Area Bounded by a Curve and the y-axis

Area Bounded by a Curve and the y-axis

Secondary 4
TGM Original Questions

Regions can also be measured against the \(y\)-axis. The roles of \(x\) and \(y\) swap: make \(x\) the subject, and integrate with respect to \(y\) between \(y\)-limits.

Method
  1. Express the curve as \(x\) in terms of \(y\).
  2. Read the limits off the \(y\)-axis.
  3. Evaluate \(\displaystyle\int_{c}^{d} x\,\mathrm{d}y\).
  4. Right of the axis the integral is positive; left of the axis it is negative — state the area as its magnitude.

Right of the axis: \(A=\displaystyle\int_c^d x\,\mathrm{d}y\)

Left of the axis: \(\displaystyle\int_c^d x\,\mathrm{d}y\) is negative, so \(A=-\displaystyle\int_c^d x\,\mathrm{d}y\)

Matching the axis to the variable

Boundary axisIntegralLimits
\(x\)-axis\(\displaystyle\int y\,\mathrm{d}x\) with \(y\) in terms of \(x\)\(x\)-values
\(y\)-axis\(\displaystyle\int x\,\mathrm{d}y\) with \(x\) in terms of \(y\)\(y\)-values
Exam check

Integrating with respect to \(y\) needs everything in \(y\): make \(x\) the subject before integrating, and take the limits from the \(y\)-axis, not the \(x\)-axis.

For total area, split the vertical interval wherever \(g(y)\) changes sign and add the magnitudes. More generally, integrate right boundary minus left boundary on intervals where their ordering is fixed.

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