Pearson Edexcel International Advanced Level Mathematics: Volumes of revolution, separable differential equations and areas from parametric equations (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover Pearson Edexcel International Advanced Level Mathematics (Specification Issue 3, April 2019), unit P4 sections 6.1, 6.4 and 6.5. P4 is compulsory for IAL Mathematics and Pure Mathematics and has no IAS weighting. Teaching is in the study guide and the revision notes. See also the Edexcel A-level Mathematics hub, the printable checklist and the free 10-minute diagnostics. Give exact answers unless told otherwise.

Questions

1. Find the exact volume generated when the region under y = 3/√x, from x = 1 to x = 4, makes a complete revolution about the x-axis. [3]

2. The curve y = x(4 − x) meets the x-axis at the origin and at (4, 0). The finite region between the curve and the x-axis makes a complete revolution about the x-axis. State the exact volume of the solid generated. [4]

3. The curve y = x² + 1 meets the line y = 5 at the point P in the first quadrant. The finite region R is bounded by the curve, the line y = 5 and the y-axis. Region R makes a complete revolution about the x-axis. State the exact volume of the solid generated. [5]

4. A curve has parametric equations x = ln t, y = t², t ≥ 1. The region under the curve between x = 0 and x = ln 2 makes a complete revolution about the x-axis. State the exact volume generated. [5]

5. A curve has parametric equations x = 3 cos t, y = 2 sin t, 0 ≤ t ≤ π. The curve is rotated through 360° about the x-axis. Use sin³t ≡ ¼(3 sin t − sin 3t) to find the exact volume enclosed. [5]

6. Solve dy/dx = 2xy², given that y = 1 when x = 0. Give y in terms of x. [4]

7. Given that y > 0 and y = 1 when x = 0, solve dy/dx = y cos²x, giving y in terms of x. [5]

8. A drink cools so that its temperature θ °C at time t minutes satisfies dθ/dt = −k(θ − 20), where k is a positive constant. At t = 0, θ = 100. At t = 10, θ = 60. Find the value of t when θ = 30. [7]

9. A curve has parametric equations x = 2t², y = t(4 − t²), t ≥ 0. The curve meets the x-axis at the origin and at the point A. Find the area of the finite region between the curve and the x-axis. [5]

10. A curve has parametric equations x = tan t, y = cos²t, 0 ≤ t ≤ π/4. Find the area under the curve between x = 0 and x = 1. [4]

11. The curve C passes through (0, 3) and satisfies dy/dx = −2y/(x + 1) for x > −1.

(a) Find y in terms of x. [5] (b) The region bounded by C, the axes and the line x = 2 makes a complete revolution about the x-axis. State the exact volume generated. [3]

Answers

1. y² = 9/x [1]. V = π ∫ (1 to 4) 9/x dx = 9π [ln x] (1 to 4) [1]. = 9π ln 4 = 18π ln 2 [1] [3] Examiner insight: Write the squared integrand before integrating. Rewriting 9π ln 4 as 18π ln 2 is optional, but state one clear exact form.

2. y² = x²(4 − x)² = x⁴ − 8x³ + 16x² [1]. V = π ∫ (0 to 4) (x⁴ − 8x³ + 16x²) dx = π [x⁵/5 − 2x⁴ + 16x³/3] (0 to 4) [1]. At x = 4: 1024/5 − 512 + 1024/3 [1]. V = 512π/15 [1] [4] Examiner insight: Expand y² fully and integrate term by term. Integrating y and squaring the result is the commonest slip.

3. x² + 1 = 5 gives x = 2 at P [1]. Cylinder of radius 5, length 2: π × 25 × 2 = 50π [1]. (x² + 1)² = x⁴ + 2x² + 1 [1]. π ∫ (0 to 2) (x⁴ + 2x² + 1) dx = π [x⁵/5 + 2x³/3 + x] (0 to 2) = 206π/15 [1]. V = 50π − 206π/15 = 544π/15 [1] [5] Examiner insight: Subtract the two volumes. Squaring the difference (5 − y) would give a different, incorrect result, so keep the two boundaries separate.

4. dx/dt = 1/t [1]. x = 0 gives t = 1 and x = ln 2 gives t = 2 [1]. V = π ∫ (1 to 2) t⁴ × (1/t) dt [1] = π ∫ (1 to 2) t³ dt = π [t⁴/4] (1 to 2) [1] = π (4 − 1/4) = 15π/4 [1] [5] Examiner insight: State the change of limits explicitly. Leaving 0 and ln 2 in a t integral is an error that is easy to spot.

