Lesson Solved problems on volume of spheres

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<H2>Solved problems on volume of spheres</H2>

In this lesson you will find typical solved problems on volume of spheres.

The theoretical base for these problems is the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Volume/Volume-of-spheres.lesson>Volume of spheres</A>&nbsp; under the topic &nbsp;<B>Volume, metric volume</B>&nbsp; of the section &nbsp;<B>Geometry</B>&nbsp; in this site.

<H3>Problem 1</H3>Find the volume of a sphere if its radius is of &nbsp;3 cm.

<B>Solution</B>

The volume of the sphere is

{{{4/3}}}{{{pi*r^3}}} = {{{4/3}}}{{{pi*3^3}}} = {{{36*pi}}} = {{{36*3.14159}}} = 113.097 {{{cm^3}}} (approximately).

<B>Answer</B>. &nbsp;The volume of the sphere is &nbsp;113.097 {{{cm^3}}} (approximately).

<H3>Problem 2</H3>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 &nbsp;(<B>Figure 1</B>)&nbsp; if both the cylinder diameter and the hemisphere diameter are of &nbsp;10 cm, &nbsp;and the cylinder height is of &nbsp;20 cm.

<TABLE>
  <TR>
  <TD>
<B>Solution</B>

The volume of the composite body under consideration is the sum of the volume&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

of the cylinder &nbsp;{{{pi}}}{{{r^2}}}{{{h}}}&nbsp; and the volume of the hemisphere &nbsp;{{{2/3}}}{{{pi}}}{{{r^3}}}.

So, &nbsp;the volume of the composite body is equal to

{{{V}}} = {{{pi}}}{{{r^2}}}{{{h}}} + {{{2/3}}}{{{pi}}}{{{r^3}}} = {{{pi}}}*({{{r^2*h + (2/3)*r^3}}}) = {{{pi*(5^2*20 + (2/3)*5^3)}}} =

= 3.14159*583.33 = 1832.59{{{cm^3}}} (approximately).

</TD>
  <TD>
{{{drawing( 210, 195,  -3.5, 3.5, -1.0, 5.5,

ellipse( 0.5, 3.5, 3.0, 1.0),
            ellipse( 0.5, 0.0, 3.0, 1.0),

line( -1, 3.5, -1, 0.0),
              line(  2, 3.5,  2, 0.0),

arc ( 0.5, 3.5, 3.00, 3.00, 180, 360),
              arc ( 0.5, 3.5, 3.06, 3.06, 180, 360)
)}}}

<B>Figure 1</B>. &nbsp;To the &nbsp;<B>Problem 2</B>

</TD>
  </TR>
</TABLE>

<B>Answer</B>. &nbsp;The volume of the composite body under consideration is &nbsp;1832.59 {{{cm^3}}} (approximately).

<H3>Problem 3</H3>Find the volume of a composite body comprised of a cone and a hemisphere attached center-to-center to the cone base &nbsp;(<B>Figure 2</B>)&nbsp; if both the cone base diameter and the hemisphere diameter are of &nbsp;10 cm&nbsp; and the cone height is of &nbsp;5 cm.

<TABLE>
  <TR>
  <TD> 
<B>Solution</B>

of the cone &nbsp;{{{1/3}}}{{{pi}}}{{{r^2}}}{{{h}}} &nbsp; and the volume of the hemisphere &nbsp;{{{2/3}}}{{{pi}}}{{{r^3}}}.

So, &nbsp;the total volume of the composite body is equal to

{{{S}}} = {{{1/3}}}{{{pi}}}{{{r^2}}}{{{h}}}  + {{{2/3}}}{{{pi}}}{{{r^3}}} = {{{1/3}}}{{{pi*r^2}}}*{{{(h + 2r)}}} =

= {{{1/3}}}{{{pi*5^2}}}*({{{5 + 2*5)}}}) = {{{1/3}}}{{{pi*25}}}*{{{15}}} = {{{125*pi}}} = 125*3.14159 =

= 392.7 {{{cm^3}}} (approximately).

</TD>
  <TD>
{{{drawing( 195, 195,  -3.0, 3.5, -1.0, 5.5,

ellipse( 0.5, 3.5, 3.0, 1.0),

line( -1.0, 3.5,  0.5, 0.0),
              line(  2.0, 3.5,  0.5, 0.0),

arc ( 0.5, 3.5, 3.00, 3.00, 180, 360),
              arc ( 0.5, 3.5, 3.06, 3.06, 180, 360)
)}}}

<B>Figure 2</B>. To the <B>Problem 3</B>

</TD>
 </TR>
</TABLE>

<B>Answer</B>. &nbsp;The volume of the composite body under consideration is &nbsp;392.7 {{{cm^3}}} (approximately).

