Prisms and Cylinders

From unit cubes to base area

Volume counts the space a solid occupies. A cubic unit is the volume of a cube with edges one unit long. A box with a $4$ cm by $3$ cm base holds $12$ unit cubes in one $1$ cm layer. Two such layers hold $24$ cubes.

The layer count gives the product. The $12$ is the base area measured in square centimeters; multiplying by the $2$ cm height gives cubic centimeters.

Fractional lengths follow the same idea with smaller cubes. For example, subdividing each centimeter into halves gives cubes of volume $1/8$ cubic centimeter. Finer subdivisions extend the measurement to any positive lengths.

A right prism can have a triangular or other polygonal base. Pair copies of a triangular base to make a parallelogram. The familiar cut-and-slide rearrangement turns that parallelogram into a rectangle of equal area. Extend these same cuts through the prism height: they rearrange the paired prism into a rectangular box. Each triangular prism is half the pair, so its volume is its own base area times the height. Splitting any polygonal base into triangles and adding their volumes gives the same rule for a right prism.

Write $B$ for the area of one base and $h$ for the perpendicular height, the distance between the parallel base planes. For every prism,

The oblique case needs a section argument, which comes next. The height is not automatically an edge that joins the bases.

Why leaning preserves volume

Sliding base-parallel layers changes their positions without changing their areas. Consider a prism with base area $12$ square units and perpendicular height $6$ units. Predict its volume after the top shifts $2$ units, then $4$ units sideways. What happens to a section halfway up?

In each position, the halfway section has area $12$. So does a section at one third of the height, or at any other height. Each section is just a translated copy of the base. All three volumes are $72$ cubic units.

This illustrates Cavalieri’s principle: suppose two ordinary solids lie between the same two parallel planes. If every plane parallel to those boundary planes cuts the solids in sections of equal area, the solids have equal volumes. Equivalently, corresponding section areas must agree at every common perpendicular height, across the full height.

Imagine approximating the solids by thinner and thinner layers. Equal section areas give matching layer-volume approximations. Refining the layers makes the discrepancies as small as desired. This is the same kind of shrinking-error reasoning used for circle area. A finite stack is an illustration; the principle concerns every height, not just the few cuts drawn.

An oblique prism and a right prism with equal base areas and equal perpendicular heights meet this test: every base-parallel section has area $B$. Therefore both have volume $Bh$. The sideways shift and the slanted edge length do not enter that volume.

Equal base area and equal height alone do not justify a comparison of arbitrary solids. Their sections between the bases must also be checked.

Cylinders use the same volume idea

A circular cylinder also repeats its base area through the perpendicular height. For radius $2$ cm and height $7$ cm, every base-parallel section is a disk of area $4\pi$ square centimeters. Compare it with a rectangular prism of the same height and base area $4\pi$ square centimeters. Their sections match in area at every height, even though the section shapes differ. Both volumes are $28\pi$ cubic centimeters.

Nothing in that comparison depended on those particular measurements. A circular cylinder of positive radius $r$ and perpendicular height $h$ has

This works for right and oblique circular cylinders. For the oblique cylinder, base-parallel sections are translated disks, and $h$ still measures the perpendicular distance between base planes. Here and below, a circular cylinder means one with circular bases.

Capacity is the greatest volume a container can hold. Use its inside dimensions. A liter equals $1{,}000$ cubic centimeters, or one cubic decimeter. A milliliter equals one cubic centimeter. Wall thickness, when relevant, makes inside and outside dimensions differ.

A right prism unfolds into rectangles

Surface area measures the boundary, in square units. The lateral area excludes the two chosen bases; total surface area includes every surface in the stated boundary.

A closed right rectangular prism with a $3$ cm by $2$ cm base and height $4$ cm unfolds as follows. Each side rectangle has one edge from the base perimeter and the other equal to the height.

The four rectangle areas total $40$ square centimeters. That is the base perimeter $10$ cm multiplied by the common height $4$ cm. The two bases add $12$ square centimeters, giving $52$ in total.

For any right prism, the lateral rectangles all have height $h$. Their widths add to the base perimeter $P$. Thus

The restriction to a right prism matters: in an oblique prism, lateral faces are generally parallelograms, and their face altitudes need not equal the solid’s perpendicular height. Compute those actual face areas instead of applying this shortcut.

Unroll a right cylinder and count its ends

A right cylinder’s curved side unrolls into a rectangle. With radius $3$ cm and height $8$ cm, the rectangle is $6\pi$ cm wide and $8$ cm high. The width wraps once around the circular base; it is the circumference, not the diameter.

The side area is $48\pi$ square centimeters. Each end adds $9\pi$ square centimeters. Replacing $3$ and $8$ by arbitrary positive $r$ and $h$ gives, for a right circular cylinder,

These surface shortcuts require a right cylinder. An oblique cylinder’s side does not unroll into that rectangle, even though its volume is still $\pi r^2h$.

Common mistakes