Lesson Solved problems on volume of spheres

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> Lesson Solved problems on volume of spheres      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   


This Lesson (Solved problems on volume of spheres) was created by by ikleyn(52787) About Me : View Source, Show
About ikleyn:

Solved problems on volume of spheres


In this lesson you will find typical solved problems on volume of spheres.
The theoretical base for these problems is the lesson  Volume of spheres  under the topic  Volume, metric volume  of the section  Geometry  in this site.


Problem 1

Find the volume of a sphere if its radius is of  3 cm.

Solution

The volume of the sphere is

4%2F3pi%2Ar%5E3 = 4%2F3pi%2A3%5E3 = 36%2Api = 36%2A3.14159 = 113.097 cm%5E3 (approximately).

Answer.  The volume of the sphere is  113.097 cm%5E3 (approximately).


Problem 2

Find the volume 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  10 cm,  and the cylinder height is of  20 cm.

Solution

The volume of the composite body under consideration is the sum of the volume                              
of the cylinder  pir%5E2h  and the volume of the hemisphere  2%2F3pir%5E3.

So,  the volume of the composite body is equal to

V = pir%5E2h + 2%2F3pir%5E3 = pi*(r%5E2%2Ah+%2B+%282%2F3%29%2Ar%5E3) = pi%2A%285%5E2%2A20+%2B+%282%2F3%29%2A5%5E3%29 =

= 3.14159*583.33 = 1832.59cm%5E3 (approximately).


Figure 1.  To the  Problem 2

Answer.  The volume of the composite body under consideration is  1832.59 cm%5E3 (approximately).


Problem 3

Find the volume 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  10 cm  and the cone height is of  5 cm.

Solution

The volume of the composite body under consideration is the sum of the volume                              
of the cone  1%2F3pir%5E2h   and the volume of the hemisphere  2%2F3pir%5E3.

So,  the total volume of the composite body is equal to

S = 1%2F3pir%5E2h + 2%2F3pir%5E3 = 1%2F3pi%2Ar%5E2*%28h+%2B+2r%29 =

= 1%2F3pi%2A5%5E2*(5+%2B+2%2A5%29) = 1%2F3pi%2A25*15 = 125%2Api = 125*3.14159 =

= 392.7 cm%5E3 (approximately).


Figure 2. To the Problem 3

Answer.  The volume of the composite body under consideration is  392.7 cm%5E3 (approximately).


Problem 4

Find the volume 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  10 cm.

Solution

The volume of the composite body under consideration is the sum of the volume                              
of the cube  a%5E3  plus the volume of the hemisphere  2%2F3pir%5E3,  where r = 10%2F2 = 5 cm.

So,  the volume of the given composite body is equal to

V = a%5E3 + 2%2F3pi%28a%2F2%29%5E3 = 10%5E3 + 2%2F33.14159%2A5%5E3 = 1000 + 261.8 = 1261.8 cm%5E3 (approximately).



Figure 3. To the Problem 4

Answer.  The volume of the composite body under consideration is  1261.8 cm%5E3 (approximately).


Problem 5

Find the volume of the sphere inscribed in a cone if the base diameter of the cone is of  30 cm  and the height of the cone is of  36 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  Problem 5      




  Figure 4b. To the solution
        of the  Problem 5
For our isosceles triangle,  we have l = sqrt%28%2830%2F2%29%5E2+%2B+36%5E2%29 = 39 cm  for its lateral side length,  P = 39+%2B+39+%2B+30 = 108 cm  for the perimeter  and  S = 1%2F230%2A36 = 540 cm%5E2  for the area.
Therefore,  the radius of the inscribed circle is  r = 2%2A540%2F108 = 10 cm  in accordance with the formula above.
Hence,  the volume of the sphere inscribed in the cone is  4%2F3pir%5E3 = 4%2F3pi%2A10%5E3 = 1333.33%2Api = 1333.33*3.14159 = 4188.79 cm%5E3.

Answer.  The volume of the sphere inscribed in the cone is  4188.79 cm%5E3 (approximately).


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

Lessons on volume of prisms

Volume of prisms
Solved problems on volume of prisms
Overview of lessons on volume of prisms                    

Lessons on volume of pyramids

Volume of pyramids
Solved problems on volume of pyramids
Overview of lessons on volume of pyramids

Lessons on volume of cylinders

Volume of cylinders
Solved problems on volume of cylinders
Overview of lessons on volume of cylinders                

Lessons on volume of cones

Volume of cones
Solved problems on volume of cones
Overview of lessons on volume of cones                    

Lessons on volume of spheres

Volume of spheres
Solved problems on volume of spheres
Overview of lessons on volume of spheres


To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.


This lesson has been accessed 9373 times.