Rotate the region between a bell curve and a line · source reference

Printed question

(i) Sketch $y=2/[1+(x-2)^2]$ and $y=1$ on the same diagram, showing clearly the coordinates of the stationary point and intersection points. [2]

(ii) Find the exact area of region R enclosed by the curves and $x=1,x=3$. [3]

(iii) Find the volume generated when R is rotated $360^\circ$ about the $y$-axis. [4]

[2011/MI/Prelim/II/2]

Printed working

(i)

The stationary point is $(2,2)$; intersections $(1,1)$ and $(3,1)$; region R lies above $y=1$ and beneath the bell curve.

(ii)

$$\text{Area R}=\int_1^3\left[\frac2{1+(x-2)^2}-1\right]dx=\left[\frac21\tan^{-1}\frac{x-2}1-x\right]_1^3=(\pi/2-3)-(-\pi/2-1)=\pi-2.$$

(iii)

$$y=\frac2{1+(x-2)^2}\Rightarrow x=2\pm\sqrt{2/y-1}.$$ $$V=\pi\int_1^2(2+\sqrt{2/y-1})^2dy-\pi\int_1^2(2-\sqrt{2/y-1})^2dy=\pi\int_1^2 8\sqrt{2/y-1}dy=14.3.$$

Teacher notes

Area R $=\int_1^3(y_1-y_2)dx=\int_1^3(f(x)-g(x))dx$, where $y=f(x)$ is above or intersects $y=g(x)$ for $1\le x\le3$.

Volume (rotation about $y$-axis) $=\pi\int_c^d x^2dy$. The right branch $x=2+\sqrt{2/y-1}$ is the outer radius; the left branch $x=2-\sqrt{2/y-1}$ is the inner radius. The original diagram marks the level $y=2$, centre $x=2$, horizontal base $y=1$, and the rotation arrow.

Printed answer key

(i) labelled native graph below; (ii) $\pi-2$ units$^2$; (iii) $14.3$ units$^3$.

Source diagrams in native form