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

These notes condense Pearson Edexcel International Advanced Level Mathematics (Specification Issue 3, April 2019), Pure Mathematics 4 sections 6.1, 6.4 and 6.5. P4 is compulsory for IAL Mathematics and Pure Mathematics and has no IAS weighting. For full explanations read the study guide, then test yourself with the practice questions. See also the Edexcel A-level Mathematics hub, the printable checklist and the free 10-minute diagnostics.

Key results

Spec	Result	Notes
6.1	V = π ∫ y² dx	Rotation through 360° about the x-axis; limits x = a to x = b
6.1	V = π ∫ y² (dx/dt) dt	Parametric form; limits are values of t
6.1	Not required: π ∫ x² dy	The specification asks for y² dx only
6.4	∫ 1/g(y) dy = ∫ f(x) dx	For dy/dx = f(x) g(y)
6.4	General solution has a constant; particular solution has it found	Use the given pair of values
6.5	Area = ∫ y dx = ∫ y (dx/dt) dt	Limits are values of t

You are not expected to sketch a curve from its parametric equations.

Method boxes

Volume of revolution, cartesian

1. Write y² in full. Expand brackets before integrating.
2. Integrate with respect to x between the given limits.
3. Keep π outside, and give the answer as a multiple of π unless a decimal is asked for.

Volume between a curve and a line

1. Find where the curve meets the line to get the limits.
2. Work out V for the outer boundary, then subtract V for the inner boundary.
3. For a horizontal line y = k the inner volume is the cylinder π k² × length.

Volume or area, parametric

1. Find dx/dt.
2. Convert limits: solve x(t) = a and x(t) = b, or y(t) = 0 when the curve meets the axis.
3. Substitute into π ∫ y² (dx/dt) dt or ∫ y (dx/dt) dt, then integrate in t.
4. If x decreases as t increases, take the limits so that x goes from smaller to larger.

Separable differential equation

1. Check the equation factorises as f(x) × g(y).
2. Separate: g-terms with dy, f-terms with dx.
3. Integrate both sides and add one constant c.
4. Use the given values to find c, then rearrange to y = … if asked.

Small worked reminders

Squaring a bracket

y = x + 1 for 0 ≤ x ≤ 2:

V = π ∫ (0 to 2) (x + 1)² dx = π [ (x + 1)³/3 ] (0 to 2) = π (27/3 − 1/3)

V = 26π/3

Exponential equation

dy/dx = 3y, y = 2 when x = 0:

ln y = 3x + c, so y = A e^(3x); A = 2

y = 2e^(3x)

Square-root equation

dy/dx = x/y, y = 3 when x = 0:

y dy = x dx
y²/2 = x²/2 + c
9/2 = c

y² = x² + 9, and y = √(x² + 9) because y = 3 is positive at x = 0.

Parametric area, limits from x

x = t², y = 2t + 1, t > 0, between x = 1 and x = 9:

t = 1 to t = 3,  dx/dt = 2t
Area = ∫ (1 to 3) (2t + 1) 2t dt = ∫ (4t² + 2t) dt = [4t³/3 + t²] (1 to 3) = 45 − 7/3

Area = 128/3

Parametric volume with logarithms

x = ln t, y = t, from x = 0 to x = ln 3:

t = 1 to t = 3,  dx/dt = 1/t
V = π ∫ (1 to 3) t² × (1/t) dt = π ∫ (1 to 3) t dt = π [t²/2] (1 to 3) = π (9/2 − 1/2)

V = 4π

A model with a constant to find

A hot drink cools so that dθ/dt = −k(θ − 20), with θ = 80 at t = 0 and θ = 50 at t = 6.

∫ 1/(θ − 20) dθ = ∫ −k dt
ln(θ − 20) = −kt + c;  θ − 20 = 60 e^(−kt)
t = 6:  30 = 60 e^(−6k), so e^(−6k) = ½ and k = (ln 2)/6

k = (ln 2)/6, about 0.116

Checking a solution

Differentiate your answer and put it back in. For y = 6/(x + 1): dy/dx = −6/(x + 1)², and −y/(x + 1) = −6/(x + 1)². The two sides agree, and y = 6 at x = 0.

