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

This guide follows the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (Issue 3, April 2019) and covers Pure Mathematics 4 (P4) integration sections 6.1, 6.4 and 6.5. P4 is a compulsory unit for IAL Mathematics and Pure Mathematics; the specification gives its IAS weighting as N/A, so it is not part of IAS. Substitution, integration by parts and partial fractions (6.2 and 6.3) are taught elsewhere, so every integral here is a standard one.

Use it with the revision notes and the practice questions. See the Edexcel A-level Mathematics hub, the printable checklist and the free 10-minute diagnostics. Parametric differentiation is in parametric and implicit differentiation, and exponential and reciprocal integrals are in integration of exponential and reciprocal functions.

What this unit covers

Spec	What you must be able to do
6.1	Find a volume of revolution about the x-axis using π∫y² dx, including when the curve is given by parametric equations
6.4	Solve first-order differential equations with separable variables, giving general and particular solutions
6.5	Use integration to find the area under a curve given by parametric equations

The specification says π∫y² dx is required, but not π∫x² dy. You are not expected to sketch a curve from its parametric equations.

6.1 Volume of revolution

Spin the region bounded by a curve, the x-axis and the vertical lines x = a and x = b through 360° about the x-axis. Each thin slice of width δx is almost a disc of radius y, so its volume is about πy² δx. Adding the slices gives

V = π ∫ y² dx, with limits x = a to x = b.

Square the whole of y before you integrate. For y = 3x the integrand is 9x², not 3x².

Worked example 1

Find the exact volume of the solid made by turning the region under y = e^(2x), for 0 ≤ x ≤ ½, a full turn about the x-axis.

y² = e^(4x)
V = π ∫ (0 to ½) e^(4x) dx
  = π [ ¼ e^(4x) ] (0 to ½)
  = ¼ π (e² − 1)

V = π(e² − 1)/4

Worked example 2

The region under y = 1/√(2x + 1) between x = 0 and x = 4 is spun a full turn around the x-axis. What is the exact volume of the solid?

y² = 1/(2x + 1)
V = π ∫ (0 to 4) 1/(2x + 1) dx
  = π [ ½ ln(2x + 1) ] (0 to 4)
  = π (½ ln 9 − ½ ln 1)
  = π ln 3

Here ½ ln 9 = ln 3, because ½ ln 9 = ln 9^(½).

V = π ln 3

A region bounded by a curve and a line

If the region lies between a curve and a horizontal line, rotate the outer boundary and subtract the inner one. For a line y = k that is a cylinder of radius k, so subtract πk² times its length.

Worked example 3

The curve y = 6/x meets the line y = 3 at the point (2, 3). The finite region R is enclosed by the curve, the line y = 3 and the line x = 1. R is spun a full turn around the x-axis. Find the volume of the solid.

Curve above the line for 1 ≤ x ≤ 2.
Volume from the curve:  π ∫ (1 to 2) 36/x² dx = π [−36/x] (1 to 2)
                        = π (−18 + 36) = 18π
Cylinder: radius 3, length 1:  π × 3² × 1 = 9π
V = 18π − 9π

V = 9π

6.1 Volume from parametric equations

If x = x(t) and y = y(t), then dx = (dx/dt) dt, so

V = π ∫ y² (dx/dt) dt, with limits in t.

Three steps: differentiate x, write y² in terms of t, and change the limits from x-values to t-values.

Worked example 4

A curve has parametric equations x = 2t, y = sin t. Find the exact volume of the solid formed by spinning the region under the curve, for 0 ≤ t ≤ π/2, a full turn around the x-axis.

dx/dt = 2
V = π ∫ (0 to π/2) sin²t × 2 dt
  = 2π ∫ (0 to π/2) ½ (1 − cos 2t) dt        [cos 2t = 1 − 2sin²t]
  = π [ t − ½ sin 2t ] (0 to π/2)
  = π (π/2 − 0)

V = π²/2

Worked example 5

A curve has equations x = t², y = t + 2, where t > 0. The region under the curve between x = 1 and x = 4 is spun a full turn around the x-axis. Find the volume of the solid.

x = 1 gives t = 1; x = 4 gives t = 2.
dx/dt = 2t
V = π ∫ (1 to 2) (t + 2)² × 2t dt
  = 2π ∫ (1 to 2) (t³ + 4t² + 4t) dt
  = 2π [ t⁴/4 + 4t³/3 + 2t² ] (1 to 2)
  = 2π (68/3 − 43/12)
  = 2π × 229/12

V = 229π/6

If dx/dt is negative on part of the range, x runs backwards as t increases. Keep the limits in the order that matches x increasing. Reversing them gives a negative volume.

