Applications of integration: NJC 2024 Promo Q2

National Junior College2024 PromoPromo Q2●○○ Foundational5 marks

What this question tests

Areas and volumes by integration.

Question

The region bounded by $y = x\sqrt{\sin x}$ , the $x$ -axis, and the lines $x = 0$ and $x = \pi$ is rotated through $2\pi$ radians about the $x$ -axis. Find the exact volume of the solid generated. [5]

Show full worked solution▾

(Applications of Integration)

Suggested Solution

$\begin{aligned} \int y^2\,\mathrm{d} x &=\int x^2\sin x\,\mathrm{d} x\\ &=-x^2\cos x-\int2x(-\cos x)\,\mathrm{d} x\\ &=-x^2\cos x+2\int x(\cos x)\,\mathrm{d} x\\ &=-x^2\cos x+2\left[x\sin x-\int\sin x\,\mathrm{d} x\right]\\ &=-x^2\cos x+2[x\sin x+\cos x]\\ &=-x^2\cos x+2x\sin x+2\cos x\\ &=(2-x^2)\cos x+2x\sin x+c. \end{aligned}$ Volume required $\begin{aligned} &=\pi\int_0^\pi y^2\,\mathrm{d} x\\ &=\pi\left[(2-x^2)\cos x+2x\sin x\right]_0^\pi\\ &=\pi[(2-\pi^2)(-1)-(2)1]\\ &=\pi^3-4\pi. \end{aligned}$

Answers

Volume $= \pi^3 - 4\pi$

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