Geometry - Sphere
- Definition of Sphere
- Properties of Sphere
- The Surface Area of a Sphere
- The Volume of a Sphere
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Definition of Sphere
Any perfectly round object around us can be called a sphere. Technically, a sphere is a three-dimensional object just like a cube, cuboid, cone, etc. So, we can define a sphere as a three-dimensional object that is perfectly round.
It can be visualized as a congruent object. The congruency can be understood by all the points lying on the surface of the sphere which are equidistant from one fixed point known as the center of the sphere.
Various examples of a sphere are planets (earth, mars, mercury, etc.), balls (basketball, cricket ball, soccer ball, etc.), marbles, soap bubbles, etc.
Properties of Sphere
- Roundness: The roundness of a sphere means that it does not have sharp edges, corners, or flat surfaces.
- Curved Surface: The surface of the sphere is completely curved and continuous without any boundaries.
- Center: The sphere has a fixed point called the center around which all the mass and matter are aggregated.
- Radius: Around the center, the point up to which the mass is distributed is called the radius of the sphere. Therefore, the radius is the distance between the center and the surface of the sphere.
- Diameter: The diameter is measured along the line made by joining any two points on the surface of the sphere such that the line passes through the center of the sphere. This also means that the length of the diameter of a sphere is twice the radius of the sphere.
So, we can define the relation of radius and diameter as follows:
\[d=2r\]
The Surface Area of a Sphere
The surface area of a sphere is the area of the curved surface of the sphere. The surface area of the sphere is also called the total surface area or the outer surface area. The surface area of a sphere is determined by the radius of the sphere. The formula to calculate the surface area of a sphere is:
Curved surface or total surface area of the sphere `=4\pi r^2`
Where, `r` is the radius of the sphere.
`(\text{pi})` is given as `22/7` or approximately `3.14`.
The area of a sphere is always measured in square units, for example `\text{cm}^2`, `\text{m}^2`, `\text{inch}^2` which directly depends on the unit of the radius of the sphere.
The surface area of a sphere calculation has numerous applications such as the area required to cover a spherical body, the surface area of the earth or other celestial bodies, for economic calculations such as paint cost for painting a spherical dome, etc.
The Volume of a Sphere
The volume of a sphere is defined as the volume needed to fill up the sphere, or the volume of liquid contained by the sphere. So, the volume of a sphere is defined as the capacity of the sphere or the space occupied by the sphere. Mathematically, it can be expressed as follows:
Volume or capacity of a sphere `=4/3 \pi r^3`
where, `r` is the radius of the sphere.
`(\text{pi})` is given as `22/7` or approximately `3.14`.
The units of measurements of volume are always in cubic units such as `\text{cm}^3`, `\text{m}^3`, `\text{inch}^3` which directly depends upon the unit of radius.
The volume of sphere calculation has numerous applications which include the capacity of the liquid containers, the volume of spherical balloons containing gas, the volume of celestial bodies, etc.
Solved Examples
Example `1`: A tennis ball is wrapped around with a melton cloth. The radius of a tennis ball is measured as `3.35` cm. Find the amount of cloth required to cover the tennis ball. (Consider `\pi=3.14`)
Solution:
Here, the radius of a tennis ball is given as `3.35` cm. Therefore, `r=3.35` cm
\(\begin{aligned}
\text{Curved surface area of the tennis ball} &=4 \pi r^2\\&= 4 \times 3.14 \times (3.35)^2 \\
&= 4 \times 3.14 \times 11.2225 \\
&\approx 140.95 \, \text{cm}^2
\end{aligned}\)
Therefore, the amount of melton cloth required is approximately `140.95` `\text{cm}^2`.
