Volume of a Combination of Solids: Class 10 Examples

Vol­ume for­mu­las and units#

The for­mu­las#

A cone uses the ver­ti­cal height $h$, not the slant height. The slant height is needed only for sur­face area.

Units and capac­ity#

Vol­ume is mea­sured in cubic units. The con­ver­sions you will use most are:

$$\begin{aligned} 1000 \text{ cm}^3 &= 1 \text{ litre} \\ 1 \text{ m}^3 &= 1\,000\,000 \text{ cm}^3 = 1000 \text{ litres} \end{aligned}$$

Capac­ity means the vol­ume of the space inside a con­tainer, usu­ally given in litres.

Unless a ques­tion says oth­er­wise, take $\displaystyle \pi = \frac{22}{7}$.

The key results#

State­ment#

1. If a solid is made by join­ing solids $A$ and $B$, then $V = V_A + V_B$.
2. If a solid $B$ is cut out of a solid $A$, the remain­ing vol­ume is $V = V_A - V_B$.
3. When a solid is fully sunk in water in a con­tainer, the water rises by a vol­ume equal to the vol­ume of the solid.

Why these are true#

Vol­ume mea­sures the space filled by mate­r­ial. When two solids are joined, they do not over­lap: each keeps all its own mate­r­ial, and none of it dis­ap­pears into the join. So the space filled by the whole is the space filled by one part plus the space filled by the other. There is no "hid­den face" cor­rec­tion, because a face has area but no thick­ness, so it has no vol­ume.

When a piece is cut out, exactly that piece's mate­r­ial is lost, so its vol­ume is sub­tracted.

For dis­place­ment: water can­not occupy the space taken by the sunken solid, so the water level goes up until the extra space above the old level equals the solid's vol­ume. If the con­tainer is a cylin­der of radius $R$ and the level rises by $x$, the extra space is a cylin­der of vol­ume $\pi R^2 x$. So

$$\displaystyle \pi R^2 x = V_{\text{solid}} \quad \Rightarrow \quad x = \frac{V_{\text{solid}}}{\pi R^2}.$$

Sur­face area think­ing ver­sus vol­ume think­ing#

A method in steps#

1. Split the object into basic solids and sketch it.
2. Find every radius and height. Halve diam­e­ters. Sub­tract the ends to find a cylin­der's length.
3. Decide for each piece whether it is added (joined on) or sub­tracted (cut out).
4. Write one expres­sion for the vol­ume and take out com­mon fac­tors such as $\pi r^2$.
5. Sub­sti­tute, sim­plify and write the unit in cubic units.
6. Con­vert to litres if capac­ity is asked, or divide by a base area if a rise in level is asked.

Worked exam­ples#

Exam­ple 1: A cylin­der with hemi­spher­i­cal ends#

A solid is a cylin­der of radius 3 cm and length 10 cm with a hemi­sphere of radius 3 cm fixed at each end. Find its vol­ume.

The two hemi­spheres together have the vol­ume of one sphere.

$$\displaystyle \begin{aligned} V &= \pi r^2 h + \frac{4}{3}\pi r^3 \\ &= \pi \times 9 \times 10 + \frac{4}{3}\pi \times 27 \\ &= 90\pi + 36\pi = 126\pi \\ &= 126 \times \frac{22}{7} = 396 \text{ cm}^3 \end{aligned}$$

Answer: 396 $\text{cm}^3$.

Exam­ple 2: A cylin­der with a con­i­cal top#

A solid has a cylin­dri­cal part of radius 7 cm and height 10 cm, with a cone of the same radius and height 6 cm on top. Find its vol­ume.

$$\displaystyle \begin{aligned} V &= \pi r^2 h_1 + \frac{1}{3}\pi r^2 h_2 = \pi r^2\left(h_1 + \frac{h_2}{3}\right) \\ &= \frac{22}{7} \times 49 \times (10 + 2) \\ &= 154 \times 12 = 1848 \text{ cm}^3 \end{aligned}$$

Why this works: both pieces share the fac­tor $\pi r^2$, so tak­ing it out keeps the arith­metic short.

Answer: 1848 $\text{cm}^3$.

Exam­ple 3: Hol­low­ing out a cone#

From a solid wooden cylin­der of radius 7 cm and height 15 cm, a cone of the same radius and the same height is carved out. Find the vol­ume of wood left.

$$\displaystyle \begin{aligned} \text{Cylinder} &= \frac{22}{7} \times 49 \times 15 = 2310 \text{ cm}^3 \\ \text{Cone} &= \frac{1}{3} \times 2310 = 770 \text{ cm}^3 \\ \text{Wood left} &= 2310 - 770 = 1540 \text{ cm}^3 \end{aligned}$$

Answer: 1540 $\text{cm}^3$.

Exam­ple 4: Rise in water level#

A cylin­dri­cal ves­sel of radius 5 cm con­tains water. A solid toy, made of a cone of radius 3 cm and height 4 cm on a hemi­sphere of radius 3 cm, is dropped in and is fully under water. By how much does the water level rise?

