Volume of Sphere

A Sphere is a three-dimensional circle. Just like a circle, in a sphere, all the points on the surface are at equal distance from a fixed point, “the Centre”. The equal distance from the centre to its surface is known as the radius(r), and the line passing through the centre of the sphere is called the diameter of the sphere. In the geometrical plane, it occupies three axes, namely the x,y, and z-axis. Volume is an important physical property of a sphere, which we will discuss in this article.

1.0Definition of Volume of a Sphere

The Volume of a sphere can be defined as the amount of three-dimensional space enclosed within the surface of a sphere. In simple words, the Volume of a sphere is the capacity of a sphere, which can be calculated in cubic units, like m³, cm³, in³ etc. The volume of a sphere formula can be written as: 

Volume of Sphere = $\frac{4}{3}\pi r^3$

Here, 

$\pi =\frac{22}{7}$

r = Radius of the sphere 

2.0Volume of Sphere and Key Concepts

Volume of Sphere with Diameter: 

If in a given sphere, the diameter d of the sphere is given, the radius can be written as r = d2. (r = \frac{d}{2})  Hence, the volume of the sphere can be rewritten as: 

Volume of sphere with Diameter:

$\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$

Or 

Volume of sphere with Diameter:

$\frac{1}{6}\pi d^3$

The Volume of Sphere with Area: 

If the total surface area A of a sphere is given, then the volume of the sphere can be found by finding the value of the radius “r” first, then putting it in the volume of the sphere formula. Mathematically, this can be represented as: 

Surface Area of Sphere = $4\pi r^2$

$r=\sqrt{\frac{A}{4\pi }}$

Volume of sphere = $\frac{4}{3}\pi \sqrt{\frac{A}{4\pi }}$

3.0Derivation of Volume of Sphere

The volume of a sphere can be derived using integral calculus by using two different types of integrations that are: 

• The volume of a sphere by integration
• The Volume of a sphere by triple integration 

Let’s derive the volume of a sphere formula using both of these methods. 

Volume of a Sphere by Integration

Imagine a sphere as a bunch of numerous thin circular disks, which are piled one over another, as shown in the figure. By the diagram, it is clear that all these discs will have a different diameter with a collinear centre. Now, select any of these thin discs with radius r and thickness “dy” located at the y distance from the x-axis. Hence, according to the concept of integration, the volume of the sphere can be expressed as the product of the area of the circle and its thickness dy. Mathematically, this can be written as: 

$dV=\pi r^{2}dy$

By using Pythagoras theorem, r can be expressed in the form of its thickness (y) and the Radius of the sphere, say R, as: 

$dV=\pi (R^2-y^2)dy$

Now, integrating the total volume of the sphere, 

$V={\int }_{y=-R}^{y=+R}dv$

$V={\int }_{y=-R}^{y=+R}\pi (R^2-y^2)dy$

$V=\pi {\left[R^{2}y-\frac{y^3}{3}\right]}_{y=-R}^{y=+R}$

Substituting the limits, 

$V=\pi \left[\left(R^3-\frac{R^3}{3}\right)-\left(-R^3+\frac{R^3}{3}\right)\right]$

$V=\pi \left[2R^3-\frac{2R^3}{3}\right]$

$V=\frac{\pi }{3}\left[6R^3-2R^3\right]$

$V=\frac{\pi }{3}(4R^3)$

Hence, the Volume of the sphere will be: 

$V=\frac{4}{3}\pi R^3$

The Volume of a Sphere by Triple Integration

To derive the volume of a sphere with the triple integration method, we will use spherical coordinates by using the standard equation of a sphere. The standard equation of a sphere of radius r with origin as the centre can be written as: 

$x^2+y^2+z^2=r^2$

Here, x, y, and z are related to the spherical coordinates of the sphere r$\mathring{,}\theta ,\varphi$ as:  

$x=r\sin \theta \cos \varphi$

$y=r\sin \theta \sin \varphi$

$z=r\cos \theta$

Where, 

• r is represented as the radial distance from the origin (which will be constant for a sphere),
• θ is the polar angle (or colatitude) measured from the positive z-axis, ranging from 0 to π,
• ϕ is the azimuthal angle (or longitude) measured in the xy-plane from the positive x-axis, ranging from 0 to 2π. 

The volume of spherical coordinates is given by: 

$dV=r^2\sin \theta drd\theta d\varphi$

Now, using triple integration to find the volume of a sphere: 

$V={\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^r{r}^2\sin \theta drd\theta d\varphi$ ……(1)

Step 1: Integrating with respect to r: 

${\int }_0^r{r}^{2}dr=\frac{r^3}{3}$

Step 2: Integrate with respect to $\theta$

${\int }_0^{2\pi }\sin \theta d\theta =2$

Step 3: Integrate with respect to$\varphi$

${\int }_0^{\pi }d\varphi =2\pi$

Putting values of Steps 1, 2, & 3 in equation (1) 

$V=(2\pi )(2)\left(\frac{r^3}{3}\right)$

$V=\frac{4\pi r^3}{3}$

$V=\frac{4}{3}\pi r^3$

4.0Volume of Sphere Examples: Numericals

Problem 1: Find the volume of a sphere whose surface area is 154 cm². 

Solution:

Surface area of sphere = 154 

$4\pi r^2=154$

$r^2=\frac{154\times 7}{4\times 22}=\frac{49}{4}$

$r=\sqrt{\frac{49}{4}}=\frac{7}{2}$

Now, 

Volume of Sphere = $\frac{4}{3}\pi r^3$

Volume of Sphere = $\frac{4}{3}\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^3$

Volume of Sphere = $\frac{22\times 4\times 7\times 7\times 7}{3\times 7\times 2\times 2\times 2}$

Volume of Sphere =36.67 $c{m}^3$

Problem 2: A spherical water tank has a diameter of 10 meters. How much water can it hold in terms of volume?

Solution: Given, Diameter = 10m 

radius will be $\frac{10}{2}$ or 5 meters

Volume of Sphere = $\frac{4}{3}\pi r^3$

Volume of Sphere = $\frac{4}{3}\times \frac{22}{7}\times 5^3$

Volume of Sphere = $523.6m^3(approx)$

Problem 3: A hollow sphere has an outer radius of 6 cm & an inner radius of 4 cm. What is the volume of the material making up the hollow sphere? 

Solution: Let Outer radius (R) = 6cm and inner radius(r) = 4cm 

The volume of material = Volume of outer surface – Volume of the inner surface 

The volume of material = $\frac{4}{3}\pi R^3-\frac{4}{3}\pi r^3=\frac{4}{3}\pi (R^3-r^3)$

The volume of material = $\frac{4}{3}\pi (R^3-r^3)=\frac{4}{3}\pi (6^3-4^3)$

$V=\frac{4}{3}\pi (6-4)\left(6^2+6\times 4+4^2\right)$

$V=\frac{4}{3}\times \frac{22}{7}\times (2)(76)=636.2c{m}^3$