Answer: (a) \(A=2\pi\displaystyle \int_{4/3}^{12/5}\sqrt{1+u^2}\,du\).

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

For a curve \(y=f(x)\) rotated about the \(x\)-axis, the surface area is

\(A=2\pi\displaystyle \int y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\).

Here \(y=e^x\), so \(\dfrac{dy}{dx}=e^x\). Therefore

\(A=2\pi\displaystyle \int_{\ln(4/3)}^{\ln(12/5)} e^x\sqrt{1+e^{2x}}\,dx\).

(a) Let \(u=e^x\). Then \(du=e^x\,dx\), so \(dx=\dfrac{du}{u}\). The limits become:

• when \(x=\ln\dfrac{4}{3}\), \(u=\dfrac{4}{3}\),
• when \(x=\ln\dfrac{12}{5}\), \(u=\dfrac{12}{5}\).

So

\(A=2\pi\displaystyle \int_{4/3}^{12/5} u\sqrt{1+u^2}\cdot \frac{1}{u}\,du\)

and hence

\(A=2\pi\displaystyle \int_{4/3}^{12/5}\sqrt{1+u^2}\,du\).

(b) Now use \(u=\sinh v\). Then

\(du=\cosh v\,dv\),

and since \(1+\sinh^2 v=\cosh^2 v\),

\(\sqrt{1+u^2}=\sqrt{1+\sinh^2 v}=\cosh v\).

The limits change as follows:

• if \(u=\dfrac{4}{3}\), then \(\sinh v=\dfrac{4}{3}\). With \(\cosh v=\dfrac{5}{3}\), we get \(e^v=\cosh v+\sinh v=3\), so \(v=\ln 3\);
• if \(u=\dfrac{12}{5}\), then \(\sinh v=\dfrac{12}{5}\). With \(\cosh v=\dfrac{13}{5}\), we get \(e^v=\cosh v+\sinh v=5\), so \(v=\ln 5\).

Therefore

\(A=2\pi\displaystyle \int_{\ln 3}^{\ln 5} \cosh^2 v\,dv\).

Using \(2\cosh^2 v=\cosh 2v+1\),

\(A=\pi\displaystyle \int_{\ln 3}^{\ln 5}(\cosh 2v+1)\,dv\).

Integrating,

\(A=\pi\left[\frac{1}{2}\sinh 2v+v\right]_{\ln 3}^{\ln 5}\).

Now evaluate:

when \(v=\ln 5\),

\(\sinh 2v=\dfrac{e^{2v}-e^{-2v}}{2}=\dfrac{25-1/25}{2}=\dfrac{312}{25}\),

so \(\dfrac{1}{2}\sinh 2v=\dfrac{156}{25}\).

When \(v=\ln 3\),

\(\sinh 2v=\dfrac{9-1/9}{2}=\dfrac{40}{9}\),

so \(\dfrac{1}{2}\sinh 2v=\dfrac{20}{9}\).

Hence

\(A=\pi\left(\frac{156}{25}+\ln 5-\frac{20}{9}-\ln 3\right)\)

\(=\pi\left(\frac{156}{25}-\frac{20}{9}+\ln\frac{5}{3}\right)\).

Now

\(\frac{156}{25}-\frac{20}{9}=\frac{1404-500}{225}=\frac{904}{225}\).

Therefore

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