5. dx/dt = −3 sin t [1]. x runs from −3 (t = π) to 3 (t = 0), so V = π ∫ (π to 0) 4 sin²t × (−3 sin t) dt [1]. Reversing the limits: V = 12π ∫ (0 to π) sin³t dt [1]. ∫ sin³t dt = ¼ [−3 cos t + ⅓ cos 3t] (0 to π) = ¼ (8/3 + 8/3) = 4/3 [1]. V = 12π × 4/3 = 16π [1] [5] Examiner insight: A negative volume tells you the limits are the wrong way round. Check that x is increasing through the range you integrate over.

6. ∫ y⁻² dy = ∫ 2x dx [1]. −1/y = x² + c [1]. x = 0, y = 1: c = −1 [1]. y = 1/(1 − x²) [1] [4] Examiner insight: Integrating y⁻² gives −1/y. Include the constant at the integration step, before substituting the values.

7. ∫ 1/y dy = ∫ cos²x dx [1]. cos²x = ½(1 + cos 2x) [1]. ln y = x/2 + ¼ sin 2x + c [1]. x = 0, y = 1: c = 0 [1]. y = e^(x/2 + ¼ sin 2x) [1] [5] Examiner insight: The double-angle identity is needed to integrate cos²x. The answer is requested for y, so take the exponential at the end.

8. ∫ 1/(θ − 20) dθ = ∫ −k dt [1]. ln(θ − 20) = −kt + c [1]. t = 0, θ = 100: ln 80 = c, so θ − 20 = 80e^(−kt) [1]. t = 10: 40 = 80e^(−10k) [1], so k = (ln 2)/10 [1]. θ = 30: 10 = 80e^(−kt), so kt = ln 8 [1]. t = ln 8 × 10/ln 2 = 30 minutes [1] [7] Examiner insight: Subtract 20 before taking the logarithm, as the equation is in terms of θ − 20. Keep k exact until the end to avoid rounding errors.

9. y = 0 for t > 0 gives t = 2, so the limits are t = 0 to t = 2 [1]. dx/dt = 4t [1]. Area = ∫ (0 to 2) (4t − t³) × 4t dt = ∫ (16t² − 4t⁴) dt [1]. = [16t³/3 − 4t⁵/5] (0 to 2) [1]. = 128/3 − 128/5 = 256/15 [1] [5] Examiner insight: Finding where the curve meets the axis gives the upper limit in t. Substituting a point’s x-value into a t integral is the usual error.

10. dx/dt = sec²t [1]. y dx/dt = cos²t × sec²t = 1 [1]. x = 0 gives t = 0 and x = 1 gives t = π/4 [1]. Area = ∫ (0 to π/4) 1 dt = π/4 [1] [4] Examiner insight: Simplify the integrand before integrating. Seeing that cos²t × sec²t = 1 makes the integral trivial.

11. (a) ∫ 1/y dy = ∫ −2/(x + 1) dx [1]. ln y = −2 ln(x + 1) + c [1]. x = 0, y = 3: c = ln 3 [1]. ln y = ln 3 − ln (x + 1)² [1]. y = 3/(x + 1)² [1] (b) y² = 9/(x + 1)⁴. V = π ∫ (0 to 2) 9(x + 1)⁻⁴ dx [1] = 9π [−1/(3(x + 1)³)] (0 to 2) [1] = 3π (1 − 1/27) = 26π/9 [1] [8] Examiner insight: Part (b) uses the equation from part (a). If part (a) is wrong, a correct method in (b) still shows what you know, so attempt it.

Where marks are usually lost

• Squaring y wrongly, or integrating y before squaring.
• Forgetting dx/dt in a parametric integral.
• Leaving x-limits in an integral with respect to t.
• Reversing limits when x decreases, giving a negative volume.
• Squaring a difference when rotating a region between a curve and a line, instead of subtracting two volumes.
• Missing the constant of integration, or adding it after substituting values.
• Mishandling ln: y = e^x + c in place of y = A e^x.
• Not rearranging to y = … when the question asks for y.
• Giving a decimal when the question asks for an exact value.

Next steps

• Condense the topic with the revision notes.
• Re-read worked methods in the study guide.
• See the Edexcel A-level Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics to find weak spots.
• Book a free trial class at the trial page for help with this unit.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited. Unit P4, section 6: 6.1, 6.4 and 6.5.