(r\), we are looking for how the volume changes as the radius changes slightly. This gives us \(\frac{dV}{dr} = 4\pi r^2\).
The derivative \(4\pi r^2\) shows that for a very small change in radius, the change in volume depends directly on the square of the radius and is proportional to the thickness. This derivative is essential in understanding the growth of volume as the sphere expands.

Surface Area of a Sphere

The surface area of a sphere can be directly related to its volume's derivative, which is fascinating! For a sphere of radius \(r\), the surface area is given by \(A = 4\pi r^2\). This formula represents the flow of volume across the sphere's outer shell.
Interestingly, the derivative of the volume formula, \(4\pi r^2\), matches exactly the formula for the surface area of the sphere. This coincidence is not mere chance—it shows how changes in volume at the boundary translate directly to the surface area.
Thus, understanding the surface area is vital in discovering how slight changes impact the sphere's immediate volume, which is crucial for computations involving shells.