# Surface Area Of A Sphere

A Sphere is a three-dimensional solid object having a round structure, just like a circle. The difference between a sphere and a circle is that, a circle is in 2-dimension, whereas, a sphere is a 3-dimensional object.

From a visual perspective, it has a three-dimensional structure that forms by rotating a disc that is circular with one of the diagonals. Let us consider an instance where spherical ball faces are painted. To paint the whole surface, the paint quantity required has to be known beforehand. Hence area of every face has to be known to calculate the paint quantity for painting the same. We define this term as the total surface area. Surface area of a sphere is equal to the areas of entire face surrounding it.

Area of a Sphere- It is the amount of region covered by the surface of a sphere.

Mathematical Formula for Surface Area of a Sphere-

 $4\pi r^{2}$

Total Surface Area- It is the sum of total area of all the surfaces be it curved surface or flat surface.

Curved Surface Area- It is the sum of the area of curved surface (leaving the flat bases).

In case of a Sphere, we only have curved surfaces (no flat surface).

Therefore the Total surface area = Curved surface area

 Example- Calculate the cost required to paint a football which is in shape of a sphere having radius of 7 cm. If the painting cost of football is INR 2.5/$cm^{2}$. (Take ? = 22/7)Solution- We know, Total surface area of a sphere = $4\pi r^{2}$ = 4 × (22/7) × 7 × 7 $= 616 \; cm^{2}$ Total cost of painting the container = 2.5 × 616 = Rs. 1540 Example- Calculate the curved surface area of a sphere having radius equals to 3.5 cm.(Take ? = 22/7) Solution- We know, Curved surface area = Total surface area = $4\pi r^{2}$ = 4 × (22/7) × 3.5 × 3.5 $= 616 \; cm^{2}$ $= 154 \; cm^{2}$<

This was just a brief introduction to the calculation of surface area of a sphere. To solve NCERT solutions related to calculation of surface area of a sphere, visit our site BYJU’S.

#### Practise This Question

When we rotate a semi-circle along the axis parallel to its plane. What shape do we obtain?