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Volumes of revolution

Rotating a region under a curve through 360° about the x-axis makes a solid. Its volume is found by integrating the areas of thin circular slices.

The formula

V = π ∫a^{b} y2 dx for rotation through 360° about the x-axis.

Each slice is a disc of radius y and thickness δx, with volume πy2 δx. Write down the formula before substituting.

Square y first

Square the whole of y before integrating: if y = 2x + 1, then y2 = 4x2 + 4x + 1. If y = sqrt(x), then y2 = x. If y = sec x, then y2 = sec2 x, which integrates to tan x.

Composite solids

For the region between a curve and a line, find each volume separately and subtract, or use π ∫ (y_12 - y_22) dx. Simple shapes such as cones and cylinders can be found with formulae.

Worked example

The region under y = sqrt(x) from x = 0 to x = 4 is rotated 360° about the x-axis. Find the volume.

  1. V = π ∫0^{4} (sqrt(x))2 dx = π ∫0^{4} x dx
  2. = π × (16/2 - 0)

Answer: 8π

Key idea: V = π ∫ y2 dx about the x-axis. Square y before integrating; subtract volumes for regions between graphs.

Check yourself

  1. Which integral gives the volume when the region under y = f(x) from a to b is rotated 360° about the x-axis?
    • ∫a^{b} πy dx
    • 2π∫a^{b} y dx
    • π(∫a^{b} y dx)2
    • π∫a^{b} y2 dx
    Show answer

    π∫a^{b} y2 dx. Each slice is a disc with area πy2.

  2. The region under y = x2 from x = 0 to x = 1 is rotated 360° about the x-axis. Find the volume, to 3 decimal places.
    Show answer

    0.628. V = π ∫0^{1} x4 dx = π/5 = 0.628.

  3. Rotating the line segment y = 2x, 0 ≤ x ≤ 3, through 360° about the x-axis makes a cone. Use integration to find its volume, to 3 significant figures.
    Show answer

    113. V = π ∫0^{3} 4x2 dx = π(4/3)(27) = 36π = 113. (This is a cone of radius 6, height 3.)

  4. Find the volume of the solid formed when the area under y = ex, from x = 0 to x = 1, is turned through 360° about the x-axis. Give 3 significant figures.
    Show answer

    10.0. V = π ∫0^{1} e2x dx = (π/2)(e2 - 1) = 10.

  5. Show that rotating the region under y = sec x, between x = 0 and x = π/4, through 360° about the x-axis gives a volume of π. Give the volume to 2 decimal places.
    Show answer

    3.14. V = π ∫0^{π/4} sec2 x dx = π(tan(π/4) - 0) = π = 3.14.

  6. R is bounded by y = 1/x, the x-axis, x = 1 and x = 3. R is rotated 360° about the x-axis. Find the volume generated, to 3 decimal places.
    Show answer

    2.094. V = π ∫1^{3} x-2 dx = π(1 - 1/3) = 2π/3 = 2.094.

  7. Which integral gives the volume of a sphere of radius 3, made by rotating y = sqrt(9 - x2) about the x-axis?
    • π∫0^{3} (9 - x2) dx
    • π∫-3^{3} sqrt(9 - x2) dx
    • π∫-3^{3} (9 - x2) dx
    • 2π∫-3^{3} (9 - x2) dx
    Show answer

    π∫-3^{3} (9 - x2) dx. y2 = 9 - x2, and the semicircle runs from x = -3 to x = 3.

  8. The region between y = x and y = x2 for 0 ≤ x ≤ 1 is rotated 360° about the x-axis. Find the volume, to 3 decimal places.
    Show answer

    0.419. V = π ∫0^{1} (x2 - x4) dx = π(1/3 - 1/5) = 2π/15 = 0.419.

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Last updated 7 November 2026. Spotted a mistake? Tell us.