Surface Area and Volume of Composite Solids
Theory
The surface area of a composite solid is the total area of the faces on the outside, and its volume is found by adding the parts joined together and subtracting any part cut out. This Year 10A maths page covers solids built from prisms, cylinders, cones, pyramids and hemispheres, holes and hollows, costs, and problems written with a pronumeral. Curriculum: VC2M10M01, VC2M10AM01 / AC9M10M01.
A composite solid is made by joining two or more simple solids, or by cutting one solid out of another.
Its volume is the volumes of the parts that are joined added together, minus any part cut out, such as a cone-shaped hollow or a bowl carved into the top.
Its surface area is the area of everything a coat of paint would touch. When two solids are glued together, the faces pressed against each other are hidden and are not counted.
When a smaller shape sits on a larger face (a cylinder on a cube, one tier of a cake on another), the circle it covers is lost but its own top adds the same area back. So the visible tops add up to the top face of the larger solid.
A hole or groove removes some flat area and adds a new curved inside surface. Always subtract what is cut away and add what is exposed.
The curved surface of a hemisphere (half a sphere) is \(2\pi r^2\). Two equal hemispheres on the ends of a cylinder make one whole sphere, \(4\pi r^2\).
Surface area of a composite solid:
Pieces you will need:
| Part | Surface | Area |
|---|---|---|
| Cylinder | curved wall | \(2\pi rh\) |
| Cone | curved surface | \(\pi rl\), with \(l^2 = r^2 + h^2\) |
| Hemisphere | curved surface (half a sphere) | \(2\pi r^2\) |
| Square pyramid | four triangles | \(4 \times \dfrac{1}{2} \times b \times s\) |
| Any prism | all the side rectangles | perimeter of the end \(\times\) length |
| Half a cylinder | curved part | \(\pi rh\) |
Volume of a composite solid:
Volumes you will combine:
| Solid | Volume |
|---|---|
| Prism / cylinder | \(Ah\) / \(\pi r^2 h\) |
| Pyramid / cone | \(\dfrac{1}{3}Ah\) / \(\dfrac{1}{3}\pi r^2 h\) |
| Sphere | \(\dfrac{4}{3}\pi r^3\) |
| Hemisphere | \(\dfrac{2}{3}\pi r^3\) |
| Half-cylinder | \(\dfrac{1}{2}\pi r^2 h\) |
How to find the surface area of a composite solid
- Split the solid into simple parts and sketch each one.
- List every surface on the outside. Cross out faces that are joined, covered or open.
- Find any missing lengths: slant heights with Pythagoras' theorem, radii from diameters.
- Calculate each area and add, keeping \(\pi\) or surds exact until the end.
Special cases
- Shape on a face (cylinder on a cube, pyramid on a cube, stacked tiers): the covered patch and the new top cancel for circles on flat faces; count only five faces of a cube under a pyramid.
- Hole through a solid: subtract the two openings, add the inside wall.
- Groove cut along a face: subtract the strip of the face that is removed and the two end pieces; add the curved surface of the groove.
- Something added versus the same shape hollowed out: the outside of a cone and the inside of the same cone have equal area, \(\pi rl\).
- Algebraic versions: write every surface in terms of the pronumeral, collect like terms, then solve. A quadratic may need factorising.
- Costs: include only the surfaces that are painted, glazed or covered, then multiply by the rate.
Volume
- Split the solid and decide which parts are added and which are cut out.
- Calculate each volume with \(\pi\) kept exact, then combine with a common denominator, such as \(30\pi + \dfrac{16\pi}{3} = \dfrac{106\pi}{3}\).
- Capacity: \(1\text{ cm}^3 = 1\text{ mL}\). Round once, at the end.
