Surface Area and Volume
Extend measurement into three dimensions
Look around the room you are in right now. Every object you see takes up space. The coffee mug on your desk, the box your headphones came in, the basketball in the corner, the ice cream cone you are thinking about getting later. All of these are three-dimensional objects, and understanding how to measure them is one of the most practical skills geometry offers. How much wrapping paper do you need for a gift? That is surface area. How much water does your fish tank hold? That is volume. In this lesson, we are going to explore the formulas and reasoning behind measuring these 3D shapes, building on the area and perimeter concepts you already know from two dimensions.
Core Concepts
From 2D to 3D: The Big Picture
In two dimensions, we measure the space inside a shape with area (square units) and the distance around it with perimeter (linear units). When we step into three dimensions, we get two new measurements:
Surface area is the total area of all the surfaces (faces) of a 3D object. Imagine you wanted to paint every outside surface of a box or wrap a present. The amount of paint or wrapping paper you need is determined by the surface area. Surface area is measured in square units (square inches, square centimeters, etc.), just like regular area, because we are still measuring flat surfaces. We are just measuring several of them and adding them up.
Volume is the amount of space inside a 3D object. If you filled a container with water, the volume tells you how much water it holds. Volume is measured in cubic units (cubic inches, cubic centimeters, cubic feet, etc.) because we are measuring three-dimensional space, essentially counting how many unit cubes would fit inside.
Prisms: The Familiar Shapes
A prism is a 3D figure with two congruent, parallel bases connected by rectangular faces. The bases can be any polygon: triangles, rectangles, pentagons, hexagons, you name it. The shape of the base gives the prism its name.
Think of a prism as taking a 2D shape and “stretching” it upward. A rectangular prism (like a cereal box) is a rectangle stretched into 3D. A triangular prism (like a tent or a Toblerone box) is a triangle stretched into 3D.
Volume of a Prism: $$V = Bh$$
where $B$ is the area of the base and $h$ is the height (the perpendicular distance between the bases).
This formula makes intuitive sense: you find the area of the base, then multiply by how many “layers” of that base you are stacking up. A rectangular prism with a base of 20 square inches and a height of 5 inches has a volume of $20 \times 5 = 100$ cubic inches.
For a rectangular prism specifically, since the base is a rectangle with area $l \times w$: $$V = lwh$$
Surface Area of a Prism:
To find the surface area, add up the areas of all the faces: $$SA = 2B + Ph$$
where $B$ is the area of one base, $P$ is the perimeter of the base, and $h$ is the height.
The $2B$ accounts for the two bases (top and bottom). The $Ph$ gives you the lateral area, which is the area of all the rectangular side faces combined. Think of it this way: if you unrolled all the side faces into one big rectangle, it would have a width equal to the perimeter of the base and a height equal to the height of the prism.
For a rectangular prism: $$SA = 2lw + 2lh + 2wh = 2(lw + lh + wh)$$
Cylinders: Circular Prisms
A cylinder is essentially a prism with circular bases. Think of a soup can, a drinking glass, or a water pipe. It has two congruent circular bases connected by a curved surface.
Volume of a Cylinder: $$V = \pi r^2 h$$
This follows the same logic as prisms: the area of the circular base is $\pi r^2$, and we multiply by the height $h$ to get the total volume.
Surface Area of a Cylinder: $$SA = 2\pi r^2 + 2\pi rh$$
The first term, $2\pi r^2$, is the area of the two circular bases. The second term, $2\pi rh$, is the lateral area. Here is a nice way to visualize it: imagine cutting the curved surface of a cylinder and unrolling it flat. You would get a rectangle whose width is the circumference of the base ($2\pi r$) and whose height is $h$. So the lateral area is $2\pi r \times h = 2\pi rh$.
Pyramids: Coming to a Point
A pyramid is a 3D figure with one polygonal base and triangular faces that meet at a single point called the apex. The Egyptian pyramids are the most famous example (with square bases), but pyramids can have triangular, pentagonal, or any polygonal base.
