Year 10AC9M10M01

Surface Area and Volume of Composite Solids

Solve problems involving the surface area and volume of composite objects using appropriate units.

A composite solid is two shapes joined together. Volume is easy — add them. Surface area is where marks are lost, because the joined faces disappear.

Method

Volume adds

V_{\text{total}} = V_1 + V_2

Split the solid, find each volume, add. Subtract instead if a piece has been removed.

Surface area does not add

SA_{\text{total}} = SA_1 + SA_2 - 2 \times \text{joined face}

The two faces that touch are inside the solid. Neither one is surface any more.

Prism volume is always the same idea

V = A_{\text{cross-section}} \times \text{length}

Area of the end face, times how long it is. True for every prism and cylinder.

Cone and pyramid are a third

V = \tfrac{1}{3} A_{\text{base}} h

A cone is a third of the cylinder that would contain it. A pyramid is a third of its prism.

Watch the units

1 \text{ m}^3 = 1\,000\,000 \text{ cm}^3

Areas scale by $k^2$ and volumes by $k^3$. Convert before you calculate, never after.

e.g.

\text{A cylinder of radius } 3 \text{ cm and height } 10 \text{ cm with a hemisphere on top}

Split the solid and name the parts. $\text{cylinder} + \text{hemisphere}$
Write the formula for each volume before substituting anything. $V_{\text{cyl}} = \pi r^2 h \qquad V_{\text{hem}} = \tfrac{2}{3}\pi r^3$
Substitute and work each one out separately. $V_{\text{cyl}} = \pi(3)^2(10) = 282{,}74 \qquad V_{\text{hem}} = \tfrac{2}{3}\pi(3)^3 = 56{,}55$
Add for the total volume, with cubic units. $V = 339{,}3 \text{ cm}^3$
Now surface area. List every face, then cross out the ones that are hidden. $\text{curved cylinder} + \text{base circle} + \text{curved hemisphere}$
The top circle of the cylinder is covered by the hemisphere, so it is not counted. $\text{omit one } \pi r^2$
Add the faces that remain, with square units. $2\pi(3)(10) + \pi(3)^2 + 2\pi(3)^2 = 273{,}3 \text{ cm}^2$

For each one, do the volume first, then list the faces before adding.

\text{A cube of side } 6 \text{ cm with a } 6\times6\times2 \text{ cm block on top. Find the volume.}

Answer

216 + 72 = 288 \text{ cm}^3

\text{For that same solid, how many of the cube's six faces are fully visible?}

Answer

5. \text{ The top face is partly covered, so only the exposed ring counts.}

\text{A cylinder } r = 4,\ h = 9 \text{ with a cone of the same radius and height } 6 \text{ on top. Find the volume.}

Answer

\pi(16)(9) + \tfrac{1}{3}\pi(16)(6) = 452{,}4 + 100{,}5 = 552{,}9 \text{ cm}^3

\text{A } 10\times10\times10 \text{ cm cube has a } 4\times4\times10 \text{ cm hole drilled straight through. Find the volume.}

Answer

1000 - 160 = 840 \text{ cm}^3

\text{Two cubes of side } 5 \text{ cm are glued face to face. Find the total surface area.}

Answer

2(150) - 2(25) = 250 \text{ cm}^2

\text{Convert } 2{,}5 \text{ m}^3 \text{ to cm}^3

Answer

2{,}5 \times 1\,000\,000 = 2\,500\,000 \text{ cm}^3