Volume Problem Solving

To solve problems on this page, you should be familiar with the following:

Volume - Cuboid

Volume - Sphere

Volume - Cylinder

Volume - Pyramid

This wiki includes several problems motivated to enhance problem-solving skills. Before getting started, recall the following formulas:

Volume of sphere with radius $r:$ $\frac43 \pi r^3$
Volume of cube with side length $L:$ $L^3$
Volume of cone with radius $r$ and height $h:$ $\frac13\pi r^2h$
Volume of cylinder with radius $r$ and height $h:$ $\pi r^2h$
Volume of a cuboid with length $l$ , breadth $b$ , and height $h:$ $lbh$

Volume Problem Solving - Basic

This section revolves around the basic understanding of volume and using the formulas for finding the volume. A couple of examples are followed by several problems to try.

Find the volume of a cube of side length $10\text{ cm}$ .

$\begin{aligned} (\text {Volume of a cube}) & = {(\text {Side length}})^{3}\\ & = {10}^{3}\\ & = 1000 ~\big(\text{cm}^{3}\big).\ _\square \end{aligned}$

Find the volume of a cuboid of length $10\text{ cm}$ , breadth $8\text{ cm}$ . and height $6\text{ cm}$ .

$\begin{aligned} (\text {Area of a cuboid}) & = l × b × h\\ & = 10 × 8 × 6\\ & = 480 ~\big(\text{cm}^{3}\big).\ _\square \end{aligned}$

I made a large ice cream cone of a composite shape of a cone and a hemisphere. If the height of the cone is 10 and the diameter of both the cone and the hemisphere is 6, what is the volume of this ice cream cone?

The volume of the composite figure is the sum of the volume of the cone and the volume of the hemisphere.

Recall the formulas for the following two volumes: $V_{\text{cone}} = \frac13 \pi r^2 h$ and $V_{\text{sphere}} =\frac43 \pi r^3$ . Since the volume of a hemisphere is half the volume of a a sphere of the same radius, the total volume for this problem is

$\frac13 \pi r^2 h + \frac12 \cdot \frac43 \pi r^3.$

With height $h =10$ , and diameter $d = 6$ or radius $r = \frac d2 = 3$ , the total volume is $48\pi. \ _\square$

Find the volume of a cone having slant height $17\text{ cm}$ and radius of the base $15\text{ cm}$ .

Let $h$ denote the height of the cone, then

$\begin{aligned} (\text{slant height}) &=\sqrt {h^2 + r^2}\\ 17&= \sqrt {h^2 + 15^2}\\ 289&= h^2 + 225\\ h^2&=64\\ h& = 8. \end{aligned}$

Since the formula for the volume of a cone is $\dfrac {1}{3} ×\pi ×r^2×h$ , the volume of the cone is

$\frac {1}{3}×3.14× 225 × 8= 1884 ~\big(\text{cm}^{2}\big). \ _\square$

Find the volume of the following figure which depicts a cone and an hemisphere, up to $2$ decimal places. In this figure, the shape of the base of the cone is circular and the whole flat part of the hemisphere exactly coincides with the base of the cone (in other words, the base of the cone and the flat part of the hemisphere are the same).

Use $\pi=\frac{22}{7}.$

$\begin{aligned} (\text{Volume of cone}) & = \dfrac {1}{3} \pi r^2 h\\ & = \dfrac {1 × 22 × 36 × 8}{3 × 7}\\ & = \dfrac {6336}{21} = 301.71 \\\\ (\text{Volume of hemisphere}) & = \dfrac {2}{3} \pi r^3\\ & = \dfrac {2 × 22 × 216}{3 × 7}\\ & = \dfrac {9504}{21} = 452.57 \\\\ (\text{Total volume of figure}) & = (301.71 + 452.57) \\ & = 754.28.\ _\square \end{aligned}$

Try the following problems.

Volume - Problem Solving - Intermediate

This section involves a deeper understanding of volume and the formulas to find the volume. Here are a couple of worked out examples followed by several "Try It Yourself" problems:

$12$ spheres of the same size are made from melting a solid cylinder of $16\text{ cm}$ diameter and $2\text{ cm}$ height. Find the diameter of each sphere.

Use $\pi=\frac{22}{7}.$

The volume of the cylinder is

$\pi× r^2 × h = \frac {22×8^2×2}{7}= \frac {2816}{7}.$

Let the radius of each sphere be $r\text{ cm}.$ Then the volume of each sphere in $\text{cm}^3$ is

$\dfrac {4×22×r^3}{3×7} = \dfrac{88×r^3}{21}.$

Since the number of spheres is $\frac {\text{Volume of cylinder}}{\text {Volume of 1 sphere}},$

\[\begin{align} 12 &= \dfrac{2816×21}{7×88×r^3}\\ &= \dfrac {96}{r^3}\\
r^3 &= \dfrac {96}{12}\\ &= 8\\ \Rightarrow r &= 2. \end{align}\]

Therefore, the diameter of each sphere is $2\times r = 2\times 2 = 4 ~(\text{cm}). \ _\square$

Find the volume of a hemispherical shell whose outer radius is $7\text{ cm}$ and inner radius is $3\text{ cm}$ , up to $2$ decimal places.