Volume of a Pyramid: $$V = \frac{1}{3}Bh$$
where $B$ is the area of the base and $h$ is the height (perpendicular distance from base to apex).
Notice that a pyramid’s volume is exactly one-third the volume of a prism with the same base and height. This might seem surprising, but it is one of those beautiful results in geometry. You could actually demonstrate this by filling a pyramid-shaped container with water and pouring it into a prism-shaped container of the same base and height. It takes exactly three pyramids to fill the prism.
Surface Area of a Pyramid: $$SA = B + \frac{1}{2}Pl$$
where $B$ is the base area, $P$ is the perimeter of the base, and $l$ is the slant height (the height of each triangular face, measured from the base edge to the apex along the face).
Be careful not to confuse the height $h$ (perpendicular from base to apex, used in volume) with the slant height $l$ (along the triangular face, used in surface area).
Cones: Circular Pyramids
A cone is to a pyramid what a cylinder is to a prism: it is a pyramid with a circular base. Think of an ice cream cone, a traffic cone, or a party hat.
Volume of a Cone: $$V = \frac{1}{3}\pi r^2 h$$
Just like with pyramids, a cone’s volume is one-third of the cylinder with the same base and height.
Surface Area of a Cone: $$SA = \pi r^2 + \pi rl$$
The first term is the base area, and the second term is the lateral area. The $l$ here is the slant height, which can be found using the Pythagorean theorem if you know the radius and height: $$l = \sqrt{r^2 + h^2}$$
Spheres: Perfect Roundness
A sphere is the set of all points in 3D space that are a fixed distance (the radius) from a center point. Basketballs, globes, and oranges are all approximately spherical.
Volume of a Sphere: $$V = \frac{4}{3}\pi r^3$$
Surface Area of a Sphere: $$SA = 4\pi r^2$$
A nice way to remember the surface area formula: the surface area of a sphere equals exactly four times the area of a circle with the same radius. Imagine four circular discs of radius $r$ laid flat. They have the same total area as the curved surface of the sphere.
Composite Solids: Combining Shapes
Many real-world objects are not simple prisms, cones, or spheres. Instead, they are composite solids made by combining basic shapes. A silo might be a cylinder with a hemisphere on top. A pencil is a cylinder with a cone at the tip. An ice cream cone with a scoop on top is a cone plus a hemisphere.
To find the volume or surface area of a composite solid:
- Break it into simpler shapes that you know how to measure
- Calculate each part separately
- Add or subtract as needed
When finding surface area, be careful not to count surfaces that are hidden inside the composite solid. If a cone sits on top of a cylinder, the top circle of the cylinder and the base circle of the cone are not part of the exterior surface.
Cavalieri’s Principle: A Powerful Insight
Cavalieri’s Principle states: If two solids have the same height and the same cross-sectional area at every level, then they have the same volume.
Imagine two stacks of coins. One stack is straight up, and the other is leaning (like a leaning tower of coins). As long as both stacks have the same number of identical coins, they have the same total volume, even though one looks different. The leaning stack has the same cross-sectional area at every height as the straight stack.
This principle explains why the formula $V = Bh$ works for oblique prisms (prisms that lean to one side). Even though the shape looks different, at every horizontal level, the cross-section has the same area as the base.
Similar Solids: Scaling in 3D
Two solids are similar if they have the same shape but different sizes. All corresponding angles are equal, and all corresponding lengths are proportional.
