In geometry, the sphere's volume is the space enclosed by its surface. In simple terms, it calculates how much material (gas, liquid, or solid) can be held within the sphere. This understanding is essential for solving geometry-related problems in American competitive exams such as the SAT, ACT, GRE, and GMAT, where questions will typically test your capacity to utilize volume formulas under real-life circumstances. Mastery of the volume of a sphere can put you ahead of others in terms of solving problems quickly and precisely, enhancing your overall test score.

What is a Sphere?

A sphere is a perfectly symmetrical geometrical shape defined as the set of all points in three-dimensional space that are equidistant from a given point called the centre. The sphere is a three-dimensional analog of the circle and has many interesting properties that make it a fascinating object of study. One of the most important properties of the sphere is its volume, which is given by the formula V = (4/3)πr³, where r is the radius of the sphere. 

There are no faces, edges, or vertices on a sphere. Our Earth, the ball and marbles you play with, are all spherical. Some of the examples of spheres in daily life are shown in the picture below. 

Types of Spheres

There are two main types of spheres: solid and hollow spheres. The solid sphere is a filled, three-dimensional object with a finite volume and surface area and no empty space. On the other hand, the hollow sphere is a spherical shell with a finite thickness enclosing an empty space. 

What is the Volume of a Sphere?

The interior space of a three-dimensional object or shape is known as volume. Volume is not a property of two-dimensional shapes. A sphere's volume is the space enclosed by the surface of the sphere. The capacity of a sphere is determined by its volume.

The sphere is hollow if there is room inside. The sphere will be solid if the filler material is inside. Cubic metres, cubic feet, cubic inches, and other equivalent measures are used to measure volume. The symbols denoting it are \(cm^3, m^3, in^3\), etc.

The Greek philosopher Archimedes developed the formula for a sphere's volume more than two thousand years ago. The volume of a sphere is exactly two-thirds that of its circumscribed cylinder, which is the smallest cylinder that can hold the sphere.

Volume of a Sphere Formula

The formula shown below is used mathematically to determine a sphere's volume:

The formula for a sphere's volume is \(\frac{4}{3}{\pi}r^3\), where r is the sphere's radius.

Steps to Calculate the Volume of a Sphere

Here are the steps to calculate the volume of the sphere. 

Step 1: Carefully read the information provided in the question.

Step 2: Determine whether a radius, diameter, surface area, or circumference value is being provided.

Step 3: Determine the sphere's radius. Divide the diameter by two to determine the radius if it is known. If the surface area is known, use the formula 4r to calculate the radius from the surface area of a sphere. Use the formula 2r to determine the radius if the circumference is known.

Step 4: Carefully read through the units. Convert all units that are interchangeable into a single form.

Step 5: Get the cube of the radius, i.e., \(r^3\).

Step 6: Divide the \(r^3\) value by.

Step 7: Add \(\frac{4}{3}\) to the value obtained in Step 6.

Step 8: The final value will be the required volume of a sphere.

Let's take an example:

Find the volume of a sphere of radius 11.2 cm 

Solution: 

Given: Radius of the sphere, r = 11.2 cm.

Formula: The formula to calculate the volume of a sphere is given by: V = \(\frac{4}{3}\pi r^3\)

\(\begin{align*}V &= \frac{4}{3}\pi r^3 \&= \frac{4}{3}\pi (11.2)^3 \&= \frac{4}{3}\pi (1419.9) \&= \frac{4}{3} \times 3.14159 \times 1419.9 \&\approx7241.96 \text{ cm}^3\end{align*}\)

Therefore, the volume of the sphere with a radius 11.2 cm is approximately 7241.96 cubic centimeters.

Derivation of Volume of a Sphere

The formula for the volume of a sphere can be derived from the integration method or the volumes of the cone and cylinder.

Integration Method

Consider a sphere with numerous thin spherical discs arranged over one another as depicted in the pictorial representation. Since the sphere is made with thin circular discs placed collinearly, their diameters will vary over the entire length of the sphere. As a result, the volume will also change over the whole diameter of the sphere.

Think of a sphere with many thin spherical discs layered on top of one another, as seen in the illustration. The sizes of the thin circular discs that make up the sphere will differ along its whole length since they are arranged collinearly. As a result, the volume will fluctuate along the sphere's whole diameter.

Pythagorean formula. 

Therefore, \(dV = r^2\) dy can be used to calculate the volume of the disc element. 

