IB Maths AA HL 5.17 Areas and volumes of revolution

[Maximum number: 6]

Consider the function defined by $f(x)=\frac{k x-5}{x-k}$ , where $x \in \mathbb{R} \backslash\{k\}$ and $k^{2} \neq 5$ .

The region bounded by the x-axis, the curve y=f(x), and the lines x=5 and x=7 is rotated through $2 \pi$ about the x-axis. Find the volume of the solid generated, giving your answer in the form $\pi(a+b \ln 2)$ , where $a, b \in \mathbb{Z}$ .

METHOD 1
volume $=\pi \int_{5}^{7}\left(\frac{3 x-5}{x-3}\right)^{2} d x$

EITHER
attempt to express $\frac{3 x-5}{x-3}$ in the form $p+\frac{q}{x-3}$ $\frac{3 x-5}{x-3}=3+\frac{4}{x-3}$

OR
attempt to expand $\left(\frac{3 x-5}{x-3}\right)^{2}$ or $(3 x-5)^{2}$ and divide out
$\left(\frac{3 x-5}{x-3}\right)^{2}=9+\frac{24 x-56}{(x-3)^{2}}$

THEN

\begin{array}{ll} \left(\frac{3 x-5}{x-3}\right)^{2}=9+\frac{24}{x-3}+\frac{16}{(x-3)^{2}} & \boldsymbol{A} \mathbf{1} \\ \text { volume }=\pi \int_{5}^{7}\left(9+\frac{24}{x-3}+\frac{16}{(x-3)^{2}}\right) \mathrm{d} x \\ =\pi\left[9 x+24 \ln (x-3)-\frac{16}{x-3}\right]_{5}^{7} \\ =\pi[(63+24 \ln 4-4)-(45+24 \ln 2-8)] & \boldsymbol{A} \mathbf{1} \\ =\pi(22+24 \ln 2) & \boldsymbol{A} \mathbf{1} \end{array}

METHOD 2

\begin{aligned} & \text { volume }=\pi \int_{5}^{7}\left(\frac{3 x-5}{x-3}\right)^{2} \mathrm{~d} x \\ & \text { substituting } u=x-3 \Rightarrow \frac{\mathrm{~d} u}{\mathrm{~d} x}=1 \\ & \begin{aligned} 3 x-5 & =3(u+3)-5=3 u+4 \\ \text { volume } & =\pi \int_{2}^{4}\left(\frac{3 u+4}{u}\right)^{2} \mathrm{~d} u \\ = & \pi \int_{2}^{4} 9+\frac{16}{u^{2}}+\frac{24}{u} \mathrm{~d} u \\ = & \pi\left[9 u-\frac{16}{u}+24 \ln u\right]_{2}^{4} \end{aligned} \end{aligned}

Note: Ignore absence of or incorrect limits seen up to this point.

IB Maths AA HL Ahl 5 17 Hl Areas and Volumes of Revolution Questions

Exam points

IB Maths AA HL Ahl 5 17 Hl Areas and Volumes of Revolution Questions question 1