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Applications of integration: RVHS 2023 Promo Q7

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River Valley High School2023 PromoPromo Q7●●○ Standard8 marks

What this question tests

Areas and volumes by integration.

Question

Without the use of a calculator, show that $\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\tan x|\,\mathrm{d} x = \ln 2$ . [3]
Hence, find the exact volume of the solid formed from completely rotating region bounded by the graph of $y = |\tan x| + 1$ , the lines $x = -\dfrac{\pi}{4}$ and $x = \dfrac{\pi}{4}$ , about the $x$ -axis. [5]

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(i)

$River Valley High School 2023 Promo Q7: solution diagram$

$\begin{aligned} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\tan x|\,\mathrm{d} x &= \int_{-\frac{\pi}{4}}^{0} -\tan x\,\mathrm{d} x + \int_{0}^{\frac{\pi}{4}} \tan x\,\mathrm{d} x\\ &= 2\int_{0}^{\frac{\pi}{4}} \tan x\,\mathrm{d} x\\ &= 2\Big[\ln(\sec x)\Big]_{0}^{\frac{\pi}{4}}\\ &= 2\left(\ln\frac{1}{\cos\frac{\pi}{4}} - \ln\frac{1}{\cos 0}\right)\\ &= 2\left(\ln\sqrt{2} - \ln 1\right)\\ &= \ln 2. \end{aligned}$

(ii) $\begin{aligned} \text{Volume} &= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left(|\tan x| + 1\right)^2\,\mathrm{d} x\\ &= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\tan x|^2 + 2|\tan x| + 1\,\mathrm{d} x\\ &= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left(\tan^2 x + 1\right)\,\mathrm{d} x + \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 2|\tan x|\,\mathrm{d} x\\ &= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x\,\mathrm{d} x + 2\pi\ln 2\\ &= \pi\Big[\tan x\Big]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} + 2\pi\ln 2\\ &= \pi\left(\tan\frac{\pi}{4} - \tan\left(-\frac{\pi}{4}\right)\right) + 2\pi\ln 2\\ &= 2\pi(1 + \ln 2)\ \text{units}^3. \end{aligned}$

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