When you scale a solid by a factor of $k$:
- All linear measurements (edges, heights, radii) multiply by $k$
- All surface areas multiply by $k^2$
- All volumes multiply by $k^3$
This is a crucial concept for understanding how size affects capacity. If you double the dimensions of a box (scale factor $k = 2$):
- The surface area becomes $2^2 = 4$ times as large
- The volume becomes $2^3 = 8$ times as large
This explains why large animals have trouble cooling off (their volume grows faster than their surface area) and why small creatures can lift proportionally more weight.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| Surface area | Total area of all surfaces | $SA$ |
| Volume | Space inside a 3D figure | $V$ |
| Prism | 3D figure with two congruent parallel bases | Rectangular, triangular |
| Cylinder | Prism with circular bases | $V = \pi r^2 h$ |
| Pyramid | Base with triangular faces meeting at apex | $V = \frac{1}{3}Bh$ |
| Cone | Circular base with apex | $V = \frac{1}{3}\pi r^2 h$ |
| Sphere | All points equidistant from center in 3D | $V = \frac{4}{3}\pi r^3$ |
| Lateral area | Area of sides (not bases) | $LA$ |
| Slant height | Distance along a face from base to apex | $l$ |
| Apex | The point where triangular faces meet | Top of pyramid or cone |
| Base ($B$) | The area of the base of a solid | $B = \pi r^2$ for a cylinder |
| Cross-section | The shape formed by cutting a solid with a plane | Circle for a cylinder cut horizontally |
Examples
A shipping box has length 18 inches, width 12 inches, and height 10 inches. Find its volume.
Solution:
For a rectangular prism, we use: $$V = lwh$$
Substituting our values: $$V = 18 \times 12 \times 10 = 2160 \text{ cubic inches}$$
The box can hold 2,160 cubic inches of material.
Practical note: This is the same as 1.25 cubic feet (since $1 \text{ ft}^3 = 1728 \text{ in}^3$).
A gift box is a cube with edges of 6 inches. How much wrapping paper is needed to cover it (ignoring overlap)?
Solution:
A cube has 6 identical square faces. Each face has area: $$A_{\text{face}} = s^2 = 6^2 = 36 \text{ square inches}$$
Total surface area: $$SA = 6s^2 = 6 \times 36 = 216 \text{ square inches}$$
You need at least 216 square inches of wrapping paper.
Alternative approach: You could also use the general prism formula. For a cube, $B = s^2$, $P = 4s$, and $h = s$: $$SA = 2B + Ph = 2s^2 + 4s \cdot s = 2s^2 + 4s^2 = 6s^2$$
Same result!
A cylindrical water tank has a radius of 4 feet and a height of 10 feet. Find its volume and surface area. (Use $\pi \approx 3.14$)
Solution:
Volume: $$V = \pi r^2 h = 3.14 \times 4^2 \times 10 = 3.14 \times 16 \times 10 = 502.4 \text{ cubic feet}$$
The tank holds about 502.4 cubic feet of water.
Surface Area: $$SA = 2\pi r^2 + 2\pi rh$$ $$SA = 2(3.14)(4^2) + 2(3.14)(4)(10)$$ $$SA = 2(3.14)(16) + 2(3.14)(40)$$ $$SA = 100.48 + 251.2 = 351.68 \text{ square feet}$$
The total surface area is about 351.68 square feet.
Breaking it down:
- The two circular bases contribute $100.48$ square feet ($2\pi r^2$)
- The curved lateral surface contributes $251.2$ square feet ($2\pi rh$)
An ice cream cone has a radius of 2 inches and a height of 5 inches. What is the maximum volume of ice cream it can hold (assuming the ice cream fills the cone completely)?
Solution:
Using the cone volume formula: $$V = \frac{1}{3}\pi r^2 h$$ $$V = \frac{1}{3}(3.14)(2^2)(5)$$ $$V = \frac{1}{3}(3.14)(4)(5)$$ $$V = \frac{1}{3}(62.8)$$ $$V = 20.93 \text{ cubic inches}$$
The cone holds about 20.93 cubic inches of ice cream.
Comparison: A cylinder with the same radius and height would hold $\pi r^2 h = 62.8$ cubic inches. The cone holds exactly one-third of that, as expected.
