27 exam-style questions on Areas and volumes of revolution — topic 5.17 of Calculus in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
The region enclosed by the curve y = x², the x-axis and the line x = 2 is rotated through 2π radians about the x-axis. Calculate the volume of the solid formed.
Answer: 32π/5
V = π∫₀² (x²)² dx = π∫₀² x⁴ dx = π·[x⁵/5]₀² = 32π/5. A common error is integrating x² without squaring first, which gives 8π/3.
The region bounded by the curve y = x², the y-axis and the line y = 4 is rotated through 360° about the y-axis. Determine the volume of the solid generated.
Answer: 8π
About the y-axis, V = π∫ x² dy. Since y = x², x² = y, so V = π∫₀⁴ y dy = π·[y²/2]₀⁴ = 8π.
A region bounded by a curve and the y-axis, between y = c and y = d, is rotated through 2π about the y-axis. Which integral gives the volume of the solid of revolution?
Answer: π∫ x² dy from y = c to y = d
For rotation about the y-axis the disc radius is x, so V = π∫ x² dy with x written as a function of y. Mixing up the variable of integration or the radius is a classic error.
The region under the curve y = 1/x from x = 1 to x = 3 is rotated through 2π about the x-axis. Calculate the exact volume of the solid formed.
Answer: 2π/3
V = π∫₁³ 1/x² dx = π[−1/x]₁³ = π(1 − 1/3) = 2π/3. Integrating 1/x instead of 1/x² (giving π ln 3) is the standard trap: you must square y first.
Calculate the area of the region enclosed between the curve x = y² and the y-axis for 0 ≤ y ≤ 2.
Answer: 8/3
The area between a curve and the y-axis is ∫ x dy, so A = ∫₀² y² dy = [y³/3]₀² = 8/3.
The region under the curve y = e^x from x = 0 to x = 1 is rotated through 2π about the x-axis. Determine the exact volume of the solid.
Answer: π(e² − 1)/2
V = π∫₀¹ e^(2x) dx = π·[e^(2x)/2]₀¹ = π(e² − 1)/2. Squaring e^x gives e^(2x), and its antiderivative carries a factor of 1/2.
That was 6 of 27 Areas and volumes of revolution questions.
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