# Volume of a Sphere

A **sphere** is a perfectly round geometrical 3-dimensional object. It can be characterized as the set of all points located distance \(r\) (radius) away from a given point (center). It is perfectly symmetrical, and has no edges or vertices.

**Note:** Earth is not a sphere! As mentioned above sphere has no edges or vertices. But earth is slightly flattened on the poles, which makes it's shape \(\text{un-sphere-ish}\). It's shape is given a special name; \(\text{Geoid}\)

A sphere with radius \(r\) has volume \( \frac{4}{3} \pi r^3 \) and surface area \( 4 \pi r^2 \).

Interesting fact: Of all shapes with the same surface area, the sphere has the largest volume.

## What is the volume of a sphere of radius \(5\)?

The volume of a sphere of radius 5 is \( \frac{4}{3} \pi \times 5^3 = \frac{ 500 } { 3} \pi = 166 \frac{2}{3} \pi \). \( _\square \)

## If the surface area of a sphere is \(144\pi,\) what is the volume of the sphere?

Observe that the surface area of the sphere can be rewritten as \[144\pi=4\pi \times 6^2.\] Then, since the surface area of a sphere with radius \(r\) is \( 4 \pi r^2 ,\) it follows that the radius of the sphere in this problem is \(r=6.\) Hence, its volume is \[\frac{4}{3} \pi r^3 =\frac{4}{3} \pi \times 6^3 =288\pi. \ _\square\]

## You have a gold sphere whose volume is \(\frac{4\pi}{3} \text{ cm}^3.\) If you want the gold sphere to be twice as large in size, how much more gold do you have to bring to a jeweler in addition to the gold sphere you currently have?

From the formula \( V=\frac{4}{3} \pi r^3 \) for the volume of a sphere with radius \(r,\) you know that the radius of your gold sphere is \(r=1 \text{ cm}.\) Since you want the radius to be \(2 \text{ cm},\) the amount of gold required to make the new, bigger sphere is \( \frac{4}{3} \pi \times 2^3 =\frac{32\pi }{3}\left(\text{cm}^3\right).\) Hence, the additional amount of gold required is \[\frac{32 \pi}{3}\text{ cm}^3-\frac{4\pi}{3}\pi \text{ cm}^3=\frac{28\pi}{3} \text{ cm}^3. \ _\square\]

## The volume of sphere \(a\) is \(\frac{1}{27}\) times that of sphere \(b.\) The surface area of \(a\) is how many times the surface area of \(b?\)

Let \(R_a\) and \(R_b\) denote the radii of spheres \(a\) and \(b,\) respectively. Then from the formula \( V=\frac{4}{3} \pi r^3 \) for the volume of a sphere with radius \(r,\) we have \[\frac{4}{3} \pi {R_a}^3=\frac{1}{27}\cdot \frac{4}{3} \pi {R_b}^3,\] which implies \(R_a=\frac{1}{3}R_b.\)

Therefore, from the formula \( S=4 \pi r^2 \) for the surface area of a sphere with radius \(r,\) we conclude that the surface area of sphere \(a\) is \(\left(\frac{1}{3}\right)^2=\frac{1}{9}\) times the surface area of sphere \(b.\) \(_\square\)

An artist has delicately taken out a conical portion from a spherical watermelon with radius \(R=5,\) as shown below:

Instead of a flat circular base, this special cone has a spherical cap (retaining its peel) with cross-sectional base radius \(r = 3.\)

What is the ratio of the volume of the original whole sphere to that of this spherical cone?

**Cite as:**Volume of a Sphere.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/volume-sphere/