1. Using washers, the volume is given by the integral
[tex]\displaystyle\pi\int_1^{e^2}((2+2)^2-(\ln x+2)^2)\,\mathrm dx[/tex]
[tex]=\boxed{\displaystyle\pi\int_1^{e^2}(20+4\ln x-(\ln x)^2)\,\mathrm dx}[/tex]
We're using washers whose centers depend on the value of [tex]x[/tex], hence we integrate with respect to
2. The area of the given region is given by the integral
[tex]\displaystyle\int_0^2\sin^{-1}\frac x2\,\mathrm dx[/tex]
To compute the integral, first consider the substitution [tex]u=\frac x2[/tex], or [tex]2u=x[/tex] so that [tex]2\,\mathrm du=\mathrm dx[/tex]. Then [tex]x\to0\implies u\to0[/tex] and [tex]x\to2\implies u\to1[/tex], so the integral is equivalently
[tex]\displaystyle2\int_0^1\sin^{-1}u\,\mathrm du[/tex]
Integrate by parts, taking
[tex]f=\sin^{-1}u\implies\mathrm df=\dfrac{\mathrm du}{\sqrt{1-u^2}}[/tex]
[tex]\mathrm dg=\mathrm du\implies g=u[/tex]
so that
[tex]\displaystyle2\int_0^1\sin^{-1}u\,\mathrm du=2\left(u\sin^{-1}u\bigg|_0^1-\int_0^1\frac u{\sqrt{1-u^2}}\,\mathrm du\right)[/tex]
[tex]\sin^{-1}0=0[/tex] and [tex]\sin^{-1}1=\frac\pi2[/tex], so the area is
[tex]\displaystyle\pi-2\int_0^1\frac u{\sqrt{1-u^2}}\,\mathrm du[/tex]
For the remaining integral, substitute [tex]w=1-u^2[/tex], so that [tex]\mathrm dw=-2u\,\mathrm du[/tex]. Then [tex]u\to0\implies w\to1[/tex] and [tex]u\to1\implies w\to0[/tex]:
[tex]\displaystyle\pi-\int_0^1\frac{\mathrm dw}{\sqrt w}[/tex]
(notice that the integral is improper)
[tex]\displaystyle\pi-\lim_{t\to0^+}2\sqrt w\bigg|_t^1[/tex]
[tex]\displaystyle\pi-2\left(1-\lim_{t\to0^+}\sqrt t\right)=\boxed{\pi-2}[/tex]
