Questions on volumes of revolution

Question 1

Write down the formula for the volume generated when the region under y = f(x) between x = a and x = b is rotated fully about the x-axis, and state why the y is squared.

Accepted Answer

V = π∫y² dx from a to b. Each strip sweeps a disc of radius y and thickness dx, and a disc's volume is π times radius squared times thickness. B1 for the formula, B1 for the disc explanation.

Question 2

The region under y = 3 from x = 0 to x = 4 is rotated about the x-axis. Find the volume by integration, and check it against a known solid.

Accepted Answer

V = π∫9 dx from 0 to 4 = 36π. The solid is a cylinder of radius 3 and length 4, and πr²h = π × 9 × 4 = 36π agrees. M1 for π∫9 dx, A1 for 36π, B1 for the cylinder check.

Question 3

Write down the formula for the volume generated when the region to the left of x = g(y) between y = c and y = d is rotated fully about the y-axis, and state which limits it uses.

Accepted Answer

V = π∫x² dy from c to d. B1 for the formula. Each strip is now horizontal, of thickness dy, so the limits run up the y-axis rather than along the x-axis. B1 for the y-limits. The x in the formula must be rewritten in terms of y before the integral can be evaluated.

Question 4

The region under y = x + 1 from x = 0 to x = 2 is rotated about the x-axis. Find the exact volume.

Accepted Answer

V = π∫(x + 1)² dx from 0 to 2. Expanding the whole bracket gives (x + 1)² = x² + 2x + 1, so V = π[x³/3 + x² + x] from 0 to 2 = π(8/3 + 4 + 2) = 26π/3. M1 for expanding (x + 1)² in full and integrating, A1 for the integrand, A1 for 26π/3. Squaring only the two written terms of the bracket, x² + 1, and dropping the cross term 2x gives a different, wrong integrand.

Question 5

The region under y = x² from x = 0 to x = 2 is rotated fully about the x-axis. Find the exact volume.

Accepted Answer

V = π∫x⁴ dx from 0 to 2 = π[x⁵/5] = 32π/5. M1 for squaring y, A1 for the integrand, M1 for integrating, A1 for 32π/5. The integrand is y² = x⁴, not x²; integrating y before squaring gives a different value.

Question 6

The region bounded by y = x³, the y-axis and y = 8 is rotated about the y-axis. Find the exact volume.

Accepted Answer

Radius is x with x³ = y, so x² = y^2/3 and V = π∫y^2/3 dy from 0 to 8 = π[3y^5/3/5] = π × 3 × 32/5 = 96π/5. M1 for x² in terms of y, A1 for the integrand, M1 for integrating with the y-limits, A1 for 96π/5.

Question 7

The region between y = x and y = x² from x = 0 to x = 1 is rotated about the x-axis. Find the volume of the solid formed.

Accepted Answer

Outer radius x, inner radius x²: V = π∫(x² − x⁴) dx from 0 to 1 = π(1/3 − 1/5) = 2π/15. M1 for the difference of squares, A1 for the integrand, M1 for integrating, A1 for 2π/15. Subtract the squares, never square the difference: π∫(x − x²)² dx describes a different and wrong solid.

Question 8

Explain why rotating the region between two curves uses π∫(outer² − inner²) dx rather than π∫(outer − inner)² dx.

Accepted Answer

Each cross-section is an annulus, a disc of the outer radius with a disc of the inner radius removed, of area π(outer² − inner²). Squaring the difference instead computes a disc whose radius is the gap between the curves, which is the volume from rotating a different region, the one under y = outer − inner, about the same axis. B1 for the annulus, B1 for its area, B1 for what squaring the difference computes instead.

Question 9

The region under x = y³ from y = 0 to y = 1 is rotated about the y-axis. Find the exact volume.

Accepted Answer

Radius is x with x = y³, so x² = y⁶ and V = π∫y⁶ dy from 0 to 1 = π[y⁷/7] = π/7. M1 for x² = y⁶, A1 for the integrand, M1 for integrating, A1 for π/7.

Question 10

The region bounded by the curve x = y², the line x = 1 and the x-axis, between y = 0 and y = 1, is rotated about the y-axis. Find the exact volume.

Accepted Answer

On [0, 1], x = 1 lies to the right of x = y², so the outer radius is 1 and the inner radius is y². V = π∫(1 − y⁴) dy from 0 to 1 = π(1 − 1/5) = 4π/5. M1 for the two radii and the difference of squares, A1 for the integrand, M1 for integrating, A1 for 4π/5. The solid is a cylinder of radius 1 with a hole bored out along its axis, widening from a point at the origin to the full radius at y = 1.

Question 11

The line x = 3y/4 from y = 0 to y = 4 is rotated about the y-axis. Find the volume by integration, and check it against the volume of a cone.

Accepted Answer

x² = 9y²/16, so V = π∫9y²/16 dy from 0 to 4 = π × (9/16) × (64/3) = 12π. M1 for x² in terms of y, A1 for the integrand, M1 for integrating. At y = 4, x = 3, so the solid is a cone of base radius 3 and height 4: πr²h/3 = π × 9 × 4/3 = 12π agrees. A1 for the cone check.