Example `2`: While experimenting with condensation, Amy is required to create water droplets of a diameter of `1` mm. Calculate the surface area of each such water droplet. (Consider `\pi=22/7`)
Solution:
Here, the diameter of the water droplet is `1` mm i.e. `d=1` mm, Therefore, the radius is
`r=d/2=1/2=0.5` mm
\(\begin{aligned}
\text{Curved Surface Area of the Water Droplet} &= 4 \pi r^2 \\
&= 4 \times \frac{22}{7} \times (0.5)^2 \\
&=\frac{22}{7} \, \text{mm}^2 \\
&\approx 3.14 \, \text{mm}^2
\end{aligned}\)
Therefore, the curved surface area of the water droplet is approximately `3.14` `\text{mm}^2`.
Example `3`: Jacob is playing with a spherical container. He wants to measure the amount of orange juice the container can hold. Calculate the amount of orange juice in liters if the radius of the container is `3.5` cm. (Consider `\pi=3.14`)
Solution:
Here, the radius of the container is `3.5` cm i.e. `r=3.5` cm.
\(\begin{align}
\text{Volume of the Container} &= \frac{4}{3} \pi r^3 \\
&= \frac{4}{3} \times 3.14 \times (3.5)^3 \\
&= \frac{4}{3} \times 3.14 \times 42.875 \\
&\approx 179.5 \, \text{cm}^3
\end{align}\)
The volume of orange juice in liters `=` `\text{Volume of juice in cm}^3/1000=179.5/1000=0.1795` `l`.
Practice Problems
Q`1`. A spherical tank has a diameter of `12` m. Calculate its volume. (Consider `\pi=22/7`)
- `1000` `\text{m}^3`
- `905.14` `\text{m}^3`
- `1200.9` `\text{m}^3`
- `809` `\text{m}^3`
Answer: b
Q`2`. A rubber balloon is filled with helium gas. After pumping the balloon with helium gas, its diameter is measured as `30` cm. Calculator its surface area. (Consider `\pi=3.14`)
- `3000` `\text{cm}^2`
- `2900` `\text{cm}^2`
- `2826` `\text{cm}^2`
- `2167.8` `\text{m}^2`
Answer: c
Q`3`. A glass sphere with one side opening is painted with red paint. If it is painted inside out i.e. both internally and externally, calculate the total surface area to be painted. The radius of the glass ball is 20 cm. (Consider `\pi=3.14`)
- `10048` `\text{cm}^2`
- `1197.4` `\text{cm}^2`
- `2900` `\text{cm}^2`
- `1300` `\text{cm}^2`
Answer: a
Q`4`. A spherical tank of water has a diameter of `20` meters. If it can fill up to `85%` of its capacity, find the amount of water that it can contain. (Consider `\pi=3.14`)
- `4188.79` `\text{m}^3`
- `3000.45` `\text{m}^3`
- `3358.67` `\text{m}^3`
- `2280.9` `\text{m}^3`
Answer: c
Frequently Asked Questions
Q`1`. Is circumference related to the sphere?
Answer: No, circumference is not related to the sphere. Circumference is commonly related to two-dimensional shapes. Although we can calculate the circumference of a sphere by the formula as shown below.
Circumference of sphere `=2\pi r`, where `r` is the radius of the sphere.
Q`2`. Is there any other round shape?
Answer: Yes, there exist shapes like ellipsoids, ovoids, etc. which are round but not perfectly round. These shapes are similar to spheres, but their properties are completely different.
Q`3`. Is a hemisphere different from a sphere?
Answer: Yes, a hemisphere is made from a sphere. If a sphere is cut by a plane along the diameter of the sphere with any orientation, then it will be divided into two equal parts. These two parts obtained are called hemispheres. A hemisphere is shown in the image below.
Note that the diameter and radius of a hemisphere are the same as that of a sphere.
Q`4`. Does a hollow sphere exist?
Answer: Yes, hollow spheres exist. When the sphere material is removed from the inside and is used to fill some liquid or gas, it is called a hollow sphere. A spherical tank is an example of a hollow sphere. Below is an image of a hollow sphere.