\((R^2 – y^2 ) dy = dV\) 

By integrating the above equation, the total volume of the sphere will be given by:

\(V = \int_{y=-R}^{Y=+R}dV\)

\(V = \int_{y=-R}^{Y=+R}{\pi}(R^2 – y^2 )dy\)

\(V={\pi}[R^2y-(\frac{y^3}{3})]^{y=+R}_{y=-R}\)

Substituting the limits we get,

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

Simplifying the above equation, we get,

\(V={\pi}[2R^3-\frac{2R^3}{3}]\)

\(V=\frac{\pi}{3}[6R^3-2R^3]\)

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

Therefore, the final formula for the volume of a sphere is given by \(V = \frac{4}{3}{\pi}r^3\).

From Volumes of Cone and Cylinder

Do you know a relationship exists between the sphere, cylinder, and cone? Their volumes are related, specifically! The volume of a cylinder is calculated by adding the volumes of a cone and a sphere. Mathematically,

\(V_{cone} + V_{sphere} = V_{cylinder}\)

As a result, we may calculate a sphere’s volume using the formula \(V_sphere\) = \(V_{cylinder} – V_{cone}\)

\(V_{cone}\) is equal to \(r^2h\), where h is the cone’s slant height.

\(V_{cylinder}\) is equal to \(\frac{1}{3}\) of \(r^2h\), where h is the cylinder’s height.

\(V_{sphere}\) is equal to \(\frac{1}{3} r^2h – r^2h\), which equals \(\frac{2}{3} r^2h\).

If we observe a sphere, we can see that the height equals the sphere’s diameter. Therefore, \(h = 2r\).

Putting the value of h in the final equation, we get,

\(V_{sphere} = \frac{2}{3}{\pi}r^2(2r) = \frac{4}{3}{\pi}r^3\), which is the volume of a sphere.

Volume of a Hollow Sphere

The hollow sphere's volume and a sphere's volume are related. R denotes the outer radius of a hollow sphere, whereas the interior radius is denoted by r. The volume of the hollow sphere is equal to that of the inner sphere subtracted from the volume of the outer sphere. It can be calculated as –

\(V_{hollow} = \frac{4}{3}{\pi}R^3 – \frac{4}{3}{\pi}r^3\)

It can also be written as \(V_{hollow} = \frac{4}{3}{\pi}(R^3 – r^3)\). The unit of volume of the hollow sphere is cubic metres.

Volume of a Sphere of Unknown Radius

Generally, if the volume of the sphere is unknown, there will be some other data given in the problem. Under such conditions, the formula here comes in handy to find the volume. The general formulas for the sphere are given below: 

Diameter of Sphere: \(D=2r\)

Surface Area of Sphere: \(A=4{\pi}r^2\)

You can use these formulas to find the radius of the sphere and then find its volume. 

Applications of the Volume of a Sphere in the Real-world

The volume of a sphere is used in various ways in the actual world. We don't require a volume of a sphere calculator to calculate it repeatedly if we know its formula. The following are some situations where the volume formula is regularly applied:

• Many industries employ the volume formula when producing items like balls, globes, bearings, bubbles, etc.
• Earth, the sun and the other planets are spherical. The formula for the volume of a sphere helps to calculate its volume. 
• Calculating how much air a hot air balloon needs to maintain leaking is useful.
• Any material stored in a bowl can be measured using the volume of a sphere.

Interesting Facts about Spheres

Here are some interesting facts about spheres

• A sphere is the only three-dimensional shape that is completely symmetrical in all directions. It has no faces, corners, or edges.
• The word "sphere" comes from the Greek word "sphaira," which means "ball" or "globe."
• Spheres have the smallest surface area for a given volume of any shape, so bubbles and soap bubbles are often spherical.
• All the sphere’s surface points are the same distance “r” from the center.
• A perfectly round drop of liquid is also a sphere due to the surface tension of the liquid.
• The Earth is approximately a sphere, although it is not perfect due to its rotation and the irregularities in its surface.
• The study of spheres and their properties is known as spherical geometry, an important branch of mathematics in many fields, including astronomy, physics, and computer graphics.

Volume of a Sphere Solved Examples

So far, we have seen the general formula of the volume of the sphere. We have also seen the three types of spheres and the formulae to find their volume. Now let’s see some solved examples of the same.

Solved Example 1: Find the volume of a sphere of radius 7cm.

Solution: Given data:

Radius of the sphere, r = 7 cm.

Formula: The formula to calculate the volume of a sphere is given by: V = \(\frac{4}{3}\pi\) r^3 

\(\begin{align*}V &= \frac{4}{3}\pi r^3 \&= \frac{4}{3}\pi (7)^3 \&= \frac{4}{3}\pi (343) \&= \frac{4}{3} \times 3.14159 \times 343 \&\approx 1436.76 \text{ cm}^3\end{align*}\)

Therefore, the volume of the sphere with radius 7 cm is approximately 1436.76 cubic centimeters.

