Prism and cylinder: surface area and volume

What is it about?

A (right) prism has two equal, parallel bases and rectangles as its lateral faces; a cylinder has a circle as its base.

• Volume: cut the solid into layers $1\,\text{cm}$ thick. Each layer has volume $G \cdot 1\,\text{cm}$; a solid $4\,\text{cm}$ high consists of 4 layers: $V = G \cdot h$ (cylinder: $V = \pi r^2 h$).
• Surface area: roll the solid once along the floor. The prints of the lateral faces form a rectangle of length $u$ (perimeter of the base) and width $h$. Lateral area: $M = u \cdot h$; surface area: $O = 2 \cdot G + M$, where $G$ is the base area (cylinder: $O = 2\pi r^2 + 2\pi r h$).
• Limit: a regular $n$-gon gets closer and closer to a circle as $n$ grows – the prism turns into a cylinder.

Try it

1. Press Stack layers and watch the calculation in the picture.
2. Choose Surface: roll out and press Roll out. Roll a hexagonal prism
3. Increase $n$ and turn on the comparison cylinder. Almost a cylinder

Tasks

Task 1: Cylinder

Find the volume and surface area of a cylinder with $r = 2\,\text{cm}$ and $h = 4\,\text{cm}$. Load task

Task 2: Hexagonal prism

In a regular hexagon each side is as long as the radius of the circumcircle. Find the lateral surface of a hexagonal prism with $r = 2\,\text{cm}$ and $h = 4\,\text{cm}$. Load task

Task 3: Almost a cylinder

What is the smallest number of vertices $n$ for which the prism has more than $99\,\%$ of the cylinder’s volume? Try it in the simulation. Load task

Ideas for the classroom

• “Stack layers” justifies $V = G \cdot h$, “Roll out” justifies $M = u \cdot h$. The calculation in the picture can be switched off with calc=0.
• Common stumbling blocks: confusing $u \cdot h$ with $G \cdot h$ (and cm² with cm³), using the diameter instead of the radius, counting only one base, expecting the volume to double when the radius doubles.