Solve Geometry Applications: Volume and Surface Area

In this section, we finish our study of geometry applications by finding the volume and surface area of some three-dimensional figures. As always, we use the Problem Solving Strategy for Geometry Applications.

Find volume and surface area of rectangular solids

The amount of paint needed to cover the outside of a box is its surface area — a square measure of the total area of all the sides. The amount of space inside the box is its volume — a cubic measure.

Each box is in the shape of a rectangular solid, with dimensions length, width, and height. Consider a rectangular solid with length $4$ units, width $2$ units, and height $3$ units. Breaking it into layers makes it easy to see: the top layer has $8$ cubic units, the middle layer has $8$ cubic units, and the bottom layer has $8$ cubic units, for $24$ cubic units altogether. Notice that $24$ is the length times the width times the height.

The volume, $V$, of any rectangular solid is the product of the length, width, and height: $V = LWH$. We can also write the formula in terms of the area of the base: the area of the base, $B$, is $L \cdot W$, so substituting gives $V = Bh$.

To find the surface area, think about finding the area of each face. A rectangular solid has $6$ faces, and for each face you see there is an identical opposite face that doesn’t show, so:

Example. For a rectangular solid with length $14$ cm, height $17$ cm, and width $9$ cm, find (a) the volume and (b) the surface area.

(a) Let $V =$ volume. Substituting into $V = LWH$:

The volume is $2{,}142$ cubic centimeters.

(b) Let $S =$ surface area. Substituting into $S = 2LH + 2LW + 2WH$:

The surface area is $1{,}034$ square centimeters.

Example. A rectangular crate has a length of $30$ inches, width of $25$ inches, and height of $20$ inches. Find its (a) volume and (b) surface area.

(a) $V = LWH = 30 \cdot 25 \cdot 20 = 15{,}000$. The volume is $15{,}000$ cubic inches.

(b) $S = 2LH + 2LW + 2WH = 2(30 \cdot 20) + 2(30 \cdot 25) + 2(25 \cdot 20) = 3{,}700$. The surface area is $3{,}700$ square inches.

Volume and surface area of a cube

A cube is a rectangular solid whose length, width, and height are all equal. Substituting $s$ for the length, width, and height into the rectangular solid formulas gives $V = s^3$ and $S = 6s^2$.

Example. A cube is $2.5$ inches on each side. Find its (a) volume and (b) surface area.

(a) $V = s^3 = (2.5)^3 = 15.625$. The volume is $15.625$ cubic inches.

(b) $S = 6s^2 = 6 \cdot (2.5)^2 = 37.5$. The surface area is $37.5$ square inches.

Example. A notepad cube measures $2$ inches on each side. Find its (a) volume and (b) surface area.

(a) $V = 2^3 = 8$. The volume is $8$ cubic inches.

(b) $S = 6 \cdot 2^2 = 24$. The surface area is $24$ square inches.

Find the volume and surface area of spheres

A sphere is the shape of a basketball, like a three-dimensional circle. Just like a circle, the size of a sphere is determined by its radius, the distance from the center of the sphere to any point on its surface. We’ll approximate $\pi$ with $3.14$.

Example. A sphere has a radius $6$ inches. Find its (a) volume and (b) surface area.

(a) $V = \tfrac{4}{3}\pi r^3 \approx \tfrac{4}{3}(3.14)6^3 \approx 904.32$. The volume is approximately $904.32$ cubic inches.

(b) $S = 4\pi r^2 \approx 4(3.14)6^2 \approx 452.16$. The surface area is approximately $452.16$ square inches.

Example. A globe of Earth is in the shape of a sphere with radius $14$ inches. Find its (a) volume and (b) surface area, rounded to the nearest hundredth.

(a) $V \approx \tfrac{4}{3}(3.14)(14)^3 \approx 11{,}488.21$. The volume is approximately $11{,}488.21$ cubic inches.

(b) $S \approx 4(3.14)(14)^2 \approx 2461.76$. The surface area is approximately $2461.76$ square inches.

