Applications of Integration: Area; vol. \(x\)-axis; parabola — VJC 2025 H2 Math Prelim Paper 1
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Question
In the diagram below, the region \(R\) is bounded by the curve \(C\) with equation \(x = 6 - (y-2)^2\), the lines \(y = 8\), \(y = 2 - x\) and the \(y\)-axis. The region \(S\) is bounded by \(C\) and the line \(y = 2 - x\).

(a) Find the exact area of region \(R\).
(b) Find the volume of the solid of revolution formed when region \(S\) is rotated through \(360^\circ\) about the \(x\)-axis, leaving your answer to 2 decimal places.
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(a) Intersection of \(C\) and \(y = 2 - x\): \[\begin{aligned} 2 - y &= 6 - (y-2)^2\\ 2 - y &= 6 - y^2 + 4y - 4\\ y^2 - 5y &= 0 \end{aligned}\] giving \(y = 0\) or \(y = 5\), so \(x = 2\) or \(x = -3\).
Intersection of \(C\) with the \(y\)-axis (\(x = 0\)): \[ 0 = 6 - (y-2)^2 \;\Longrightarrow\; y = 2 \pm \sqrt{6}. \] For region \(R\) (bounded by \(C\), \(y = 8\), \(y = 2 - x\), \(y\)-axis), integrate w.r.t. \(y\): \[\begin{aligned} \text{Area of } R &= -\!\int_{2+\sqrt{6}}^{5}\!\!\bigl[6 - (y-2)^2\bigr]\,\mathrm{d}y \;-\; \int_{5}^{8}(2 - y)\,\mathrm{d}y\\ &= -\!\left[6y - \frac{(y-2)^3}{3}\right]_{2+\sqrt{6}}^{5} \;-\; \left[2y - \frac{y^2}{2}\right]_{5}^{8}. \end{aligned}\] Evaluating the first bracket at the endpoints: \[\begin{aligned} \left[6y - \tfrac{(y-2)^3}{3}\right]_{2+\sqrt{6}}^{5} &= \left(30 - \tfrac{27}{3}\right) - \left(6(2+\sqrt{6}) - \tfrac{(\sqrt{6})^3}{3}\right)\\ &= (30 - 9) - (12 + 6\sqrt{6} - 2\sqrt{6})\\ &= 21 - 12 - 4\sqrt{6} = 9 - 4\sqrt{6}. \end{aligned}\] The second bracket: \(\left[2y - \tfrac{y^2}{2}\right]_{5}^{8} = (16 - 32) - (10 - 12.5) = -16 + 2.5 = -13.5\).
Combining (with the negative sign): \[ \text{Area of }R = -(9 - 4\sqrt{6}) - (-13.5) = -9 + 4\sqrt{6} + 13.5 = \tfrac{9}{2} + 4\sqrt{6}. \] (b)
\[\begin{aligned} x &= 6-(y-2)^2 \\ y &= 2 \pm \sqrt{6-x} \end{aligned}\] \[\begin{aligned} \text{Volume of solid} &= \pi\int_{-3}^{6}\!\left(2+\sqrt{6-x}\right)^2\mathrm{d}x - \pi\int_{-3}^{2}\!\left(2-x\right)^2\mathrm{d}x - \pi\int_{2}^{6}\!\left(2-\sqrt{6-x}\right)^2\mathrm{d}x \\ &= 327.25 \text{ units}^3 \end{aligned}\]