6.4 Separable differential equations

A first-order equation is separable if dy/dx = f(x) g(y). Move all the y terms to one side and all the x terms to the other, then integrate both sides:

∫ 1/g(y) dy = ∫ f(x) dx

Put one constant c on one side only. The result with c is the general solution. Substituting a given pair of values to find c gives the particular solution.

Worked example 6

Solve dy/dx = 3x²y, given that y = 2 when x = 0. Assume y > 0.

∫ 1/y dy = ∫ 3x² dx
ln y = x³ + c
y = e^(x³ + c) = A e^(x³)        where A = e^c
x = 0, y = 2:  2 = A

y = 2e^(x³)

For general solutions with ln|y| you can write y = A e^(x³), where A is any non-zero constant.

Worked example 7

Solve dy/dx = e^(x − y), given that y = ln 2 when x = 0. Give y in terms of x.

dy/dx = e^x × e^(−y)
e^y dy = e^x dx
e^y = e^x + c
x = 0, y = ln 2:  2 = 1 + c, so c = 1
e^y = e^x + 1

y = ln(e^x + 1)

Worked example 8

For x > 0, solve x dy/dx = 2y + 1, given that y = 2 when x = 1.

dy/dx = (2y + 1)/x
∫ 1/(2y + 1) dy = ∫ 1/x dx
½ ln(2y + 1) = ln x + c
x = 1, y = 2:  ½ ln 5 = c
ln(2y + 1) = 2 ln x + ln 5 = ln(5x²)
2y + 1 = 5x²

y = (5x² − 1)/2

Check: at x = 1 this gives y = 2, and x dy/dx = 5x² = 2y + 1.

Worked example 9 (a model)

A crystal grows so that its radius r mm at time t hours satisfies dr/dt = k/r², where k > 0. At t = 0, r = 1, and at t = 7, r = 2. Find k, and the time when r = 3.

∫ r² dr = ∫ k dt
r³/3 = kt + c
t = 0, r = 1:  1/3 = c
t = 7, r = 2:  8/3 = 7k + 1/3, so k = 1/3
r = 3:  9 = t/3 + 1/3, so t = 26

k = 1/3 and t = 26 hours

6.5 Area from parametric equations

The area under a curve is ∫ y dx. With a parameter,

Area = ∫ y (dx/dt) dt, with limits in t.

Find the limits by solving for t at each given x-value, or at each point where y = 0 if the curve meets the axis.

Worked example 10

A curve has parametric equations x = t², y = t(6 − t) for t ≥ 0. Work out the area enclosed by the curve, the x-axis, and the ordinates at t = 1 and t = 3 (that is, x = 1 and x = 9).

dx/dt = 2t
Area = ∫ (1 to 3) t(6 − t) × 2t dt
     = 2 ∫ (1 to 3) (6t² − t³) dt
     = 2 [ 2t³ − t⁴/4 ] (1 to 3)
     = 2 (33.75 − 1.75)

Area = 64

Worked example 11

The curve with equations x = t² + 2t, y = 3t (t > 0) bounds a region with the x-axis, between the lines x = 3 and x = 8. Calculate its area.

t² + 2t = 3 gives t = 1;  t² + 2t = 8 gives t = 2.
dx/dt = 2t + 2
Area = ∫ (1 to 2) 3t (2t + 2) dt = 3 ∫ (1 to 2) (2t² + 2t) dt
     = 3 [ 2t³/3 + t² ] (1 to 2)
     = 3 (28/3 − 5/3)

Area = 23

Worked example 12 (x decreasing)

A curve has equations x = 5 cos t, y = 2 sin t, 0 ≤ t ≤ π. Find the area between the curve and the x-axis.

x runs from 5 (at t = 0) to −5 (at t = π). Integrate from the left-hand end, which is t = π.

dx/dt = −5 sin t
Area = ∫ (t = π to 0) 2 sin t × (−5 sin t) dt
     = 10 ∫ (0 to π) sin²t dt
     = 5 [ t − ½ sin 2t ] (0 to π)

Area = 5π

This is half the area of an ellipse with semi-axes 5 and 2, which is ½ × π × 5 × 2 = 5π.

Common errors

• Forgetting to square y, or squaring only part of it, such as writing (t + 2)² as t² + 4.
• Using π∫x² dy. The specification asks for π∫y² dx.
• Leaving limits as x-values after changing to t.
• Forgetting dx/dt, so integrating y dt in place of y (dx/dt) dt.
• Dividing by y without noting that y = 0 may be a solution.
• Putting a constant on both sides of a separated equation.
• Applying the initial condition before the constant has been introduced.
• Losing a negative sign when dx/dt is negative.

Next steps

Try the practice questions, then condense with the revision notes. For an overview of all papers, see IAL Mathematics exam preparation.

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.