Question 12

The diagram shows the region R, bounded by a straight line, the x-axis, the y-axis and the line x = 4. R is rotated through 2π about the x-axis. Use the diagram to find the equation of the straight line, and find the exact volume of the solid generated.

Accepted Answer

The line passes through (0, 1) and (4, 3), read from the diagram, so its gradient is (3 − 1)/(4 − 0) = ½. M1 for the gradient from two points on the diagram. It cuts the y-axis at 1, so the line is y = ½x + 1, and the diagram gives the limits x = 0 and x = 4. A1 for the equation and the limits. Rotating about the x-axis, V = π∫y² dx = π∫(½x + 1)² dx from 0 to 4, and the whole bracket must be squared: (½x + 1)² = x²/4 + x + 1. M1 for squaring y in full and setting up the integral. Integrating, V = π[x³/12 + x²/2 + x] from 0 to 4. A1 for the integrated form. At x = 4 the bracket is 64/12 + 8 + 4 = 16/3 + 12 = 52/3, and at x = 0 it is 0, so V = 52π/3. A1 for the exact volume. The solid is a frustum of a cone, of radii 1 and 3 and height 4, and the frustum formula πh(R² + Rr + r²)/3 = π × 4 × (9 + 3 + 1)/3 = 52π/3 agrees.

Question 13

A curve has parametric equations x = t², y = 2t for 0 ≤ t ≤ 2. The region bounded by the curve, the x-axis and the line x = 4 is rotated through 2π about the x-axis. Find the exact volume of the solid formed.

Accepted Answer

The formula V = π∫y² dx still applies, with dx rewritten in terms of t. Here dx/dt = 2t, so dx = 2t dt, and y² = 4t². The limits transform as well, since x = 0 gives t = 0 and x = 4 gives t = 2. So V = π∫4t² × 2t dt from 0 to 2 = π∫8t³ dt = π[2t⁴] from 0 to 2 = 32π. M1 for dx = 2t dt, B1 for the new limits, M1 for y² = 4t², A1 for π∫8t³ dt, M1 for integrating, A1 for 32π. As a check, eliminating t gives y² = 4x, and π∫4x dx from 0 to 4 = π[2x²] = 32π as well.

Question 14

A thin vertical strip of a region under y = f(x), at position x and thickness δx, sweeps out a solid close to a disc of radius y and thickness δx when rotated about the x-axis. Using this, explain why V = π∫_a^b y² dx is the volume generated, and hence find the exact volume when y = √x is rotated about the x-axis between x = 0 and x = 9.

Accepted Answer

A disc of radius y and thickness δx has volume πy²δx, since a disc's volume is π × radius² × thickness. B1 for the disc volume. Summing πy²δx over every strip from a to b and letting δx → 0 turns the sum of disc volumes into the definite integral π∫y² dx, exactly as a sum of thin rectangles becomes an integral for area. B1 for the limit of the sum becoming the integral. For y = √x, y² = x, so V = π∫x dx from 0 to 9. M1 for y² = x. This gives π[x²/2] from 0 to 9 = π × 81/2 = 81π/2. M1 for setting up the integral with the given limits, A1 for integrating, A1 for 81π/2.

Question 15

The region enclosed between the lines y = x and y = 2 − x, for 0 ≤ x ≤ 2, is rotated about the x-axis. The two lines cross inside this interval. Explain why the volume integral must be split there, and find the exact volume of the solid formed.

Accepted Answer

The lines meet where x = 2 − x, at x = 1, where both equal 1. For x < 1, 2 − x > x, so 2 − x is the outer radius and x the inner; for x > 1 the two swap, x becoming the outer radius. A single bracket π∫(outer² − inner²) dx needs one formula for outer and inner throughout, so where they swap the integral has to be split into two pieces, each with its own correct assignment. B1 for identifying x = 1 as the crossing point with the reason the split is needed. On [0, 1]: π∫[(2 − x)² − x²] dx = π∫(4 − 4x) dx = π[4x − 2x²] from 0 to 1 = 2π. M1 for the first bracket, A1 for 2π. On [1, 2]: π∫[x² − (2 − x)²] dx = π∫(4x − 4) dx = π[2x² − 4x] from 1 to 2 = 2π. M1 for the second bracket, A1 for 2π. The total volume is 2π + 2π = 4π. A1 for the total. By symmetry about x = 1 the two pieces are equal.

Question 16

A bowl is modelled by rotating the curve y = x² (0 ≤ x ≤ 3, units in cm) about the y-axis, so the bowl stands on its point with a vertical axis. Find the volume of water needed to fill it to a depth of 5 cm, and state one assumption of the model.

Accepted Answer

About the y-axis, radius is x with x² = y, and the bowl reaches height 3² = 9 cm, above the 5 cm fill depth, so the curve covers the full depth asked for. M1 for x² = y with the y-axis formula. V = π∫x² dy from 0 to 5 = π∫y dy from 0 to 5 = π[y²/2] from 0 to 5 = 25π/2 cm³. M1 for the integral with limits 0 to 5, A1 for integrating, A1 for 25π/2. The model assumes the bowl's inner surface follows y = x² exactly, with no thickness to the material and no meniscus at the water's surface. B1 for a stated assumption.