Applications of Integration: Area; vol. about \(y\)-axis — HCI 2025 H2 Math Prelim Paper 2

(a) Sketch of the region \(A\) bounded by \(y=\sqrt{x+1}\), \(y=\sqrt{7-2x}\), the \(x\)-axis and the \(y\)-axis. The two curves intersect at \((2,\sqrt{3})\).

Figure 1 — Region \(A\) (shaded). Curves meet at \((2,\sqrt{3})\).

Method 1: Integrate with respect to \(x\). \[\begin{aligned} \text{Area} &= \int_{0}^{2} y\,\mathrm{d}x + \int_{2}^{3.5} y\,\mathrm{d}x \\ &= \int_{0}^{2} \sqrt{x+1}\,\mathrm{d}x + \int_{2}^{3.5} \sqrt{7-2x}\,\mathrm{d}x \\ &= \left[\frac{2(x+1)^{3/2}}{3}\right]_{0}^{2} - \left[\frac{(7-2x)^{3/2}}{3}\right]_{2}^{3.5} \\ &= \left(\frac{2\sqrt{27}}{3}-\frac{2}{3}\right) - \left(-\frac{\sqrt{27}}{3}\right) \\ &= 2\sqrt{3}-\frac{2}{3}+\sqrt{3} \\ &= 3\sqrt{3}-\frac{2}{3}\ \text{units}^{2}. \end{aligned}\]

Method 2: Integrate with respect to \(y\). \[\begin{aligned} \text{Area} &= \int_{0}^{\sqrt{3}} x\,\mathrm{d}y - \int_{1}^{\sqrt{3}} x\,\mathrm{d}y \\ &= \int_{0}^{\sqrt{3}} \frac{7-y^{2}}{2}\,\mathrm{d}y - \int_{1}^{\sqrt{3}} (y^{2}-1)\,\mathrm{d}y \\ &= \left[\frac{7y}{2}-\frac{y^{3}}{6}\right]_{0}^{\sqrt{3}} - \left[\frac{y^{3}}{3}-y\right]_{1}^{\sqrt{3}} \\ &= \left(\frac{7\sqrt{3}}{2}-\frac{3\sqrt{3}}{6}\right) - \left[\left(\frac{3\sqrt{3}}{3}-\sqrt{3}\right)-\left(\frac{1}{3}-1\right)\right] \\ &= 3\sqrt{3}-\frac{2}{3}\ \text{units}^{2}. \end{aligned}\]

(b) From \(y=\sqrt{7-2x}\), \(\ x=\dfrac{7-y^{2}}{2}\). From \(y=\sqrt{x+1}\), \(\ x=y^{2}-1\). \[\begin{aligned} \text{Vol} &= \pi\int_{0}^{\sqrt{3}} \left(\frac{7-y^{2}}{2}\right)^{2}\,\mathrm{d}y - \pi\int_{1}^{\sqrt{3}} (y^{2}-1)^{2}\,\mathrm{d}y \\ &= 47.38326 \\ &\approx 47.4\ \text{units}^{3}\ (3\,\text{s.f.}) \end{aligned}\]