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Pure · Integration

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Areas under parametric curves

Finding the area under a curve given parametrically as ∫ y (dx/dt) dt with the limits in t — and combining it with tangents/normals to find regions bounded by the curve, a line and an axis.

Pure · Integration

Areas under parametric curves

Finding the area under a curve given parametrically as ∫ y (dx/dt) dt with the limits in t — and combining it with tangents/normals to find regions bounded by the curve, a line and an axis.

Why it works

The area under a curve is ∫y dx\displaystyle\int y\,dx. When the curve is given parametrically — x=f(t)x = f(t), y=g(t)y = g(t) — both yy and dxdx need to be put in terms of tt before you can integrate. Since dx=dxdt dtdx = \dfrac{dx}{dt}\,dt, the area becomes ∫y dx=∫t1t2y dxdt dt,\int y\,dx = \int_{t_1}^{t_2} y\,\frac{dx}{dt}\,dt, where t1t_1 and t2t_2 are the parameter values at the two ends of the region — not the xx-values. This is just substitution: you replace dxdx by dxdt dt\dfrac{dx}{dt}\,dt and change the limits from xx to tt.123452468xyTake the curve x=t2, y=4tx = t^2,\ y = 4t (for t≥0t \ge 0). To find the area between it and the xx-axis from x=0x = 0 to x=4x = 4, note these are t=0t = 0 and t=2t = 2. With dxdt=2t\dfrac{dx}{dt} = 2t, ∫y dxdt dt=∫024t⋅2t dt=∫028t2 dt=[83t3]02=643.\int y\,\frac{dx}{dt}\,dt = \int_0^2 4t\cdot 2t\,dt = \int_0^2 8t^2\,dt = \left[\tfrac83 t^3\right]_0^2 = \frac{64}{3}.

A few things to watch:
  • Limits are tt-values. Convert each xx-limit to its tt using x=f(t)x = f(t) before
integrating; forgetting this is the commonest slip.
  • Keep the dxdt\dfrac{dx}{dt} factor. Integrating ∫y dt\int y\,dt without it gives the
wrong thing entirely — it isn't an area.
  • Watch the sign. If xx decreases as tt increases (so dxdt<0\dfrac{dx}{dt} < 0) the
integral comes out negative; take the modulus for the area, as usual.

Bringing in a tangent or normal. The hardest exam questions bound a region with a curve, a straight line (a tangent or normal), and an axis. The trick is to split the region into the part under the curve (a parametric-area integral) and a triangle under the line, and add or subtract them. Find the line with parametric differentiation (dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}), find where it meets the axis, then assemble the pieces.