Lesson Surface area of spheres

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Surface area of spheres


A  sphere  is a surface in a  3D  space all points of which are  equidistant  from one point called the  center of the sphere.
Figure 1a  shows the sphere.

A  radius of a sphere  is a straight segment connecting the center of the sphere with a point on the sphere surface  (Figure 1b).
A  diameter of a sphere  is the straight segment passing through the center of the sphere and connecting the opposite points of the sphere  (Figure 1c).

  
      Figure 1a. A sphere
          

        
          Figure 1b. A sphere
              and its radius

        
          Figure 1c. A sphere
              and its diameter


Properties of spheres


1.  Every section of a sphere by a plane is a circle.

2.  A tangent segment to a sphere released from a point in  3D  space outside the sphere is perpendicular to the radius of the sphere drawn from its center
     to the tangent point.

3.  All tangent segment to a sphere released from one fixed point outside the sphere have the same length.


Formula for calculating the surface area of spheres


The surface area of a sphere  is  S =  4pir%5E2 = pid%5E2,
where  r  is the radius of the sphere and  d  is the diameter of the sphere.


Example 1

Find the surface area of a sphere if its radius is of  10 cm.

Solution

The surface area of the sphere is

4pi%2Ar%5E2 = 4%2A3.14159%2A10%5E2 = 3.14159%2A400 = 1256.636 cm%5E2 (approximately).

Answer.  The surface area of the sphere is  1256.636 cm%5E2 (approximately).


Example 2

Find the surface area of a composite body comprised of a right circular cylinder and a hemisphere attached center-to-center to one of the cylinder bases  (Figure 1)  if both the cylinder diameter and the hemisphere diameter are of  20 cm,  and the cylinder height is of  40 cm.

Solution

The full surface area of the composite body under consideration is the sum of                        
the lateral surface area of the cylinder  2pi%2Ar%2Ah,  the area of the base of the
cylinder  pi%2Ar%5E2  and the area of the hemisphere  2pi%2Ar%5E2.

So,  the total surface area of the composite body is equal to

S = 2pi%2Ar%2Ah + pi%2Ar%5E2 + 2pi%2Ar%5E2 = 2pi%2Ar%2Ah + 3pi%2Ar%5E2 = pi*(2r%2Ah+%2B+3r%5E2) =
= 3.14159*(2*10*40 + 3*10^2) = 3.14159*1100 = 3455.749 cm%5E2 (approximately).


Figure 1.  To the  Example 2

Answer.  The surface area of the composite body under consideration is  3455.749 cm%5E2 (approximately).


Example 3

Find the surface area of a composite body comprised of a cone and a hemisphere attached center-to-center to the cone base  (Figure 2)  if both the cone base diameter and the hemisphere diameter are of  20 cm  and the cone height is of  20 cm.

Solution

The full surface area of the composite body under consideration is the sum of                        
the lateral surface area of the cone  pi%2Ar%2Aslant_height  and the area of the
hemisphere  2pi%2Ar%5E2.

So,  the total surface area of the composite body is equal to

S = pi%2Ar%2Aslant_height + 2pi%2Ar%5E2 = pi%2Ar%2Asqrt%28r%5E2+%2B+h%5E2%29 + 2pi%2Ar%5E2 = pi%2Ar*(sqrt%28r%5E2%2Bh%5E2%29+%2B+2r) =
= 3.14159%2A10*(sqrt%2810%5E2+%2B+20%5E2%29+%2B+2%2A10) = 3.14159%2A10*%28sqrt%28500%29+%2B+20%29 = 3.14159%2A10*%2810sqrt%285%29+%2B+20%29 =
= 1330.8 cm%5E2 (approximately).


Figure 2. To the Example 3

Answer.  The surface area of the composite body under consideration is  1330.8 cm%5E2 (approximately).


Example 4

Find the surface area of a composite body comprised of a cube and a hemisphere attached center-to-center to one of the cube faces  (Figure 3)  if both the cube edge measure and the hemisphere diameter are of  20 cm.

Solution

The full surface area of the composite body under consideration is the sum of                        
the surface area of the six cube faces  6a%5E2  minus the area of the base of the
hemisphere  pi%2Ar%5E2  plus the area of the hemisphere  2pi%2Ar%5E2,  where r = 20%2F2 = 10 cm.

So,  the total surface area of the composite body is equal to

S = 6a%5E2 - pi%2A%28a%2F2%29%5E2 + 2pi%2A%28a%2F2%29%5E2 = 6a%5E2 + pi%2A%28a%2F2%29%5E2 = 6%2A20%5E2 + 3.14159%2A10%5E2 =
= 2714.159 cm%5E2 (approximately).



Figure 3. To the Example 4

Answer.  The surface area of the composite body under consideration is  2714.159 cm%5E2 (approximately).


Example 5

Find the surface area of the sphere inscribed in a cone if the base diameter of the cone is of  24 cm  and the height of the cone is of  16 cm  (Figure 4a).

Solution

Figure 4a  shows  3D  view of the cone with the inscribed sphere.  Figure 4b                            
shows the axial section of this cone and the inscribed sphere as the isosceles
triangle with the inscribed circle.

The radius of the inscribed sphere in the cone in the  Figure 4a  is the same
as the radius of the inscribed circle in the triangle in the  Figure 4b.  So,
instead of determining the radius of the sphere we will find the radius of the
inscribed circle.  For it,  use the formula  r = 2S%2FP,  where  r  is the radius
of the inscribed circle in a triangle,  S  is the area of the triangle and  P  is
the perimeter of the triangle.  The proof of this formula is in the lesson
Proof of the formula for the area of a triangle via the radius of the inscribed
circle  under the topic  Area and surface area  of the section  Geometry
in this site.



Figure 4a. To the  Example 5      




  Figure 4b. To the solution
        of the  Example 5
For our isosceles triangle,  we have l = sqrt%28%2824%2F2%29%5E2+%2B+16%5E2%29 = 20 cm  for its lateral side length,  P = 20+%2B+20+%2B+24 = 64 cm  for the perimeter  and  S = 1%2F216%2A12 = 96 cm%5E2  for the area.  Therefore,  the radius of the inscribed circle is  r = 2%2A96%2F64 = 3 cm  in accordance with the formula above.
Hence,  the area of the sphere inscribed in the cone is  4pi%2Ar%5E2 = 4%2A3.14159%2A3%5E2 = 113.097 cm%5E2.

Answer.  The surface area of the sphere inscribed in the cone is  113.097 cm%5E2 (approximately).


My lessons on surface area of spheres and other 3D solid bodies in this site are

Lessons on surface area of prisms

Surface area of prisms
Solved problems on surface area of prisms
Overview of lessons on surface area of prisms                  

Lessons on surface area of pyramids

Surface area of pyramids
Solved problems on surface area of pyramids
Overview of lessons on surface area of pyramids

Lessons on surface area of cylinders

Surface area of cylinders
Solved problems on surface area of cylinders
Overview of lessons on surface area of cylinders              

Lessons on surface area of cones

Surface area of cones
Solved problems on surface area of cones
Overview of lessons on surface area of cones                

Lessons on surface area of spheres

Surface area of spheres
Solved problems on surface area of spheres
Overview of lessons on surface area of spheres


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