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Surface Area of Prisms & Cylinders

Surface area is the total area of every face of a solid — the amount of wrapping paper it would take to cover it with no overlap. For prisms and cylinders it splits neatly into two bases plus the wall that wraps around them.

The wall is called the lateral area, and it is always the perimeter of the base times the height. Add the two bases and you have the full surface area. One idea covers every prism and the cylinder too.

The two-part formula

Total surface area = lateral area + 2 × (base area). The lateral area is PhP \cdot h, the base perimeter times the height. For a rectangular prism that is 2(lw+lh+wh)2(lw + lh + wh) once you expand it out.

For a cylinder the base is a circle, so the perimeter becomes the circumference 2πr2\pi r and the two bases add 2πr22\pi r^2. That gives surface area =2πrh+2πr2= 2\pi r h + 2\pi r^2. The figure marks the radius rr of the circular base and the height hh that the wall wraps.

rr
hh

Keeping the pieces straight

The most common mistake is forgetting one of the two bases, or doubling the lateral area by accident. Write the two pieces separately, then add.

Units are always squared — square inches, square centimeters — because area measures a surface, not a length or a volume.

Worked examples

Example 1: a rectangular prism

Find the surface area of a 55 by 44 by 33 rectangular prism.

Use the prism formula2(lw+lh+wh)2(lw + lh + wh)
Substitute2(54+53+43)2(5\cdot4 + 5\cdot3 + 4\cdot3)
Add inside, then double2(20+15+12)=942(20 + 15 + 12) = 94

Answer: 9494 square units

Example 2: a cylinder

Find the surface area of a cylinder with radius 33 and height 55.

Lateral area plus two bases2πrh+2πr22\pi r h + 2\pi r^2
Substitute2π(3)(5)+2π(3)22\pi(3)(5) + 2\pi(3)^2
Simplify30π+18π=48π30\pi + 18\pi = 48\pi

Answer: 48π48\pi square units

Try one yourself

55
33
44

Common questions

What is lateral area versus total surface area?

Lateral area is just the wall around the sides (PhP \cdot h), leaving out the top and bottom. Total surface area adds the two bases back in.

Why is the cylinder's wall 2πrh2\pi r h?

Unroll the wall and it is a rectangle: its height is hh and its width is the circle's circumference 2πr2\pi r. Area of that rectangle is 2πrh2\pi r \cdot h.

Do I leave answers in terms of pi?

Usually yes for exact answers. If a decimal is asked for, multiply by 3.141593.14159 at the very end so rounding does not build up.

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