2023HSCShort answer

Q12 (e)

4 marks

Question

The region, $R$ , bounded by the hyperbola

y=\frac{60}{x+5},

the line $x=10$ and the coordinate axes is shown. [Diagram of the shaded region and rotation about the $y$ -axis.]

Find the volume of the solid of revolution formed when the region $R$ is rotated about the $y$ -axis. Leave your answer in exact form.

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	4
Obtains the volume formed by revolving the hyperbola about the $y$ -axis, or equivalent merit	3
Obtains correct integral expression for the volume formed by revolving the hyperbola about the $y$ -axis, or equivalent merit	2
Writes $x$ in terms of $y$ , or recognises that the volume is the sum of two simpler volumes, or equivalent merit	1

Sample answer:

The total volume is the sum of the volume when the given hyperbola is revolved about the $y$ -axis, $V_1$ , and the volume of a cylinder, $V_2$ .

\begin{aligned} V_1&=\pi\int_4^{12}x^2\,dy,\qquad y=\frac{60}{x+5}\\ x+5&=\frac{60}{y}\\ x&=\frac{60}{y}-5\\ V_1&=\pi\int_4^{12}\left(\frac{60}{y}-5\right)^2dy\\ &=\pi\int_4^{12}\left(\frac{3600}{y^2}-\frac{600}{y}+25\right)dy\\ &=\pi\left[-\frac{3600}{y}-600\ln y+25y\right]_4^{12}\\ &=\pi(800-600\ln3) \end{aligned}

\begin{aligned} V_2&=\pi\times10^2\times4=400\pi\\ \text{Total}&=V_1+V_2\\ \therefore\quad\text{Total}&=\pi(1200-600\ln3)\text{ units}^3 \end{aligned}

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Syllabus alignment

Calculus

Applications of CalculusApply calculus to areas between curves, volumes of solids of revolution and differential equation models.

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