Problem

A curve has equation $y = e^x$ for $\ln\frac{4}{3} \le x \le \ln\frac{12}{5}$. The area of the surface generated when the curve is rotated through $2\pi$ radians about the $x$-axis is denoted by $A$.

(a) Use the substitution $u = e^x$ to show that

$$A = 2\pi\int_{\frac{4}{3}}^{\frac{12}{5}} \sqrt{1 + u^2}\,\mathrm{d}u.$$

(b) Use the substitution $u = \sinh v$ to show that

$$A = \pi\left(\frac{904}{225} + \ln\frac{5}{3}\right).$$