A storage tank consists of a cylinder with a cone on top (like a rocket shape). The cylindrical section has a radius of 3 meters and a height of 8 meters. The conical top has the same radius and a height of 4 meters. Find the total volume.
Solution:
We need to find the volume of each part and add them together.
Cylindrical section: $$V_{\text{cylinder}} = \pi r^2 h = \pi (3)^2 (8) = 72\pi \text{ cubic meters}$$
Conical section: $$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (4) = \frac{1}{3}\pi(9)(4) = 12\pi \text{ cubic meters}$$
Total volume: $$V_{\text{total}} = V_{\text{cylinder}} + V_{\text{cone}} = 72\pi + 12\pi = 84\pi$$
Using $\pi \approx 3.14$: $$V_{\text{total}} = 84 \times 3.14 = 263.76 \text{ cubic meters}$$
The tank holds about 263.76 cubic meters (or about 69,700 gallons).
Note: When finding the surface area of this composite solid, you would not include the top circle of the cylinder or the base circle of the cone, since those surfaces are inside the tank.
Two similar spheres have radii of 3 cm and 6 cm respectively.
a) What is the ratio of their surface areas? b) What is the ratio of their volumes? c) If the smaller sphere weighs 200 grams, how much does the larger sphere weigh (assuming they are made of the same material)?
Solution:
The scale factor from smaller to larger is: $$k = \frac{6}{3} = 2$$
a) Surface area ratio: Surface areas scale by $k^2$: $$\frac{SA_{\text{large}}}{SA_{\text{small}}} = k^2 = 2^2 = 4$$
The larger sphere has 4 times the surface area.
Verification:
- $SA_{\text{small}} = 4\pi(3)^2 = 36\pi$
- $SA_{\text{large}} = 4\pi(6)^2 = 144\pi$
- Ratio: $\frac{144\pi}{36\pi} = 4$
b) Volume ratio: Volumes scale by $k^3$: $$\frac{V_{\text{large}}}{V_{\text{small}}} = k^3 = 2^3 = 8$$
The larger sphere has 8 times the volume.
Verification:
- $V_{\text{small}} = \frac{4}{3}\pi(3)^3 = 36\pi$
- $V_{\text{large}} = \frac{4}{3}\pi(6)^3 = 288\pi$
- Ratio: $\frac{288\pi}{36\pi} = 8$
c) Weight comparison: Since the spheres are made of the same material, weight is proportional to volume: $$\text{Weight}_{\text{large}} = 8 \times 200 = 1600 \text{ grams}$$
The larger sphere weighs 1,600 grams (1.6 kg).
Key insight: Doubling all dimensions makes an object 8 times heavier!
Key Properties and Rules
Volume Formulas
Prism (any base): $$V = Bh$$
Rectangular Prism: $$V = lwh$$
Cube: $$V = s^3$$
Cylinder: $$V = \pi r^2 h$$
Pyramid (any base): $$V = \frac{1}{3}Bh$$
Cone: $$V = \frac{1}{3}\pi r^2 h$$
Sphere: $$V = \frac{4}{3}\pi r^3$$
Surface Area Formulas
Prism (any base): $$SA = 2B + Ph$$
Rectangular Prism: $$SA = 2(lw + lh + wh)$$
Cube: $$SA = 6s^2$$
Cylinder: $$SA = 2\pi r^2 + 2\pi rh$$
Pyramid (regular): $$SA = B + \frac{1}{2}Pl$$
Cone: $$SA = \pi r^2 + \pi rl$$
where $l = \sqrt{r^2 + h^2}$ (slant height)
Sphere: $$SA = 4\pi r^2$$
The One-Third Rule
Pyramids and cones have exactly one-third the volume of the corresponding prisms and cylinders with the same base and height:
- $V_{\text{pyramid}} = \frac{1}{3}V_{\text{prism}}$
- $V_{\text{cone}} = \frac{1}{3}V_{\text{cylinder}}$
Scaling Relationships for Similar Solids
If two similar solids have a scale factor of $k$:
- Linear dimensions: ratio is $k$
- Surface areas: ratio is $k^2$
- Volumes: ratio is $k^3$
Cavalieri’s Principle
If two solids have the same height and equal cross-sectional areas at every level, they have equal volumes.
