Areas and volumes with parametric equations

Area

Area = ∫ y dx = ∫ y (dx/dt) dt.

Replace y with its formula in t, replace dx with (dx/dt) dt, and change the x-limits into t-limits. You do not need to sketch the curve.

Volume

Volume about the x-axis = π ∫ y² dx = π ∫ y² (dx/dt) dt.

A common error is to replace dx with dt without the factor dx/dt.

Limits

Find the t-values that give the x-limits. If x decreases as t increases (for example x = cos t), the lower x-limit may correspond to the larger t-value. Keep the t-limits matched to the x-limits and the sign takes care of itself.

Worked example

x = t², y = 2t for t ≥ 0. Find the area under the curve from t = 0 to t = 2.

1. dx/dt = 2t
2. Area = ∫₀^{2} 2t × 2t dt = ∫₀^{2} 4t² dt
3. = 4(8/3)

Answer: 32/3

Check yourself

1. Which integral gives the area under a curve defined by x = f(t), y = g(t)?
   • ∫ y dt
   • ∫ x (dy/dt) dt
   • ∫ y (dx/dt) dt
   • ∫ (dy/dx) dt
   Show answer

   ∫ y (dx/dt) dt. dx = (dx/dt) dt, so ∫ y dx = ∫ y (dx/dt) dt.

2. x = t², y = t³. Find the area under the curve from t = 0 to t = 1. Give your answer as a fraction in its simplest form.
   Show answer

   2/5. dx/dt = 2t. Area = ∫₀^{1} t³ × 2t dt = ∫₀^{1} 2t⁴ dt = 2/5.

3. A curve has parametric equations x = 2t, y = t² + 1. Find the area between the curve, the x-axis and the lines x = 0 and x = 4. Give your answer as a fraction in its simplest form.
   Show answer

   28/3. x = 0 gives t = 0 and x = 4 gives t = 2. Area = ∫₀^{2} 2(t² + 1) dt = 2(8/3 + 2) = 28/3.

4. x = t² with t > 0. The area is wanted between x = 1 and x = 9. What are the t-limits?
   • t = 1 and t = 81
   • t = 1 and t = 3
   • t = 1 and t = 9
   • t = 0 and t = 3
   Show answer

   t = 1 and t = 3. t² = 1 gives t = 1 and t² = 9 gives t = 3.

5. x = t², y = 2t. The region under the curve from t = 0 to t = 1 is rotated 360° about the x-axis. Find the volume, to 3 decimal places.
   Show answer

   6.283. V = π ∫₀^{1} (2t)² × 2t dt = π ∫₀^{1} 8t³ dt = 2π = 6.283.

6. x = cos t, y = sin t for 0 ≤ t ≤ π/2. Find the area under the curve, to 3 decimal places.
   Show answer

   0.785. dx/dt = -sin t. x = 0 at t = π/2 and x = 1 at t = 0, so Area = ∫_π/2^{0} sin t(-sin t) dt = ∫₀^{π/2} sin² t dt = π/4 = 0.785.

7. For x = cos t, y = sin t, why does the area integral run from t = π/2 to t = 0?
   • Integrals must always go from larger to smaller values
   • sin t is negative between 0 and π/2
   • The curve is not continuous
   • x = 0 when t = π/2 and x = 1 when t = 0, so the t-limits follow the x-limits
   Show answer

   x = 0 when t = π/2 and x = 1 when t = 0, so the t-limits follow the x-limits. The lower x-limit (0) corresponds to t = π/2.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 4 scheme of work

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