Applications of Integration: Hyperbola; area; vol. \(x\), \(y\) — NYJC 2025 H2 Math Prelim Paper 1
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Question
The curve \(C\) has equation \(x^2 - y^2 = 16\), where \(y \ge 0\). The line \(L\) has equation \(y = -\dfrac{1}{2}x + \dfrac{11}{2}\).
(a) Find the area enclosed by \(C\), \(L\) and the line \(x = 8\).
(b) For \(x > 0\), the region bounded by \(C\), \(L\) and the \(x\)-axis is rotated about the \(x\)-axis through \(2\pi\) radians. Find the exact volume generated.
(c) The region bounded by \(C\), the line \(x = 5\) and the \(x\)-axis is rotated about the \(y\)-axis through \(2\pi\) radians. Find the exact volume generated.
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\(x^2 - y^2 = 16 \implies y = \sqrt{x^2-16}\) (since \(y \ge 0\)).
(a) The line \(L\): \(y = -\dfrac{1}{2}x + \dfrac{11}{2}\) and curve \(C\) intersect when: \[ \sqrt{x^2-16} = -\tfrac{1}{2}x + \tfrac{11}{2} \implies x^2-16 = \tfrac{1}{4}(x-11)^2 \implies 3x^2+22x-121=0 \implies x=5 \text{ or } x=-\tfrac{121}{3} \] (taking \(x=5\) as the intersection on the branch \(y\ge0\); at \(x=8\), \(y=\sqrt{48}\) from \(C\) and \(y=\tfrac{3}{2}\) from \(L\)).
Area required: \[ = \int_5^8\!\left[\sqrt{x^2-16} - \left(-\tfrac{1}{2}x+\tfrac{11}{2}\right)\right]\mathrm{d}x = 8.4723 \approx 8.47\text{ units}^2 \]
(b) Volume (rotation about \(x\)-axis): the region under \(C\) (\(x\in[4,5]\)) plus the cone formed by \(L\) (\(x\in[5,11]\)): \[ V = \pi\int_4^5(x^2-16)\;\mathrm{d}x + \tfrac{1}{3}\pi(3)^2(6) = \pi\!\left[\frac{x^3}{3}-16x\right]_4^5 + 18\pi = \pi\!\left[-\frac{115}{3}+\frac{128}{3}\right]+18\pi = \frac{67}{3}\pi\text{ units}^3 \]
(c) Volume (rotation about \(y\)-axis, region bounded by \(C\), \(x=5\), \(x\)-axis): \[ V = \pi(5)^2(3) - \pi\int_0^3(y^2+16)\;\mathrm{d}y = 75\pi - \pi\!\left[\frac{y^3}{3}+16y\right]_0^3 = 75\pi - \pi(9+48) = 75\pi - 57\pi = 18\pi\text{ units}^3 \]