Answer
(a)
(i) the area of the surface obtained by rotating the curve about the x-axis
$$
S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi y \sqrt {1+\left(\frac{2}{2y+1}\right)^{2}} dy.
$$
(ii) the area of the surface obtained by rotating the curve about the y-axis
$$
S=\int 2 \pi x d s=\int_{0}^{1} 2 \pi \ln (2y+1) \sqrt {1+\left(\frac{2}{2y+1}\right)^{2}} dy.
$$
(b) the surface areas, correct to four decimal places, by using the numerical integration, are
(i) 4.2583
(ii) 5.6053
Work Step by Step
$$
x= \ln (2y+1), \quad \quad 0 \leq y \leq 1
$$
$\Rightarrow$
$$
\frac{dx}{dy} =\frac{2}{2y+1}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(\frac{dx}{dy}\right)^{2}} dy\\
&=\sqrt {1+\left(\frac{2}{2y+1}\right)^{2}} dy
\end{aligned}
$$
(a)
(i) the area of the surface obtained by rotating the curve about the x-axis
$$
S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi y \sqrt {1+\left(\frac{2}{2y+1}\right)^{2}} dy.
$$
(ii) the area of the surface obtained by rotating the curve about the y-axis
$$
S=\int 2 \pi x d s=\int_{0}^{1} 2 \pi \ln (2y+1) \sqrt {1+\left(\frac{2}{2y+1}\right)^{2}} dy.
$$
(b) the surface areas, correct to four decimal places, by using the numerical integration, are
(i) 4.2583
(ii) 5.6053