(a) First find the intersections of [tex]y=e^{2x-x^2}[/tex] and [tex]y=2[/tex]:
[tex]2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}[/tex]
So the area of [tex]R[/tex] is given by
[tex]\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx[/tex]
If you're not familiar with the error function [tex]\mathrm{erf}(x)[/tex], then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.
(b) Find the intersections of the line [tex]y=1[/tex] with [tex]y=e^{2x-x^2}[/tex].
[tex]1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2[/tex]
So the area of [tex]S[/tex] is given by
[tex]\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx[/tex]
[tex]\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx[/tex]
which is approximately 1.546.
(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve [tex]y=e^{2x-x^2}[/tex] and the line [tex]y=1[/tex], or [tex]e^{2x-x^2}-1[/tex]. The area of any such circle is [tex]\pi[/tex] times the square of its radius. Since the curve intersects the axis of revolution at [tex]x=0[/tex] and [tex]x=2[/tex], the volume would be given by
[tex]\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx[/tex]
