Surface Area and Volume of 3D Solids

Surface area and volume measure different things

Surface area measures the outside covering of a solid. Imagine cutting along some edges and opening the surface into flat pieces: the surface area is the sum of the areas of those pieces. It is measured in square units such as $\text{cm}^2$.

Volume measures the space occupied inside a solid. It is measured in cubic units such as $\text{cm}^3$. A surface-area calculation therefore involves squared lengths, whereas a volume calculation involves three-dimensional space.

Before choosing a formula, identify the solid and label its important measurements. In particular, distinguish radius from diameter and perpendicular height from slant height.

Spheres and hemispheres

Every point on the surface of a sphere is the same distance $r$ from its centre. Its surface area and volume are

$A=4\pi r^2$

and

$V=\frac{4}{3}\pi r^3.$

Area depends on $r^2$, while volume depends on $r^3$. If the diameter $d$ is given, first find the radius using $r=\frac d2$.

A hemisphere is half a sphere, so its volume is

$V=\frac{2}{3}\pi r^3.$

Its curved surface area is $2\pi r^2$. A separate hemisphere also has a flat circular face of area $\pi r^2$, making its total surface area

$A_{\text{total}}=2\pi r^2+\pi r^2=3\pi r^2.$

Whether the circle is included depends on whether it is exposed. If the hemisphere is attached to another solid across that circle, the join is internal and does not contribute to the external surface area.

Pyramids

A pyramid has a polygonal base and triangular faces meeting at an apex. If its base area is $B$ and its perpendicular height is $h$, then

$V=\frac13Bh.$

The perpendicular height runs from the apex to the base at a right angle. In a right square-based pyramid, it meets the base at its centre. It is not a sloping edge or a line drawn down one triangular face.

The factor $\frac13$ means that the pyramid occupies one third of the volume of a prism with the same base area and perpendicular height.

For surface area, add the base and all exposed triangular faces:

$A_{\text{surface}}=B+\text{areas of exposed triangular faces}.$

Each triangular face has area

$\frac12\times\text{base of face}\times\text{perpendicular height of face}.$

That face height is a slant height: it lies in the triangular face and meets its base edge at a right angle. For a right, regular pyramid whose triangular faces have the same slant height $l$, their combined area is $\frac12Pl$, where $P$ is the perimeter of the base. Thus

$A_{\text{surface}}=B+\frac12Pl.$

If the triangular faces have different slant heights, calculate their areas separately.

Cones

A cone has a circular base. With base radius $r$ and perpendicular height $h$, its volume is

$V=\frac13\pi r^2h.$

Its curved surface uses the slant height $l$:

$A_{\text{curved}}=\pi rl.$

Including the circular base gives

$A_{\text{total}}=\pi rl+\pi r^2=\pi r(l+r).$

For a right cone, a central cross-section forms a right-angled triangle with legs $r$ and $h$ and hypotenuse $l$. Therefore,

$l^2=r^2+h^2.$

This distinction is essential: use $h$ for volume and $l$ for curved surface area.

Worked calculation: total surface area of a cone

Suppose a cone has radius $7\text{ cm}$ and perpendicular height $24\text{ cm}$. Total surface area needs the slant height, so find it first:

$l^2=24^2+7^2=576+49=625,$

so $l=25\text{ cm}$.

Now substitute $r=7$ and $l=25$ into the total surface-area formula:

$A=\pi r^2+\pi rl$

$A=\pi(7)^2+\pi(7)(25)$

$A=49\pi+175\pi=224\pi\text{ cm}^2.$

This is an exact answer. It includes both the curved surface and the exposed circular base, and the unit is squared because an area has been calculated.

Composite solids

A composite solid can be split into familiar solids, completed to form a larger familiar solid, or treated as a whole with a section removed.

For volume, add joined pieces or subtract removed pieces:

$V_{\text{joined}}=V_1+V_2+\cdots$

$V_{\text{remaining}}=V_{\text{whole}}-V_{\text{removed}}.$

For example, consider a hemisphere of radius $6\text{ cm}$ joined to a cylinder of radius $6\text{ cm}$ and height $15\text{ cm}$. The hemisphere contributes

$\frac23\pi(6)^3=144\pi\text{ cm}^3,$

and the cylinder contributes

$\pi(6)^2(15)=540\pi\text{ cm}^3.$

The total volume is therefore

$144\pi+540\pi=684\pi\text{ cm}^3,$

which is approximately $2150\text{ cm}^3$ to three significant figures. The circular join does not affect volume: the two non-overlapping volumes are simply added.

Surface area needs different reasoning because only the exposed surface counts. For the same joined solid, the shared circular face lies inside the object, so it must not be included. List each exposed region and count it once. If a section has been cut away, remember that the cut may create new exposed surfaces.

Units and accuracy

Convert all lengths to the same unit before substituting. A length conversion is applied once, an area conversion is squared and a volume conversion is cubed. Since $1\text{ m}=100\text{ cm}$,

$1\text{ m}^2=10{,}000\text{ cm}^2$

and

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

Use the calculator's full value of $\pi$ and keep unrounded intermediate results. Leave the answer in terms of $\pi$ when an exact answer is requested; otherwise round only the final result to the stated accuracy.

Core formulae

For radius $r$, perpendicular height $h$, slant height $l$ and base area $B$:

• Sphere: $A=4\pi r^2$, $V=\frac43\pi r^3$
• Hemisphere: curved area $=2\pi r^2$, total area with exposed base $=3\pi r^2$, $V=\frac23\pi r^3$
• Pyramid: $V=\frac13Bh$; surface area $=$ base $+$ exposed triangular faces
• Cone: $V=\frac13\pi r^2h$, curved area $=\pi rl$, total area $=\pi rl+\pi r^2$
• Right cone: $l^2=r^2+h^2$

Essential distinctions

• Area uses square units; volume uses cubic units.
• Cone and pyramid volume uses perpendicular height.
• Cone surface area and pyramid face areas use slant height.
• Curved surface area excludes a base; total surface area includes every exposed base.
• Halve a diameter before using it as $r$.

Composite solids

Add volumes of joined pieces and subtract removed volumes. For surface area, count only exposed regions: omit internal joins and include any new surfaces made by a cut.

Convert lengths to one unit first. Keep $\pi$ and full calculator values during working, then round once at the end.