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IB Mathematics: Applications and Interpretation · Calculus

Areas enclosed by curves; volumes of revolution practice questions.

27 exam-style questions on Areas enclosed by curves; volumes of revolution — topic 5.12 of Calculus in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.

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About this topic

Syllabus code
5.12
Subject
IB Mathematics: Applications and Interpretation
Topic
5 · Calculus
Questions
27 with worked answers
Question types
12 fill-in-the-gap questions · 8 multiple-choice questions · 4 true/false questions · 2 flashcards · 1 matching task
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Sample questions on Areas enclosed by curves; volumes of revolution

Determine the expression for the area enclosed between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to x = b.

  • ∫ from a to b of [g(x) − f(x)] dx
  • ∫ from a to b of [f(x) − g(x)] dx
  • ∫ from a to b of [f(x) + g(x)] dx
  • ∫ from a to b of [f(x)·g(x)] dx

Answer: ∫ from a to b of [f(x) − g(x)] dx

The area between curves is the integral of (upper − lower), i.e. ∫ [f(x) − g(x)] dx, which is always non-negative on [a, b].

Determine the formula for the volume when the region under y = f(x), between x = a and x = b, is rotated 2π radians about the x-axis.

  • π ∫ from a to b of f(x) dx
  • 2π ∫ from a to b of f(x) dx
  • π ∫ from a to b of [f(x)]² dx
  • π ∫ from a to b of f(x²) dx

Answer: π ∫ from a to b of [f(x)]² dx

The volume of revolution about the x-axis is V = π ∫ [f(x)]² dx; the radius of each disc is f(x), so its area is π[f(x)]².

The region bounded by y = x², the x-axis and x = 2 is rotated about the x-axis. Determine the exact volume.

  • (8/5)π
  • (32/5)π
  • 8π
  • (32/3)π

Answer: (32/5)π

V = π ∫ from 0 to 2 of (x²)² dx = π ∫ x⁴ dx = π[x⁵/5] from 0 to 2 = π·(32/5) = (32/5)π.

A region is rotated 2π about the x-axis to give a volume of revolution. Determine why the integrand [f(x)]² must be used rather than f(x).

  • Because each cross-section is a disc of area π[f(x)]²
  • Because the radius must be squared to remove negatives only
  • Because volume is always the square of area
  • Because π must be squared as well

Answer: Because each cross-section is a disc of area π[f(x)]²

Each thin slice perpendicular to the axis is a disc of radius f(x), so its area is π[f(x)]²; summing these discs gives the volume.

The curves y = x² and y = 8 − x² enclose a region. Determine the correct integral for its area.

  • ∫ from −2 to 2 of (8 − 2x²) dx
  • ∫ from −2 to 2 of (2x² − 8) dx
  • ∫ from 0 to 8 of (8 − 2x²) dx
  • ∫ from −2 to 2 of (8 − x²) dx

Answer: ∫ from −2 to 2 of (8 − 2x²) dx

The curves meet where x² = 8 − x², i.e. x = ±2. Upper minus lower is (8 − x²) − x² = 8 − 2x², integrated from −2 to 2.

The region under y = 3 (a horizontal line) from x = 0 to x = 5 is rotated 2π about the x-axis. Determine the volume.

  • 15π
  • 45π
  • 9π
  • 75π

Answer: 45π

This forms a cylinder: V = π ∫ from 0 to 5 of 3² dx = π·9·5 = 45π, matching πr²h with r = 3, h = 5.

That was 6 of 27 Areas enclosed by curves; volumes of revolution questions.

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27 Areas enclosed by curves; volumes of revolution questions — free with an account. Spaced repetition, streaks and full exam simulations included.

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