Vol­ume of the toy.

$$\displaystyle \begin{aligned} V &= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \\ &= \frac{1}{3}\pi \times 9 \times 4 + \frac{2}{3}\pi \times 27 \\ &= 12\pi + 18\pi = 30\pi \text{ cm}^3 \end{aligned}$$

Rise in level. Let the rise be $x$ cm. Then

$$\displaystyle \pi \times 5^2 \times x = 30\pi \quad \Rightarrow \quad x = \frac{30}{25} = 1.2 \text{ cm}.$$

Why this works: $\pi$ can­cels, so no value of $\pi$ is needed.

Answer: the water rises by 1.2 cm.

Exam­ple 5: Capac­ity of a tank in litres#

A water tank is a cylin­der of radius 0.6 m and height 1.2 m with a hemi­spher­i­cal bot­tom of the same radius. Find how many litres of water it can hold. Use $\pi = 3.14$.

$$\displaystyle \begin{aligned} V &= \pi r^2 h + \frac{2}{3}\pi r^3 \\ &= \pi(0.36 \times 1.2) + \frac{2}{3}\pi(0.216) \\ &= 0.432\pi + 0.144\pi = 0.576\pi \\ &= 0.576 \times 3.14 = 1.80864 \text{ m}^3 \end{aligned}$$

Since $1 \text{ m}^3 = 1000$ litres, the capac­ity is $1.80864 \times 1000 = 1808.64$ litres.

Answer: about 1808.64 litres.

Com­mon mis­takes#

• Try­ing to remove "hid­den faces" in a vol­ume ques­tion. Joined vol­umes sim­ply add.
• Using the slant height instead of the ver­ti­cal height in $\displaystyle \frac{1}{3}\pi r^2 h$.
• Using $\displaystyle \frac{4}{3}\pi r^3$ for a hemi­sphere; a hemi­sphere is $\displaystyle \frac{2}{3}\pi r^3$.
• Adding the vol­ume of a cut-out part instead of sub­tract­ing it.
• Con­vert­ing $\text{m}^3$ to litres by mul­ti­ply­ing by 100 or 1,000,000; the cor­rect fac­tor is 1000.
• In rise-in-level prob­lems, divid­ing by the solid's base area instead of the con­tain­er's base area.

Try these#

1. A toy is a cone of radius 7 cm and height 7 cm on a hemi­sphere of radius 7 cm. Find its vol­ume. Answer: 1078 $\text{cm}^3$.
2. From a cube of edge 7 cm, the largest pos­si­ble hemi­sphere is scooped out of one face. Find the vol­ume left, cor­rect to two dec­i­mal places, using $\displaystyle \pi = \frac{22}{7}$. Answer: 253.17 $\text{cm}^3$.
3. A solid made of a cylin­der (radius 2 cm, height 10 cm) with a cone (radius 2 cm, height 6 cm) on top is sunk in a cuboidal tank with a base of 20 cm by 10 cm. Find the rise in water level, using $\pi = 3.14$. Answer: 0.7536 cm.
4. Find the capac­ity in litres of a cylin­dri­cal con­tainer of radius 7 cm and height 20 cm. Answer: 3.08 litres.

Key terms#

Vol­ume

The amount of space a solid occu­pies, mea­sured in cubic units.

Capac­ity

The vol­ume of the space inside a con­tainer, often mea­sured in litres.

Litre

A unit of capac­ity equal to 1000 $\text{cm}^3$.

Com­bined solid

A solid formed by join­ing basic solids; its vol­ume is the sum of their vol­umes.

Hol­low­ing out

Remov­ing a solid from inside another; the removed vol­ume is sub­tracted.

Dis­place­ment

The rise of a liq­uid when a solid is sunk in it; the extra vol­ume equals the solid's vol­ume.

Ver­ti­cal height

The per­pen­dic­u­lar dis­tance from the ver­tex of a cone to its base, used in the vol­ume for­mula.

Com­mon ques­tions#

Why do I not sub­tract hid­den faces when find­ing vol­ume?#

A face is a sur­face with no thick­ness, so it holds no vol­ume. Join­ing two solids does not remove any mate­r­ial, so their vol­umes sim­ply add.

How are two hemi­spheres at the ends of a cylin­der han­dled?#

If they have the same radius, together they make one sphere, so add $\displaystyle \frac{4}{3}\pi r^3$ to the cylin­der's vol­ume.

How do I con­vert cubic cen­time­tres to litres?#

Divide by 1000, because 1000 $\text{cm}^3$ make one litre. For cubic metres, mul­ti­ply by 1000 to get litres.

Does the solid's shape mat­ter in a rise-in-level prob­lem?#

Only its vol­ume mat­ters, pro­vided it is fully under water. The rise equals the solid's vol­ume divided by the base area of the con­tainer.

Is a cone always one third of a cylin­der?#

Only when the cone and the cylin­der have the same base radius and the same height. Then the cone's vol­ume is exactly one third of the cylin­der's.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.