Surface Area and Volume of 3D Shapes

Calculate surface area and volume of prisms, cylinders, cones, spheres, and pyramids for the Digital SAT using reference formulas.

Three-dimensional geometry on the Digital SAT involves calculating volumes and surface areas of common solids. The SAT provides key formulas on the reference sheet, but understanding when and how to apply them is essential.

Core Concepts

Volume Formulas

Shape Volume
Rectangular prism V=lwhV = lwh
Cylinder V=πr2hV = \pi r^2 h
Cone V=13πr2hV = \frac{1}{3}\pi r^2 h
Sphere V=43πr3V = \frac{4}{3}\pi r^3
Pyramid V=13BhV = \frac{1}{3}Bh (where BB = base area)

Surface Area Formulas

Shape Surface Area
Rectangular prism SA=2(lw+lh+wh)SA = 2(lw + lh + wh)
Cylinder SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi rh
Sphere SA=4πr2SA = 4\pi r^2

Key Relationships

  • A cone has one-third the volume of a cylinder with the same base and height.
  • A pyramid has one-third the volume of a prism with the same base and height.
  • Doubling a dimension: doubling radius quadruples the area and octuples the volume.

Density and Rate Problems

Density=massvolume\text{Density} = \frac{\text{mass}}{\text{volume}}, so mass=density×volume\text{mass} = \text{density} \times \text{volume}.

Flow rate: volume=rate×time\text{volume} = \text{rate} \times \text{time}.

Strategy Tips

Tip 1: Use the Reference Sheet

Volume formulas for cones, spheres, and pyramids are provided. Don't waste time memorising — but do know how to apply them.

Tip 2: Identify the Shape First

Read the problem carefully to determine which formula applies.

Tip 3: Watch the Units

Volume is in cubic units; surface area is in square units.

Tip 4: Radius vs. Diameter

If given the diameter, halve it to get the radius before using formulas.

Worked Example: Example 1

Problem

Find the volume of a cylinder with radius 5 and height 10.

V=π(5)2(10)=250π785.4V = \pi(5)^2(10) = 250\pi \approx 785.4

Solution

Worked Example: Example 2

Problem

A cone has radius 3 and height 7. Find its volume.

V=13π(3)2(7)=21π66.0V = \frac{1}{3}\pi(3)^2(7) = 21\pi \approx 66.0

Solution

Worked Example: SAT-Style

Problem

A spherical balloon has radius 6 inches. What is its volume?

V=43π(6)3=288π904.8V = \frac{4}{3}\pi(6)^3 = 288\pi \approx 904.8 cubic inches.

Solution

Worked Example: Example 4

Problem

A rectangular box has dimensions 4 × 5 × 6. Find the surface area.

SA=2(20+24+30)=2(74)=148SA = 2(20 + 24 + 30) = 2(74) = 148 square units.

Solution

Worked Example: Example 5

Problem

Water fills a cylindrical tank (radius 3 m, height 8 m) at 2 m³/min. How long to fill it?

V=π(9)(8)=72π226.2V = \pi(9)(8) = 72\pi \approx 226.2 m³. Time =226.2/2113.1= 226.2/2 \approx 113.1 minutes.

Solution

Practice Problems

  1. Problem 1

    Volume of a sphere with radius 4.

    Problem 2

    Surface area of a cylinder: radius 3, height 10.

    Problem 3

    A pyramid has a square base of side 6 and height 9. Find the volume.

    Problem 4

    If the radius of a sphere is doubled, by what factor does the volume increase?

    Problem 5

    A cone and cylinder have the same base radius and height. The cylinder's volume is 300π300\pi. What is the cone's volume?

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Common Mistakes

  • Forgetting the 13\frac{1}{3} factor for cones and pyramids.
  • Using diameter instead of radius. Always halve the diameter first.
  • Confusing surface area and volume formulas.
  • Unit errors. Volume = cubic units, area = square units.

Key Takeaways

  • Cylinder: V=πr2hV = \pi r^2 h. Cone: V=13πr2hV = \frac{1}{3}\pi r^2 h.

  • Sphere: V=43πr3V = \frac{4}{3}\pi r^3, SA=4πr2SA = 4\pi r^2.

  • Cones and pyramids have one-third the volume of their prism/cylinder counterparts.

  • Formulas are provided on the SAT reference sheet.

  • Doubling the radius octuples the volume.

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