Find the volume and surface area of a cylinder

A cylinder is a solid figure with two parallel circles of the same size at the top and bottom, called the bases. The height $h$ of a cylinder is the distance between the two bases; for the cylinders here, the sides and height are perpendicular to the bases.

Rectangular solids and cylinders are similar because they both have two bases and a height, so the formula for volume of a rectangular solid, $V = Bh$, can also be used for a cylinder — but now the base area, $B$, is the area of a circle, $\pi r^2$:

To understand the surface area formula, think of a can of vegetables. It has three surfaces: the top, the bottom, and the piece that forms the sides. If you carefully cut the label off the side and unroll it, you’ll see that it is a rectangle. The distance around the edge of the can is the circumference of the cylinder’s base, and this is also the length $L$ of the rectangular label; the height of the cylinder is the width $W$ of the label. So the area of the label is $A = L \cdot W = 2\pi r \cdot h$. Adding the areas of the two circles to the area of the rectangle:

Example. A cylinder has height $5$ inches and radius $3$ inches. Find its (a) volume and (b) surface area.

(a) $V = \pi r^2 h \approx (3.14)3^2 \cdot 5 \approx 141.3$. The volume is approximately $141.3$ cubic inches.

(b) $S = 2\pi r^2 + 2\pi rh \approx 2(3.14)3^2 + 2(3.14)(3)5 \approx 150.72$. The surface area is approximately $150.72$ square inches.

Example. Find the (a) volume and (b) surface area of a can of soda. The radius of the base is $4$ centimeters and the height is $13$ centimeters. Assume the can is shaped exactly like a cylinder.

(a) $V \approx (3.14)4^2 \cdot 13 \approx 653.12$. The volume is approximately $653.12$ cubic centimeters.

(b) $S \approx 2(3.14)4^2 + 2(3.14)(4)13 \approx 427.04$. The surface area is approximately $427.04$ square centimeters.

Find the volume of cones

The first image many of us have when we hear the word “cone” is an ice cream cone. In geometry, a cone is a solid figure with one circular base and a vertex. The height of a cone is the distance between its base and the vertex; the cones here always have the height perpendicular to the base.

We saw that the volume of a cylinder is $V = \pi r^2 h$. If we picture a cone placed inside a cylinder with the same height and same base, the volume of the cone is less than that of the cylinder — in fact, the volume of a cone is exactly one-third of the volume of a cylinder with the same base and height:

Since the base of a cone is a circle, we substitute $\pi r^2$ for $B$:

We only find the volume of a cone in this book, not its surface area.

Example. Find the volume of a cone with height $6$ inches and radius of its base $2$ inches.

The volume is approximately $25.12$ cubic inches.

Example. Marty’s favorite gastro pub serves french fries in a paper wrap shaped like a cone. What is the volume of a conic wrap that is $8$ inches tall and $5$ inches in diameter? Round to the nearest hundredth.

Since we’re given the diameter, the radius is $2.5$ inches:

The volume of the wrap is approximately $52.33$ cubic inches.

Key terms

rectangular solid — a three-dimensional figure with six rectangular faces (length, width, height). surface area — a square measure of the total area of all the faces of a solid. volume — a cubic measure of the space enclosed by a solid. cube — a rectangular solid whose length, width, and height are all equal. sphere — a three-dimensional figure where every point on the surface is the same distance (the radius) from the center. cylinder — a solid figure with two parallel congruent circular bases connected by a curved surface perpendicular to the bases. cone — a solid figure with one circular base tapering to a single vertex.

This section is adapted from Prealgebra 2e, Section 9.6: Solve Geometry Applications: Volume and Surface Area by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: described the crate, globe, soda can, and french-fry-wrap illustrations in prose instead of hotlinking images, and omitted the end-of-chapter geometry formula summary chart (each formula already appears inline where it’s introduced); omitted the Be Prepared quiz, Manipulative Mathematics callout, Media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.