Surface Area and Volume of Composite Solids

A composite shape contains two or more basic shapes. To get the volume of the component shape, add the volume of the individual shapes that make up the composite. Similarly, to get the total surface area, add up the surface areas of the basic shapes.

Surface Area and Volume of Solid Shapes:

Solid:	Image:	Formula:
Cube		Volume = l3 Curved Surface Area = 4l2 Total Surface Area = 4l2 + 2l2 = 6l2
Cuboid		Volume = lbh Curved Surface Area = 2h(l + b) Total Surface Area = 2lb + 2bh + 2lh = 2(lb + bh + lh)
Cylinder		Volume = πr2h Curved Surface Area = 2πrh Total Surface Area = 2πrh + 2πr2 = 2πr(h + r)
Cone		Volume = \( \frac{1}{3} \scriptsize \pi r^2 h \) Curved Surface Area = πrl Total Surface Area = πrl + πr2 = πr(l + r)
Sphere		Volume = \( \frac{4}{3} \scriptsize \pi r^3 \) Curved Surface Area = 4πr2 Total Surface Area = 4πr2
HemisphereA hemisphere is half the Earth's surface. The four hemispheres are the Northern and Southern hemispheres, divided by the equator (0° latitude), and the Eastern and Western hemispheres, divided by the... More		Volume = \( \frac{2}{3} \scriptsize \pi r^3 \) Curved Surface Area = 2πr2 Total Surface Area = 2πr2 + πr2 = 3πr2

Example 5.4.1:

In the diagram above, a cylinder is surmounted by a hemisphere bowl. Calculate the volume of the solid.

Solution

Volume of the solid = Volume of cylinder + Voume of hemisphere

Volume of cylinder = πr²h
= π x 3² x 20
= π x 9 x 20
= 180 π cm³

Voume of hemisphere = \( \frac{2}{3} \scriptsize \pi r^3 \\ = \frac{2}{3} \scriptsize \: \times \: \pi \: \times \: 3^3 \\ = \frac{2}{3} \scriptsize \: \times \: \pi \: \times \: 27 \\ = \scriptsize 2 \: \times \: \pi \: \times \: 6 \\ = \scriptsize 18 \: \pi \: cm^3 \)

Volume of the solid = 180 π cm³ + 18 π cm³

Volume of the solid = 198 π cm³

Example 5.4.2:

A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times the radius of its base.

(a) Illustrate this information in a diagram.

(b) If the volume of the reservoir is \( \scriptsize 333 \frac{1}{3} \scriptsize \pi \:m^3\), calculate, correct to the nearest whole number, the :

(I) Volume of the hemisphere;
(II) Total surface area of the reservoir.

[Take π = \( \frac{22}{7} \)] (WAEC 2015)

Solution

(a)

(b) (I) Volume of the hemisphere

Step 1: The volume of the reservoir = \( \scriptsize 333 \frac{1}{3} \pi \:m^3 \) Given

But by formula:

Volume of reservoir = Volume of the hemisphere + Volume of the cone

Step 2:

• Volume of the hemisphere = \(\frac{1}{2} \: \times \: \frac{4}{3} \scriptsize \pi r^3 \)

• Volume of the hemisphere = \( \frac{2}{3} \scriptsize \pi r^3 \)

• Volume of cone = \(\frac{1}{3} \scriptsize \pi r^2 h \)

⇒ \( \scriptsize 333 \frac{1}{3} \pi   =  \normalsize  \frac{2}{3} \scriptsize \pi r^3 \: + \: \normalsize \frac{1}{3} \scriptsize \pi r^2 h \)

⇒ \( \normalsize  \frac{1000}{3} \scriptsize \pi   =  \normalsize \frac{1}{3} \scriptsize \pi  \left (\scriptsize 2r^3 \: +\: r^2h \right )\)

Let the radius of the hemisphere be x m;

⇒ \(  \frac{1000}{3} =  \normalsize \frac{1}{3} \scriptsize  \left (\scriptsize 2x^3 \: +\: x^2(6x) \right )\)

⇒ \(  \scriptsize 1000 = 2x^3 \: +\: 6x^3 \)

⇒ \(  \scriptsize 1000 = 8x^3 \)

⇒ \(  \scriptsize x^3 = \normalsize \frac{1000}{8} \)

⇒ \(  \scriptsize x^3 = 125 \)

⇒ \(  \scriptsize x = \sqrt[3]{125}\)

⇒ \(  \scriptsize x = 5\:m\)

Step 3: The volume of the hemisphere is obtained from the formula:

Volume = \( \frac{2}{3} \scriptsize \pi r^3 \)

Volume = \( \frac{2}{3} \: \times \: \frac{22}{7} \scriptsize \: \times \: 5^3 \)

= \( \frac{5500}{21} \)

= 261.905 m³

= 262 m³ (to the nearest whole number)

(II) Total surface area of the reservoir

Step 1: Total surface area of the reservoir = curved surface area of the cone + surface area of the hemisphere

Total surface area = \( \scriptsize \pi r l \: + \: \normalsize \frac{1}{2} \scriptsize \: \times \: 4 \pi r^2 \)

Total surface area = \( \scriptsize \pi r l \: + \: 2 \pi r^2 \)

Step 2: To obtain the value of  l, we consider the triangle below

• If x = 5 cm
• ; then the height 6x = 6 × 5 = 30 cm

⇒ \( \scriptsize l^2 = 30^2 \: + \: 5^2 \)

⇒ \( \scriptsize l^2 = 900 \: + \: 25 \)

⇒ \( \scriptsize l^2 = 925 \)

⇒ \( \scriptsize l = \sqrt{925} \)

⇒ \( \scriptsize l = 30.41\:m \)

Step 4: Therefore;

T.S. = \( \frac{22}{7} \left [\scriptsize  (5 \: \times \: 30.41) \: + \: (2 \: \times \: 5^2)  \right ] \)

T.S. = \( \frac{22}{7} \left [\scriptsize   152.05 \: + \: 50  \right ] \)

T.S. = \( \frac{22}{7}  \: \times  \: \frac{202.05}{1}\)

T.S. = \( \frac{4445.1}{7}  \)

T.S. = 635.01

T.S. = 635 m² (nearest whole number)