2021HSCShort answer

Q13 (c) (ii)

4 marks

Question

Region $A$ is bounded by $y=1$ and $x^2+y^2=1$ between $x=0$ and $x=1$ . Region $B$ is bounded by $y=1$ and $y=\ln x$ between $x=1$ and $x=e$ . Using part (i), or otherwise, show that the ratio of the volumes formed by rotating regions $A$ and $B$ about the $x$ -axis is $V_A:V_B=1:3$ .

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Solution

Criteria	Marks
Provides correct solution	4
Shows $V_B=\pi$ , or equivalent	3
Obtains volume $V_A$ OR writes volume $V_A$ or $V_B$ as a difference of volumes involving an integral expression	2
States volume of a relevant sphere OR hemisphere OR cylinder OR provides correct integral expression for a relevant volume	1

Sample answer:

$V_A=\pi\int_0^1\left((1)^2-(1-x^2)\right)dx=\pi\int_0^1x^2dx=\frac\pi3$

$V_B=\pi\int_1^e\left((1)^2-(\ln x)^2\right)dx=\pi(e-1)-\pi I_2$

$I_2=e-2I_1,\qquad I_1=e-I_0,\qquad I_0=\int_1^e1\,dx=e-1$
$I_2=e-2[e-(e-1)]=e-2$
$V_B=\pi(e-1)-\pi(e-2)=\pi$
$\therefore V_A:V_B=\frac\pi3:\pi=1:3$

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Syllabus alignment

Calculus

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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