Math problem set Grade 8 Math

Vellacott Dam Penstock, Trap, and Air Chamber: Cylinder, Cone, Sphere

A grade 8 problem set finding the volume of a pipe, a cone-shaped sediment trap, and a spherical air chamber at a fictional hydropower dam.

Grade: Grade 8
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

8.G.C.9 Common Core Math

Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.

How this material aligns: Problems 1, 3, 4, 5, and 8 apply the cylinder, cone, and sphere volume formulas to the penstock, sediment trap, and air chamber, and problems 6, 7, and 9 use those volumes to compare two parts, find a fill time, and explain the effect of doubling the diameter.

All materials for 8.G.C.9 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will find the volume of a cylinder, a cone, and a sphere from a stated diameter and depth.
Students will convert a volume in cubic feet to gallons.
Students will explain how doubling a diameter changes the volume of a cylinder.

Vocabulary

Penstock: The huge pipe that carries water from a reservoir down to the turbines in a hydropower plant. Falling water rushes through it with enormous force.

Also: cylinder, cone, sphere, volume, radius

Teacher overview

Vellacott Dam is a fictional hydropower dam on the Stagwater River. Water leaves the reservoir through a long steel pipe called a penstock, drops to the turbine, and passes a cone-shaped trap that catches sand along the way. A round steel air chamber above the pipe cushions the pressure surge when a valve closes. All three parts are ordinary solids, so one problem set can cover the cylinder, the cone, and the sphere. Teaching note: the single most common error here is substituting the diameter where the formula asks for the radius, which makes the penstock volume four times too large. The second most common is squaring the radius in the sphere formula instead of cubing it, so ask students to name the exponent aloud before they compute problem 4.

Vellacott Dam Penstock, Trap, and Air Chamber: Cylinder, Cone, Sphere — Grade 8 Math

Name: ____________________________ Date: ______________

Problems

Vellacott Dam is fictional. Use 3.14 for pi in every problem, and use 7.48 gallons for each cubic foot. The volume formulas are V = pi x r^2 x h for a cylinder, V = one third x pi x r^2 x h for a cone, and V = four thirds x pi x r^3 for a sphere.

The penstock is a cylinder 8 feet across and 120 feet long. Find its volume in cubic feet.
Convert the penstock volume from problem 1 to gallons.
The sediment trap is a cone 10 feet across and 9 feet deep. Find its volume in cubic feet.
The air chamber is a sphere 6 feet across. Find its volume in cubic feet.
Find the combined volume of the penstock, the sediment trap, and the air chamber in cubic feet.
How many times as large as the sediment trap is the penstock?
The crew refills the empty penstock at 1,500 cubic feet per minute. How many minutes does the refill take? Round to the nearest tenth.
A larger penstock is 16 feet across and still 120 feet long. Find its volume in cubic feet.
The diameter in problem 8 is twice the diameter in problem 1. Explain why the volume did not simply double.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

Answer: 6,028.8 cubic feet.
The radius is half of 8 feet, or 4 feet. V = 3.14 x 4^2 x 120 = 3.14 x 16 x 120 = 3.14 x 1,920 = 6,028.8 cubic feet.
Answer: 45,095.4 gallons.
6,028.8 x 7.48 = 45,095.424, which rounds to 45,095.4 gallons.
Answer: 235.5 cubic feet.
The radius is half of 10 feet, or 5 feet. V = one third x 3.14 x 5^2 x 9 = one third x 3.14 x 25 x 9 = one third x 706.5 = 235.5 cubic feet.
Answer: 113.04 cubic feet.
The radius is half of 6 feet, or 3 feet. V = four thirds x 3.14 x 3^3 = four thirds x 3.14 x 27 = four thirds x 84.78 = 113.04 cubic feet.
Answer: 6,377.34 cubic feet.
6,028.8 + 235.5 + 113.04 = 6,377.34 cubic feet.
Answer: 25.6 times.
6,028.8 divided by 235.5 = 25.6, so the penstock holds 25.6 times what the sediment trap holds.
Answer: 4.0 minutes.
6,028.8 divided by 1,500 = 4.0192 minutes, which rounds to 4.0 minutes.
Answer: 24,115.2 cubic feet.
The radius is half of 16 feet, or 8 feet. V = 3.14 x 8^2 x 120 = 3.14 x 64 x 120 = 3.14 x 7,680 = 24,115.2 cubic feet.
Answer: The volume grew four times, not two.
The radius is squared in the cylinder formula, so doubling the radius from 4 feet to 8 feet multiplies r^2 by 4. Since the length did not change, the volume also grew by a factor of 4: 6,028.8 x 4 = 24,115.2 cubic feet.

Differentiation

Support

Give students a formula card with the cylinder, cone, and sphere drawn and labeled, and with a blank line for the radius above each formula so they must record the radius before they substitute. Work problem 1 together to the end, then let students begin at problem 2.

Extension

Ask students to find the depth a cone-shaped trap would need if it kept the 10-foot width but held the same volume as the spherical air chamber. Have them show the equation they solved and check the result in the cone formula.

ELL supports

Pre-teach penstock, sediment, radius, and diameter with a cross-section sketch of the dam that labels each part and marks the radius as half of the width. Provide the sentence frame for problem 9: "When the radius doubles, the volume becomes ___ times as large because the formula ___ the radius."

Extensions

Have students measure a cylindrical container in the classroom, find its volume in cubic inches, and convert to a fraction of a gallon. Ask them to compare the number of those containers it would take to hold the volume of the fictional penstock from problem 1.