We have

$\begin{aligned} (\text {Volume of inner hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × R^3\\ & = \dfrac {1 × 4 × 22 × 27}{2 × 3 × 7}\\ & = \dfrac {396}{7}\\ & = 56.57 ~\big(\text{cm}^{3}\big) \\\\ (\text {Volume of outer hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × r^3\\ & = \dfrac {1 × 4 × 22 × 343}{2 × 3 × 7}\\ & = \dfrac {2156}{7}\\ & = 718.66 ~\big(\text{cm}^{3}\big) \\\\ (\text{Volume of hemispherical shell}) & = (\text{V. of outer hemisphere}) - (\text{V. of inner hemisphere})\\ & = 718.66 - 56.57 \\ & = 662.09 ~\big(\text{cm}^{3}\big).\ _\square \end{aligned}$

Try the following problems.

Volume Problem Solving - Advanced

Please remember this section contains highly advanced problems of volume. Here it goes:

Consider a tetrahedron with side lengths $2, 3, 3, 4, 5, 5$ . The largest possible volume of this tetrahedron has the form $\frac {a \sqrt{b}}{c}$ , where $b$ is an integer that's not divisible by the square of any prime, $a$ and $c$ are positive, coprime integers. What is the value of $a+b+c$ ?

The correct answer is: 13

Label the tetrahedron $WXYZ$ . Consider the edge of side length 2. Let the vertices be $W, X$ . Since $WXY$ and $WXZ$ are triangles, they must satisfy the triangle inequality. These are the following possibilities for the triangles: A. $\{ 3, 3 \}$ , B. $\{ 5, 5\}$ , C. $\{ 4, 5\}$ , D. $\{ 3,4\}$ . Hence, the only possibilities for $WXY$ and $WXZ$ are: A, B, or A, C, or B, D. Thus, we consider the the following cases:

Case 1. A, B. Without loss of generality, we have $WX = 2$ , $WY=3$ , $XY=3$ , $WZ=5$ , $XZ=5$ and so $YZ=4$ . Thus, $WYZ$ and $XYZ$ are both 3-4-5 triangles, hence right angled. This tetrahedron has base $WXY$ (with area $\frac {1}{2} \times 2 \times \sqrt{3^2-1^2}$ ) and height $YZ = 4$ , so has volume $\frac {1}{3} \times 4 \times \sqrt{8} = \frac {8\sqrt{2}}{3}$ .

Case 2. A, C. We have 2 subcases, depending on what $WZ, XZ$ are equal to. However, these tetrahedrons are reflections of each other, so have the same volume. Without loss of generality, we have $WX = 2$ , $WY=3$ , $XY=3$ , $WZ=4$ , $XZ=5$ and so $YZ = 5$ . Notice that this has the same base as Case 1, so let's consider the height of the tetrahedron. From $Z$ , drop a perpendicular to $WX$ , intersecting at $E$ . Since $2^2 + 4^2 > 5^2$ , $WXZ$ is an acute triangle. Thus, $WE < 4$ , which also means that the height of the tetrahedron is less than $4$ . Thus, the volume will be less than Case 1.

Case 3. B, D. Similarly, we have 2 sub cases, depending on what $WZ, XZ$ are equal to. However, these tetrahedrons are reflections of each other, so have the same volume. Without loss of generality, we have $WX = 2$ , $WY=5$ , $XY=5$ , $WZ=3$ , $XZ=4$ and so $YZ=3$ . Consider base $WZX$ with height $YZ$ , then the volume of the tetrahedron is $\frac {1}{3} [WZX] h_C \leq \frac {1}{3} \cdot ( \frac {1}{2} WX \times WZ ) \cdot YZ = 3$ . This is clearly less than Case 1.

Therefore the largest volume possible is $\frac {8\sqrt{2}}{3}$ . Hence $a + b + c = 8 + 2 + 3 = 13$ .

Contents

Volume Problem Solving - Basic

Find the volume of a cube of side length $10\text{ cm}$ .

Find the volume of a cuboid of length $10\text{ cm}$ , breadth $8\text{ cm}$ . and height $6\text{ cm}$ .

I made a large ice cream cone of a composite shape of a cone and a hemisphere. If the height of the cone is 10 and the diameter of both the cone and the hemisphere is 6, what is the volume of this ice cream cone?

Find the volume of a cone having slant height $17\text{ cm}$ and radius of the base $15\text{ cm}$ .

Volume - Problem Solving - Intermediate

Find the volume of a hemispherical shell whose outer radius is $7\text{ cm}$ and inner radius is $3\text{ cm}$ , up to $2$ decimal places.

Volume Problem Solving - Advanced

See Also

Contents

Find the volume of a cube of side length 10 cm10\text{ cm}10 cm.

Find the volume of a cuboid of length 10 cm10\text{ cm}10 cm, breadth 8 cm8\text{ cm}8 cm. and height 6 cm6\text{ cm}6 cm.

I made a large ice cream cone of a composite shape of a cone and a hemisphere. If the height of the cone is 10 and the diameter of both the cone and the hemisphere is 6, what is the volume of this ice cream cone?

Find the volume of a cone having slant height 17 cm17\text{ cm}17 cm and radius of the base 15 cm15\text{ cm}15 cm.

Find the volume of a hemispherical shell whose outer radius is 7 cm7\text{ cm}7 cm and inner radius is 3 cm3\text{ cm}3 cm, up to 222 decimal places.

Find the volume of a cube of side length $10\text{ cm}$ .

Find the volume of a cuboid of length $10\text{ cm}$ , breadth $8\text{ cm}$ . and height $6\text{ cm}$ .

Find the volume of a cone having slant height $17\text{ cm}$ and radius of the base $15\text{ cm}$ .

Find the volume of a hemispherical shell whose outer radius is $7\text{ cm}$ and inner radius is $3\text{ cm}$ , up to $2$ decimal places.