Real-World Applications
Packaging and Shipping
When companies design product packaging, they need to balance surface area (which determines material costs) with volume (which determines how much product fits inside). A sphere has the smallest surface area for a given volume, which is why bubbles are round. But spheres do not stack efficiently, so rectangular boxes are more practical for shipping. The dimensions you choose for a box can significantly affect how many fit on a shipping pallet.
Swimming Pool Capacity
Pool companies need to calculate volume to know how much water a pool holds, which affects chemical treatment requirements and heating costs. A standard rectangular pool might hold 20,000 gallons, but one with curved walls or varying depths requires breaking the shape into sections and calculating each part separately, just like our composite solid example.
Ice Cream Cones
The ice cream industry uses cone and sphere volumes constantly. How much ice cream should a machine dispense to give a customer a generous-looking scoop on a cone? Too little looks stingy; too much falls off. Understanding that a hemisphere of ice cream on top of a cone holds more than you might think helps with portion control and cost management.
Storage Tanks and Silos
Agricultural silos are often cylinders with conical or hemispherical tops. Engineers calculate volume to know storage capacity and surface area to estimate construction material needs and heat loss. A grain silo that is 20 feet in diameter and 50 feet tall holds a lot more than one that is 15 feet in diameter and 60 feet tall, even though the second one is taller.
Architecture and Construction
The surface area of a building’s exterior determines heating and cooling costs, since that is where heat escapes. Architects balance aesthetics with energy efficiency by considering the ratio of interior volume (living space) to exterior surface area (where energy is lost). Compact shapes like cubes are more energy-efficient than sprawling shapes.
Medical Dosing
Medication dosing sometimes depends on body surface area rather than weight, especially for chemotherapy drugs. Formulas estimate surface area from height and weight. Understanding the $k^2$ relationship for surface area and $k^3$ relationship for volume helps explain why children need different doses than adults even when adjusted for weight.
Self-Test Problems
Problem 1: A rectangular prism has dimensions 5 cm by 8 cm by 12 cm. Find its volume and surface area.
Show Answer
Volume: $$V = lwh = 5 \times 8 \times 12 = 480 \text{ cm}^3$$
Surface Area: $$SA = 2(lw + lh + wh)$$ $$SA = 2(5 \times 8 + 5 \times 12 + 8 \times 12)$$ $$SA = 2(40 + 60 + 96)$$ $$SA = 2(196) = 392 \text{ cm}^2$$
The volume is 480 cubic centimeters and the surface area is 392 square centimeters.
Problem 2: A cylinder has radius 5 inches and height 12 inches. Find its volume. (Use $\pi \approx 3.14$)
Show Answer
$$V = \pi r^2 h = 3.14 \times 5^2 \times 12$$ $$V = 3.14 \times 25 \times 12 = 942 \text{ cubic inches}$$
The cylinder holds 942 cubic inches.
Problem 3: A cone has radius 6 cm and height 8 cm. Find its volume and slant height. (Use $\pi \approx 3.14$)
Show Answer
Volume: $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}(3.14)(6^2)(8)$$ $$V = \frac{1}{3}(3.14)(36)(8) = \frac{1}{3}(904.32) = 301.44 \text{ cm}^3$$
Slant height: Using the Pythagorean theorem with $r = 6$ and $h = 8$: $$l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}$$
The volume is about 301.44 cubic centimeters and the slant height is 10 centimeters.