Solved Example 2: Find the volume of a sphere whose circumference is 36 units.

Solution: Given that the circumference of a sphere is 36 units.

We know the circumference of a circle is given by \(2{\pi}r\), where r is the radius.

Therefore, \(C = 2{\pi}r = 36\)

Solving this, we get \(r = \frac{36}{2\pi}\) units.

The formula gives the volume of a sphere, \(V = \frac{4}{3}{\pi}r^3\)

Putting the value of r, we get \(V = \frac{4}{3}{\pi}{\frac{36}{2\pi}}^3\)

\(V = 787.87\) cubic units.

Solved Example 3: A hollow sphere is designed by a company such that its thickness is 10 cm and 4 m inside diameter. What will be the volume of the sphere designed by the company?

Solution: Given that the inside diameter is 4 m and thickness is 10 cm, equal to 0.1 m.

Therefore, the outer diameter will be 4 + 0.1 m = 4.1 m.

The volume of a hollow sphere is denoted by: Volume = \(\frac{4}{3} {\pi}R^3 – \frac{4}{3} {\pi}r^3\), where R is the radius of the outer sphere and r is the radius of the inside sphere.

Putting the values in the above equation, we get,

\(V = \frac{4}{3} {\pi} (2.05^3 – 2^3) = \frac{4}{3} {\pi} (0.61512)\)

Therefore, \(V = 2.5766 m^3\).

Solved Example 4: Find the volume of a sphere if its surface area is 100 square metres.

Solution: We know that the surface area of a sphere is given by \(S = 4{\pi}r\), where r is the radius of a sphere.

Therefore, \(S = 4{\pi}r = 100\)

Finding the value of r, we get, \(r = 7.96 m\)

The volume of a sphere is given by \(V = \frac{4}{3} {\pi} r^3\)

Putting the value of r, we get,

\(V = \frac{4}{3} {\pi} (7.96)^3 = 2111.58 m^3\).

Solved Example 5: The surface of the sphere is 500 cm2. If you change the radius to reduce the area by 50%, then find the radius. \(({\pi}=3.14)\)

Solution:

Since the area gets reduced by 50%, we can say that

New surface area = 50% of the original area

\({\implies}4{\pi}r^2 = \frac{1}{2}\times500 \)

\({\implies}r^2 = \frac{(\frac{1}{2}\times500 )}{4{\pi}}\)

\({\implies}r^2=\frac{250}{12.56}\)

\({\implies}r^2 = 19.8945 \)

\({\implies}r = 4.46 cm\)

Solved Examples 6: Hollow spheres melt into the same small hollow sphere. The inner and outer radii of the larger sphere are 9 cm and 7 cm, respectively. If the inner and outer radii of the small spheres are 3 cm and 4 cm, respectively, how many small spheres can be formed?

Solution:

We know that the Volume of the sphere = \(\frac{4}{3}{\pi}r^3\)

Now, the volume of the bigger sphere = volume of the sphere with outer radius – the volume of the sphere with inner radius

\({\implies}Volume of the bigger sphere = \frac{4}{3}{\times}{\pi}{\times}(9)cm^3 – \frac{4}{3}{\pi}(7)cm^3\)

\({\implies}Volume of the bigger sphere = \frac{4}{3}{\times}{\pi}{\times}(729-343) = \frac{4}{3}{\times}{\pi}{\times}(386) cm^3\)

In the same way, the Volume of smaller sphere = \(\frac{4}{3}{\times}{\pi}{\times}(4)cm^3 – \frac{4}{3}{\times}{\pi}{\times}(3)cm^3\)

\({\implies}\text{ Volume of the smaller sphere }= \frac{4}{3}{\times}{\pi}{\times}(64-27) = \frac{4}{3}{\times}{\pi}{\times}(37) cm^3\)

hence, the number of spheres that can be formed = volume of the bigger sphere/ volume of the smaller sphere,

Therefore, the number of spheres that can be formed = \(\frac{\frac{4}{3}{\times}{\pi}{\times}(386)}{\frac{4}{3}{\times}{\pi}{\times}(37)}\)

The number of spheres that can be formed = 10.43 spheres.

In conclusion, understanding the volume of a sphere and its applications is essential for solving geometry problems effectively, especially in competitive exams like the SAT, ACT, GRE, and GMAT. Mastery of the formula V= \frac{4}{3}\pi r^3 not only helps in solving mathematical problems but also provides insights into real-world applications in industries, astronomy, and everyday life. With consistent practice and application of the derived formulas, you can confidently tackle sphere-related questions, improving your overall performance in exams.