<H3>Problem 4</H3>Find the volume of a composite body comprised of a cube and a hemisphere attached center-to-center to one of the cube faces &nbsp;(<B>Figure 3</B>)&nbsp; if both the cube edge measure and the hemisphere diameter are of &nbsp;10 cm.

<TABLE>
  <TR>
  <TD> 
<B>Solution</B>

of the cube &nbsp;{{{a^3}}}&nbsp; plus the volume of the hemisphere &nbsp;{{{2/3}}}{{{pi}}}{{{r^3}}}, &nbsp;where {{{r}}} = {{{10/2}}} = 5 cm.

So, &nbsp;the volume of the given composite body is equal to

{{{V}}} = {{{a^3}}} + {{{2/3}}}{{{pi}}}{{{(a/2)^3}}} = {{{10^3}}} + {{{2/3}}}{{{3.14159*5^3}}} = 1000 + 261.8 = 1261.8 {{{cm^3}}} (approximately).

</TD>
  <TD>
{{{drawing( 180, 180,  -2.5, 3.5, -0.5, 5.5,

line (  0.0, 0.0,  3.0,  0.0),
            line (  0.0, 0.0,  0.0,  3.0),
            line (  0.0, 0.0, -2.0,  0.8),

line (  0.0, 3.0,  3.0,  3.0),
            line (  3.0, 3.0,  3.0,  0.0),

line (  0.0, 3.0, -2.0,  3.8),
            line ( -2.0, 3.8,  1.0,  3.8),
            line ( -2.0, 3.8, -2.0,  0.8),
      green(line (  1.0, 3.8,  1.0,  0.8)),

line (  1.0, 3.8,  3.0,  3.0),
      green(line ( -2.0, 0.8,  1.0,  0.8)),
      green(line (  1.0, 0.8,  3.0,  0.0)),

locate( 1.2,  0.1, a),
            locate(-1.5,  0.6, a),
            locate(-0.4,  2.1, a),

ellipse( 0.52, 3.43, 2.54, 0.77),
            arc    ( 0.52, 3.43, 2.54, 2.54, 175, 354)
)}}}

<B>Figure 3</B>. To the <B>Problem 4</B>

</TD>
 </TR>
</TABLE>

<B>Answer</B>. &nbsp;The volume of the composite body under consideration is &nbsp;1261.8 {{{cm^3}}} (approximately).

<H3>Problem 5</H3>Find the volume of the sphere inscribed in a cone if the base diameter of the cone is of &nbsp;30 cm&nbsp; and the height of the cone is of &nbsp;36 cm &nbsp;(<B>Figure 4a</B>).

<TABLE>
  <TR>
  <TD> 
<B>Solution</B>

<B>Figure 4a</B>&nbsp; shows &nbsp;<B>3D</B>&nbsp; view of the cone with the inscribed sphere. &nbsp;<B>Figure 4b</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

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 &nbsp;<B>Figure 4a</B>&nbsp; is the same

as the radius of the inscribed circle in the triangle in the &nbsp;<B>Figure 4b</B>. &nbsp;So,

instead of determining the radius of the sphere we will find the radius of the

inscribed circle. &nbsp;For it, &nbsp;use the formula &nbsp;{{{r}}} = {{{2S/P}}}, &nbsp;where &nbsp;{{{r}}}&nbsp; is the radius

of the inscribed circle in a triangle, &nbsp;{{{S}}}&nbsp; is the area of the triangle and &nbsp;{{{P}}}&nbsp; is

the perimeter of the triangle. &nbsp;The proof of this formula is in the lesson

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Proof-of-the-formula-for-the-area-of-a-triangle-via-the-radius-of-the-inscribed-circle.lesson>Proof of the formula for the area of a triangle via the radius of the inscribed

circle</A>&nbsp; under the topic &nbsp;<B>Area and surface area</B>&nbsp; of the section &nbsp;<B>Geometry</B>

in this site.

