Answer
(a)
(i) the area of the surface obtained by rotating the curve about the x-axis
$$
S=\int 2 \pi y d s=\int_{0}^{2} 2 \pi \tan^{-1}(x ) \sqrt{1+\left(\frac{1}{1+x^{2}}\right)^{2}} dx
$$
(ii) the area of the surface obtained by rotating the curve about the y-axis
$$
S=\int 2 \pi x d s=\int_{0}^{2} 2 \pi x \sqrt{1+\left(\frac{1}{1+x^{2}}\right)^{2}} dx
$$
(b) the surface areas, correct to four decimal places, by using the numerical integration, are
(i) 9.7956
(ii) 13.7209
Work Step by Step
$$
y= \tan^{-1} x, \quad \quad 0 \leq x \leq 2
$$
$\Rightarrow$
$$
\frac{dy}{dx} = \frac{1}{1+x^{2}}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(\frac{dy}{dx} \right)^{2}} dx \\
&=\sqrt{1+\left(\frac{1}{1+x^{2}}\right)^{2}} dx
\end{aligned}
$$
(a)
(i) the area of the surface obtained by rotating the curve about the x-axis
$$
S=\int 2 \pi y d s=\int_{0}^{2} 2 \pi \tan^{-1}(x ) \sqrt{1+\left(\frac{1}{1+x^{2}}\right)^{2}} dx
$$
(ii) the area of the surface obtained by rotating the curve about the y-axis
$$
S=\int 2 \pi x d s=\int_{0}^{2} 2 \pi x \sqrt{1+\left(\frac{1}{1+x^{2}}\right)^{2}} dx
$$
(b) the surface areas, correct to four decimal places, by using the numerical integration, are
(i) 9.7956
(ii) 13.7209
