Question

Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y = e^x, and the line x = \ln(2 ) about the line x = \ln(2 ). Answer = 2pi(2ln(2)-1)

Answer 1

Using the shell method, set up the integral as

`\displaystyle2\pi\int_0^(\ln2)e^x(\ln 2-x)\,\mathrm dx`

For any given
`x` along the interval
`[0,\ln2]`, the corresponding shell has a height of
`e^x`, and the radius is given by the distance between
`x` and the axis of revolution
`x=\ln2`, or
`|x-\ln2|`. For
`0\le x\le\ln2`, we have
`|x-\ln2|=\ln2-x`.

Integrate by parts, taking

`u=\ln2-x\implies\mathrm du=-\mathrm dx`

`\mathrm dv=e^x\,\mathrm dx\implies v=e^x`

`\displaystyle2\pi\int_0^(\ln2)e^x(\ln2-x)\,\mathrm dx=2\pi\left(e^x(\ln2-x)\bigg|_0^(\ln2)+\int_0^(\ln2)e^x\,\mathrm dx\right)`

`=2\pi e^x(1+\ln2-x)\bigg|_0^(\ln2)`

`=2\pi\left(e^(\ln2)(1+\ln2-\ln2)-e^0(1+\ln2-0)\right)`

`=2\pi(1-\ln2)`