Must-know distinctions

• Volume uses y², area uses y. Both need dx/dt in the parametric form.
• General solution: contains a constant. Particular solution: constant found from given values.
• ln y = x + c gives y = A e^x with A = e^c. It does not give y = e^x + c.
• π ∫ y² dx needs y as a function of x. π ∫ x² dy is not required.
• A limit on x and a limit on t are different things. Convert before you substitute.
• For a region between a curve and a line, subtract volumes. Do not square the difference of the two y values.

Which method?

You are given	Do this
y in terms of x, region rotated about the x-axis	V = π ∫ y² dx
x and y in terms of t, region rotated	V = π ∫ y² (dx/dt) dt
x and y in terms of t, area under the curve	A = ∫ y (dx/dt) dt
dy/dx written as a product or quotient of x and y terms	Separate the variables and integrate both sides
A rate of change proportional to a quantity	Use the equation given, then separate
Two values of the variable that fix the constant	Substitute after integrating

Typical question set-ups

• A region under a curve rotated about the x-axis, exact answer asked.
• A region between a curve and a line, with the intersection to find.
• Parametric curve with a range of t, volume or area wanted.
• Differential equation given, particular solution wanted.
• A model for a quantity changing with time, find the constant, then predict a value.

Quick self-test

Work each out by hand before checking.

1. Find V when the region under y = x², for 0 ≤ x ≤ 1, makes a full revolution about the x-axis.
2. Write the integral for V when x = t², y = 3t, 1 ≤ t ≤ 2, then evaluate it.
3. A curve has x = 2t, y = t² for 0 ≤ t ≤ 3. Find the area between it and the x-axis.
4. Write down the general solution of dy/dx = 2y, y > 0.
5. Solve dy/dx = x/y given y = 3 when x = 0.
6. Solve dy/dx = e^(2x)/y given y = 1 when x = 0 (y > 0).
7. For x = t², t > 0, state the t-limits that match x = 4 to x = 9.
8. Find the area when x = 5 − t, y = t², 0 ≤ t ≤ 2.
9. Why must dx/dt appear when you integrate y² with respect to t?
10. Solve dy/dx = −y/(x + 1) given y = 6 when x = 0, for x > −1.
11. Find V when the region under y = 2/√x, for 1 ≤ x ≤ e, makes a full revolution about the x-axis.
12. Solve dy/dx = y² given y = 1 when x = 0, and state the value of x where y becomes undefined.

Answers

1. V = π ∫ (0 to 1) x⁴ dx = π/5.
2. V = π ∫ (1 to 2) 9t² × 2t dt = 18π [t⁴/4] (1 to 2) = 18π × 15/4 = 135π/2.
3. Area = ∫ (0 to 3) t² × 2 dt = 2 [t³/3] (0 to 3) = 18.
4. ln y = 2x + c, so y = A e^(2x), A > 0.
5. y dy = x dx, y²/2 = x²/2 + c, c = 9/2, so y = √(x² + 9).
6. y dy = e^(2x) dx, y²/2 = e^(2x)/2 + c, c = 0, so y = e^x.
7. t = 2 to t = 3.
8. x runs from 5 down to 3, so take t = 2 to t = 0: ∫ (2 to 0) t² × (−1) dt = 8/3.
9. Because dx = (dx/dt) dt. Changing the variable from x to t changes the integrand as well as the limits.
10. ln y = −ln(x + 1) + c, so y = A/(x + 1), A = 6, giving y = 6/(x + 1).
11. V = π ∫ (1 to e) 4/x dx = 4π [ln x] (1 to e) = 4π.
12. ∫ y⁻² dy = ∫ dx gives −1/y = x + c; c = −1, so y = 1/(1 − x). It is undefined at x = 1.

Quick checks before you move on

• Can you recall V = π ∫ y² dx and the parametric version without looking?
• Can you separate and integrate dy/dx = f(x) g(y) and then find the constant?
• Can you convert x-limits to t-limits for x = t² and for x = t² + 2t?
• Can you say why x decreasing as t increases needs care with limits?

Where marks are usually lost

• Squaring y wrongly: (t + 2)² written as t² + 4.
• Integrating y² dt instead of y² (dx/dt) dt.
• Keeping x-limits in a t integral.
• A negative volume or area from reversed limits when x decreases.
• Subtracting y values before squaring, not after, for a region between curve and line.
• Writing the constant on both sides, or adding it after substituting values.
• Taking the exponential of ln y = x + c as y = e^x + c.
• Stopping at e^y = … when the question asks for y in terms of x.
• Answers given as decimals when the question asks for an exact value.

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.