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)

Using the shell method, set up the integral as[tex]\displaystyle2\pi\int_0^{\ln2}e^x(\ln 2-x)\,\mathrm dx[/tex]For any given [tex]x[/tex] along the interval [tex][0,\ln2][/tex], the corresponding shell has a height of [tex]e^x[/tex], and the radius is given by the distance between [tex]x[/tex] and the axis of revolution [tex]x=\ln2[/tex], or [tex]|x-\ln2|[/tex]. For [tex]0\le x\le\ln2[/tex], we have [tex]|x-\ln2|=\ln2-x[/tex].Integrate by parts, taking[tex]u=\ln2-x\implies\mathrm du=-\mathrm dx[/tex][tex]\mathrm dv=e^x\,\mathrm dx\implies v=e^x[/tex][tex]\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)[/tex][tex]=2\pi e^x(1+\ln2-x)\bigg|_0^{\ln2}[/tex][tex]=2\pi\left(e^{\ln2}(1+\ln2-\ln2)-e^0(1+\ln2-0)\right)[/tex][tex]=2\pi(1-\ln2)[/tex]