(a) The area of the region would be given by the integral
[tex]\displaystyle\int_0^\pi y(t)\left|x'(t)\right|\,\mathrm dt = 8 \int_0^\pi \sin^2(t)\,\mathrm dt[/tex]
(b) The area of the surface of revolution would be given by
[tex]\displaystyle\int_0^\pi y(t)\sqrt{x'(t)^2+y'(t)^2}\,\mathrm dt = 4\int_0^\pi\sin(t)\sqrt{4\sin^2(t)+16\cos^2(t)}\,\mathrm dt[/tex]