Problem 4: Find the surface area and volume of a sphere with radius 7 meters. (Use $\pi \approx 3.14$)
Show Answer
Surface Area: $$SA = 4\pi r^2 = 4(3.14)(7^2) = 4(3.14)(49) = 615.44 \text{ m}^2$$
Volume: $$V = \frac{4}{3}\pi r^3 = \frac{4}{3}(3.14)(7^3) = \frac{4}{3}(3.14)(343)$$ $$V = \frac{4}{3}(1077.02) = 1436.03 \text{ m}^3$$
The surface area is about 615.44 square meters and the volume is about 1,436 cubic meters.
Problem 5: A silo consists of a cylinder with radius 5 meters and height 15 meters, topped by a hemisphere (half a sphere) of the same radius. Find the total volume. (Use $\pi \approx 3.14$)
Show Answer
Cylindrical section: $$V_{\text{cylinder}} = \pi r^2 h = 3.14 \times 5^2 \times 15 = 3.14 \times 25 \times 15 = 1177.5 \text{ m}^3$$
Hemispherical top: (half of a full sphere) $$V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3$$ $$V_{\text{hemisphere}} = \frac{2}{3}(3.14)(5^3) = \frac{2}{3}(3.14)(125) = \frac{2}{3}(392.5) = 261.67 \text{ m}^3$$
Total volume: $$V_{\text{total}} = 1177.5 + 261.67 = 1439.17 \text{ m}^3$$
The silo holds about 1,439 cubic meters.
Problem 6: A pyramid has a square base with side length 10 feet and a height of 12 feet. Find its volume.
Show Answer
First, find the base area: $$B = s^2 = 10^2 = 100 \text{ ft}^2$$
Then apply the pyramid volume formula: $$V = \frac{1}{3}Bh = \frac{1}{3}(100)(12) = \frac{1200}{3} = 400 \text{ ft}^3$$
The pyramid has a volume of 400 cubic feet.
Problem 7: Two similar rectangular prisms have corresponding edges in the ratio 2:5. If the smaller prism has a volume of 48 cubic inches, what is the volume of the larger prism?
Show Answer
The scale factor from smaller to larger is $k = \frac{5}{2} = 2.5$.
For similar solids, volumes scale by $k^3$: $$\frac{V_{\text{large}}}{V_{\text{small}}} = k^3 = (2.5)^3 = 15.625$$
Therefore: $$V_{\text{large}} = 15.625 \times 48 = 750 \text{ cubic inches}$$
The larger prism has a volume of 750 cubic inches.
Alternative approach: The ratio of volumes is $(2:5)^3 = 8:125$. If the smaller volume is 48, then: $$\frac{48}{V_{\text{large}}} = \frac{8}{125}$$ $$V_{\text{large}} = \frac{48 \times 125}{8} = 750 \text{ cubic inches}$$
Summary
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Volume measures the 3D space inside a solid (cubic units), while surface area measures the total area of all exterior surfaces (square units).
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Prisms and cylinders have volume $V = Bh$, where $B$ is the base area:
- Rectangular prism: $V = lwh$
- Cylinder: $V = \pi r^2 h$
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Pyramids and cones have volume equal to one-third of the corresponding prism or cylinder: $V = \frac{1}{3}Bh$
- Cone: $V = \frac{1}{3}\pi r^2 h$
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Spheres have special formulas:
- Volume: $V = \frac{4}{3}\pi r^3$
- Surface area: $SA = 4\pi r^2$
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Composite solids can be analyzed by breaking them into simpler shapes, calculating each part, and combining the results.
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Cavalieri’s Principle tells us that solids with equal cross-sectional areas at every height have equal volumes.
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Similar solids with scale factor $k$ have:
- Linear dimensions in ratio $k$
- Surface areas in ratio $k^2$
- Volumes in ratio $k^3$
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These concepts apply everywhere in the real world: shipping boxes, storage tanks, swimming pools, ice cream cones, and architectural design all require understanding surface area and volume.
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When working with 3D shapes, always pay attention to which measurement you need (height vs. slant height) and which units you should use (square for area, cubic for volume).