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How do you find the volume of solids, identify cross sections and solids of revolution, and apply density?

Use volume formulas for prisms, cylinders, pyramids, cones, and spheres; identify the two-dimensional cross sections of three-dimensional solids and the solids formed by rotating a region; and solve density problems combining volume with mass or population.

A NY Regents Geometry answer on volume and solids: the prism, cylinder, pyramid, cone, and sphere formulas, identifying cross sections and solids of revolution, and applying density to mass and population problems.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. The volume formulas
  3. Cross sections and solids of revolution
  4. Density problems
  5. Applying it cleanly
  6. Try this

What this topic is asking

The Regents Geometry exam (the Geometric Measurement and Dimension, G-GMD, and Modeling with Geometry, G-MG, clusters) wants you to compute the volume of common solids, identify the cross section formed by slicing a solid and the solid of revolution formed by rotating a region, and solve density problems that combine volume with mass or population. Volume and density appear in Part I and reliably in a Part III or Part IV modeling task.

The volume formulas

The Regents reference sheet provides the volume formulas, but you must recognize each solid and identify its base and height.

The key relationship is that a pyramid or cone is exactly one-third of the prism or cylinder with the same base and height. Forgetting the 13\frac{1}{3} is the most common volume error. A composite solid (such as a cylinder with a cone on top) is handled by adding or subtracting the relevant volumes.

Cross sections and solids of revolution

Two spatial-reasoning ideas appear regularly.

  • A cross section is the 2D shape revealed when a plane cuts through a solid. A horizontal slice of a cylinder is a circle; a vertical slice through its axis is a rectangle. A slice of a sphere is always a circle.
  • A solid of revolution is formed by rotating a 2D region about a line. A rectangle rotated about one side sweeps out a cylinder; a right triangle rotated about a leg sweeps out a cone; a semicircle rotated about its diameter sweeps out a sphere.

Density problems

Density is a rate, an amount per unit of volume (or area, or length). The Regents uses it to model mass per volume, population per area, or similar.

Applying it cleanly

A clarifying point worth stressing is the radius-versus-diameter trap: volume formulas use the radius, so if a problem gives a diameter you must halve it first. A cylinder of diameter 8 has radius 4, and using 8 by mistake inflates the volume fourfold (since the radius is squared). A second habit that protects credit is to track units: a volume in cubic meters multiplied by a density in kilograms per cubic meter gives kilograms, and checking that the units cancel correctly confirms you have set the density equation up the right way.

Try this

Q1. Find the volume of a cylinder with radius 3 and height 7, in terms of π\pi. [1 credit]

  • Cue. V=πr2h=π(9)(7)=63πV = \pi r^2 h = \pi (9)(7) = 63\pi.

Q2. What solid is formed by rotating a right triangle about one of its legs? [1 credit]

  • Cue. A cone.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). A cone has radius 6 cm and height 10 cm. What is its volume, in terms of π\pi? (1) 120π120\pi cm3^3 (2) 360π360\pi cm3^3 (3) 60π60\pi cm3^3 (4) 100π100\pi cm3^3
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The correct answer is (1).

The cone volume formula (on the reference sheet) is V=13πr2hV = \frac{1}{3}\pi r^2 h. Substitute r=6r = 6 and h=10h = 10: V=13π(6)2(10)=13π(36)(10)=13(360)π=120πV = \frac{1}{3}\pi (6)^2 (10) = \frac{1}{3}\pi (36)(10) = \frac{1}{3}(360)\pi = 120\pi cm3^3. Forgetting the 13\frac{1}{3} gives 360π360\pi (choice 2), the cylinder volume, which is the most common error.

Regents (style)4 marksPart III (constructed response). A cylindrical water tank has radius 2 m and height 5 m. Water has a density of about 1000 kg per cubic meter. (a) Find the volume of the tank in terms of π\pi, then as a decimal to the nearest cubic meter. (b) Find the mass of water it holds when full, to the nearest thousand kilograms.
Show worked answer →

A 4-credit question: credit for the volume formula, the evaluation, and the density step.

(a) V=πr2h=π(2)2(5)=20π≈62.83≈63V = \pi r^2 h = \pi (2)^2 (5) = 20\pi \approx 62.83 \approx 63 cubic meters.
(b) Mass == density ×\times volume =1000×20π≈1000×62.83≈62,832≈63,000= 1000 \times 20\pi \approx 1000 \times 62.83 \approx 62{,}832 \approx 63{,}000 kg. Using the diameter as the radius, or omitting the density multiplication in part (b), costs credits.

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