</TD>
  <TD> 
{{{drawing( 210, 249,  -3.5, 3.5, -3.8, 4.5,

ellipse (0.0, -2.0, 6.0, 3.0),

green(ellipse (0.0,  0.8, 3.2, 1.6)),

line ( -2.75, -1.5,  -1.375,  1.25),
            line (  2.82, -1.5,   1.410,  1.25),

line ( -1.375,  1.25,  0.0,  4.0),
            line (  1.410,  1.25,  0.0,  4.0),

green(line (0, -2, 0, 4)),
            circle(  0.0, -0.12, 1.88, 1.88)
)}}}

<B>Figure 4a</B>. To the &nbsp;<B>Problem 5</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD> 
{{{drawing( 210, 249,  -3.5, 3.5, -3.8, 4.5,

line (-3.0, -2.0, 3.0, -2.0),

line ( -3.0, -2.0,   0.0, 4.0),
            line (  3.0, -2.0,   0.0, 4.0),

green(line (0, -2, 0, 4)),
            circle(  0.0, -0.12, 1.88, 1.88),

line ( 0.0, -0.2, -1.6,  0.8),
            line ( 0.0, -0.2,  1.6,  0.8),
            line ( 0.0, -0.2,  0.0, -2.0)
)}}}

&nbsp;&nbsp;<B>Figure 4b</B>. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the &nbsp;<B>Problem 5</B>

</TD>
 </TR>
</TABLE>For our isosceles triangle, &nbsp;we have {{{l}}} = {{{sqrt((30/2)^2 + 36^2)}}} = 39 cm&nbsp; for its lateral side length, &nbsp;{{{P}}} = {{{39 + 39 + 30}}} = 108 cm &nbsp;for the perimeter &nbsp;and &nbsp;{{{S}}} = {{{1/2}}}{{{30*36}}} = 540 {{{cm^2}}}&nbsp; for the area.

Therefore, &nbsp;the radius of the inscribed circle is &nbsp;{{{r}}} = {{{2*540/108}}} = {{{10}}} {{{cm}}}&nbsp; in accordance with the formula above.

Hence, &nbsp;the volume of the sphere inscribed in the cone is &nbsp;{{{4/3}}}{{{pi}}}{{{r^3}}} = {{{4/3}}}{{{pi*10^3}}} = {{{1333.33*pi}}} = 1333.33*3.14159 = 4188.79 {{{cm^3}}}.

<B>Answer</B>. &nbsp;The volume of the sphere inscribed in the cone is &nbsp;4188.79 {{{cm^3}}} (approximately).

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

<TABLE>
  <TR>
  <TD>
<B>Lessons on volume of prisms</B>

<A HREF=http://www.algebra.com/algebra/homework/Volume/_Volume-of-prisms.lesson>Volume of prisms</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-volume-of-prisms.lesson>Solved problems on volume of prisms</A>

<A HREF=http://www.algebra.com/algebra/homework/Volume/OVERVIEW-of-LESSONS-on-volume-of-prisms.lesson>Overview of lessons on volume of prisms</A>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
<B>Lessons on volume of pyramids</B>

<A HREF=http://www.algebra.com/algebra/homework/Volume/_Volume-of-pyramids.lesson>Volume of pyramids</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-volume-of-pyramids.lesson>Solved problems on volume of pyramids</A>

<A HREF=http://www.algebra.com/algebra/homework/Volume/OVERVIEW-of-LESSONS-on-volume-of-pyramids.lesson>Overview of lessons on volume of pyramids</A>

</TD>
 </TR>
</Table><TABLE>
  <TR>
  <TD>
<B>Lessons on volume of cylinders</B>

<A HREF=http://www.algebra.com/algebra/homework/Volume/_Volume-of-cylinders.lesson>Volume of cylinders</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-volume-of-cylinders.lesson>Solved problems on volume of cylinders</A>

<A HREF=http://www.algebra.com/algebra/homework/Volume/OVERVIEW-of-LESSONS-on-Volume-of-cylinders.lesson>Overview of lessons on volume of cylinders</A>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
<B>Lessons on volume of cones</B>

<A HREF=http://www.algebra.com/algebra/homework/Volume/Volume-of-cones.lesson>Volume of cones</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-Volume-of-cones.lesson>Solved problems on volume of cones</A>

<A HREF=http://www.algebra.com/algebra/homework/Volume/OVERVIEW-of-LESSONS-on-Volume-of-cones.lesson>Overview of lessons on volume of cones</A>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
<B>Lessons on volume of spheres</B>

<A HREF=http://www.algebra.com/algebra/homework/Volume/Volume-of-spheres.lesson>Volume of spheres</A>

Solved problems on volume of spheres

<A HREF=http://www.algebra.com/algebra/homework/Volume/OVERVIEW-of-LESSONS-on-Volume-of-spheres.lesson>Overview of lessons on volume of spheres</A>

</TD>
 </TR>
</Table>

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.