Question

The volume of the sphere is given by $V = \dfrac{4}{3}\pi {R^3}$ where $R$ is the radius of the sphere.
(a) Find the rate of change of volume with respect to $R$.
(b) Find the change in volume of the sphere as the radius is increased from $20.0\,cm$ to $20.1\,cm$. Assume that the rate does not appreciably change between $R = 20.0\,cm$ to $R = 20.1\,cm$.

Answer 1

Hint: A sphere is the geometrical object in three dimensional space that is the surface of a ball. Like a circle in a two-dimensional space, a sphere is defined mathematically as the set of points that are all at the same distance $r$ from the given point in a three-dimensional space.

Complete step by step answer:
(a) Given, ${R_1} = 20\,cm$ ans ${R_2} = 20.1\,cm$.
The rate of change of volume with respect to $R$. As we know that volume of sphere
$V = \dfrac{4}{3}\pi {R^3}$
Differentiate the volume w.r.t R
$\dfrac{{dV}}{{dR}} = \dfrac{4}{3}\pi \dfrac{d}{{dR}}{(R)^3}$
$ \Rightarrow \dfrac{{dV}}{{dR}} = \dfrac{4}{3}\pi \times 3{R^2}$
Simplify
$\therefore \dfrac{{dV}}{{dR}} = 4\pi {R^2}$

Hence, the rate of change of volume w.r.t $R$ is $4\pi {R^2}$.

(b) The change in volume as we know that
$V = 4\pi {R^2}$
The change in volume will be
\[V = 4\pi {R_2}^2 - 4\pi {R_1}^2\]
$ \Rightarrow V = 4\pi ({R_2}^2 - {R_1}^2)$
Put the values
$V = 4\pi \{ {(20.1)^2} - {(20)^2}\} $
As we know that
${a^2} - {b^2} = (a + b)(a - b)$
$ \Rightarrow V = 4 \times \dfrac{{22}}{7} \times \{ (20.1 - 200)(20.1 + 20)\} $
$ \Rightarrow V = \dfrac{{88}}{7} \times (0.1) \times (40.1)$
$ \therefore V = 50.411\,c{m^3}$

Hence, the change in volume of the sphere as the radius is increased from $20.0\,cm$ to $20.1\,cm$ is $50.411\,c{m^3}$.

Note: The points on the sphere are all the same distance from a fixed point. The contours and plane sections of the sphere are circular. The sphere has constant width and constant girth. All points of a sphere are umbilicus. The sphere does not have a surface of centers.