A hemisphere is half a sphere, so its volume is \(\dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3\).
| \(V_{\text{cyl}}\) | \(=\) | \(\pi \times 2.4^2 \times 5 = 28.8\pi\) |
| \(V_{\text{hemi}}\) | \(=\) | \(\dfrac{2}{3}\pi \times 2.4^3 = 9.216\pi\) |
| \(V\) | \(=\) | \(28.8\pi + 9.216\pi\) |
| \(\) | \(=\) | \(\mathbf{38.016\pi}\text{ m}^3\) |
| \(\) | \(=\) | \(119.4307\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{119.43}\text{ m}^3\) |
The outside is the base, the curved wall and the cone's curved surface. The circle where they join is hidden.
| \(l^2\) | \(=\) | \(9^2 + 12^2 = 225, \ l = 15\) |
| \(\text{SA}\) | \(=\) | \(\pi r^2 + 2\pi rh + \pi rl\) |
| \(\) | \(=\) | \(81\pi + 2\pi(9)(15) + \pi(9)(15)\) |
| \(\) | \(=\) | \(81\pi + 270\pi + 135\pi\) |
| \(\) | \(=\) | \(\mathbf{486\pi}\text{ cm}^2\) |
| \(\) | \(=\) | \(1526.8140\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1526.81}\text{ cm}^2\) |
(i) Find the slant height \(s\) of each roof face.
(ii) Find the area of the four walls and the roof (not the floor).
(i) The perpendicular height, half the base side and the slant height form a right-angled triangle, with the slant height as the hypotenuse.
| \(s^2\) | \(=\) | \(12^2 + 5^2\) |
| \(\) | \(=\) | \(169\) |
| \(s\) | \(=\) | \(\mathbf{13}\text{ m}\) |
(ii) The top of the prism is covered by the roof, so it is not counted. The four roof faces are triangles with base \(10\) m and height \(s = 13\) m.
| \(\text{walls}\) | \(=\) | \(4 \times 10 \times 3 = 120\text{ m}^2\) |
| \(\text{roof}\) | \(=\) | \(4 \times \dfrac{1}{2} \times 10 \times 13 = 260\text{ m}^2\) |
| \(\text{total}\) | \(=\) | \(\mathbf{380}\text{ m}^2\) |
Use the slant height of a face, not the perpendicular height of the pyramid, for the area of each triangle.
(i) Show that its surface area is \(4\pi r^2\).
(ii) The surface area is \(100\pi\text{ cm}^2\). Find \(r\) and the exact perpendicular height of the cone.
(i) The flat faces are joined, so only the two curved surfaces count. The curved surface of a hemisphere is half of a sphere's \(4\pi r^2\):
| \(\text{SA}\) | \(=\) | \(2\pi r^2 + \pi r(2r)\) |
| \(\) | \(=\) | \(2\pi r^2 + 2\pi r^2 = 4\pi r^2\) |
(ii)
| \(4\pi r^2\) | \(=\) | \(100\pi\) |
| \(r^2\) | \(=\) | \(25\) |
| \(r\) | \(=\) | \(\mathbf{5} \quad (r \gt 0)\) |
| \(h^2\) | \(=\) | \(10^2 - 5^2 = 75\) |
| \(h\) | \(=\) | \(\sqrt{25 \times 3} = \mathbf{5\sqrt{3}}\text{ cm}\) |
Common pitfalls
Frequently asked questions
How do you find the surface area of a composite solid?
Split it into simple solids, list every surface that is on the outside, leave out faces that are joined or hidden, find each area and add them.
Do you include the area where two solids are joined?
No. The two faces pressed together are inside the solid, so neither is part of the surface area.
What is the surface area of a hemisphere?
The curved surface of a hemisphere is \(2\pi r^2\), half of a sphere's \(4\pi r^2\). A solid hemisphere also has a flat circle, \(\pi r^2\), so its total is \(3\pi r^2\).
Does drilling a hole increase or decrease the surface area?
It can do either. You lose the two openings but gain the inside wall of the hole. For a long, narrow hole the inside wall is usually larger, so the surface area goes up.
How do you find the volume of a composite solid?
Split it into simple solids, find the volume of each, then add the parts that are joined together and subtract any part that is cut out.
What is the volume of a hemisphere?
Half the volume of a sphere: \(\dfrac{2}{3}\pi r^3\).