Change the lower boundary at the intersection · source reference

Printed question

Curve $C_1$: $y=1/x$; curve $C_2$: $y=1/\sqrt{2-x^2}$. Region Q is bounded by $x=1/\sqrt2$ and both curves.

(a) Find the exact area of Q. [3]

(b) The region bounded by $y=5$ and both curves in the first quadrant is rotated about the $x$-axis by $2\pi$ radians. Find the volume of the solid. [3]

[2021/ASRJC/Prelim/I/4]

Printed working

(a)

Let $1/x=1/\sqrt{2-x^2}$. Then $x^2=2-x^2$, so $x=1$ since $x>0$.

$$\begin{aligned}\text{Area Q}&=\int_{1/\sqrt2}^1\frac1x dx-\int_{1/\sqrt2}^1\frac1{\sqrt{2-x^2}}dx\\&=[\ln x-\sin^{-1}(x/\sqrt2)]_{1/\sqrt2}^1\\&=(\ln1-\sin^{-1}(1/\sqrt2))-(\ln(1/\sqrt2)-\sin^{-1}(1/2))\\&=0-\pi/4-\ln(1/\sqrt2)+\pi/6=\frac12\ln2-\frac\pi{12}\quad\text{units}^2.\end{aligned}$$

Accept $\ln\sqrt2-\pi/12$ and $-\frac12\ln(1/2)-\pi/12$.

(b)

Let $5=1/x$ and $5=1/\sqrt{2-x^2}$. From GC $x=0.2$ and $x=1.4$ since $x>0$.

$$\begin{aligned}V&=\text{volume of cylinder}-\pi\int_{0.2}^1(1/x)^2dx-\pi\int_1^{1.4}(1/\sqrt{2-x^2})^2dx\\&=\pi(5)^2(1.4-0.2)-12.566-3.9158\\&=77.8\quad\text{units}^3\quad\text{(3 s.f.)}.\end{aligned}$$

Teacher notes

Area Q $=\int_{1/\sqrt2}^1(y_1-y_2)dx=\int_{1/\sqrt2}^1(f(x)-g(x))dx$, with $f$ above or intersecting $g$ on $[1/\sqrt2,1]$.

The black rectangular region rotating about the $x$-axis is a cylinder of radius $5$ units and height $(1.4-0.2)$ units. Subtract the blue region below $1/x$ on $[0.2,1]$ and red region below $1/\sqrt{2-x^2}$ on $[1,1.4]$. All three subtraction diagrams and their rotation arrows are retained as native panels.

Printed answer key

(a) $\frac12\ln2-\pi/12$ units$^2$; (b) $77.8$ units$^3$.

Source diagrams in native form