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

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