Question

Find the area of the surface generated by revolving the curve xequalsStartFraction e Superscript y Baseline plus e Superscript negative y Over 2 EndFraction in the interval 0 less than or equals y less than or equals ln 5 about the​ y-axis.

Answer 1

The area is given by the integral,

$\displaystyle\int_0^{\ln5}2\pi x(y)\sqrt{1+\left(\dfrac{\mathrm dx}{\mathrm dy}\right)^2}\,\mathrm dy$

We have

$x=\dfrac{e^y+e^{-y}}2\implies\dfrac{\mathrm dx}{\mathrm dy}=\dfrac{e^y-e^{-y}}2$

So now compute the integral:

$\displaystyle\frac\pi2\int_0^{\ln5}(e^y+e^{-y})\sqrt{4+(e^y-e^{-y})^2}\,\mathrm dy$

Substitute $u=e^y-e^{-y}$ and $\mathrm du=(e^y+e^{-y})\,\mathrm dy$:

$\displaystyle\frac\pi2\int_0^{\frac{24}5}\sqrt{4+u^2}\,\mathrm du$

Another substitution, $u=2\tan v$ and $\mathrm dv=2\sec^2v\,\mathrm dv$:

$\displaystyle\frac\pi2\int_0^{\tan^{-1}\frac{12}5}\sqrt{4+(2\tan v)^2}\,2\sec^2v\,\mathrm dv$

$\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sqrt{1+\tan^2v}\,\sec^2v\,\mathrm dv$

$\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sec^3v\,\mathrm dv$

There's a well-known formula for the integral of secant cubed, but if you don't know it off the top of your head (like me), you can integrate by parts:

$\displaystyle I=\int\sec^3v\,\mathrm dv=\sec v\tan v-\int\sec v\tan^2v\,\mathrm dv$

Expand the remaining the integral in terms of powers of secant:

$\displaystyle\int\sec v\tan^2v\,\mathrm dv=\int\sec v(\sec^2v-1)\,\mathrm dv=\int\sec^3v\,\mathrm dv-\int\sec v\,\mathrm dv$

so that

$I=\sec v\tan v-\left(I-\displaystyle\int\sec v\,\mathrm dv\right)$

$2I=\sec v\tan v+\displaystyle\int\sec v\,\mathrm dv$

$\implies I=\displaystyle\int\sec^3v\,\mathrm dv=\frac{\sec v\tan v}2+\frac12\ln|\sec v+\tan v|+C$

Coming back to the area integral, we use the formula above to get

$\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sec^3v\,\mathrm dv=\pi\left(\sec v\tan v+\ln|\sec v+\tan v|\right)\bigg|_0^{\tan^{-1}\frac{12}5}$

Next,

$\tan\left(\tan^{-1}\dfrac{12}5\right)=\dfrac{12}5$

$\sec\left(\tan^{-1}\dfrac{12}5\right)=\dfrac{13}5$

$\tan0=0$

$\sec0=1$

so the area is

$\pi\left(\dfrac{13}5\cdot\dfrac{12}5+\ln\left(\dfrac{13}5+\dfrac{12}5\right)-1\cdot0-\ln(1+0)\right)=\boxed{\left(\dfrac{156}{25}+\ln5\right)\pi}$