Answer:
See explanation
Step-by-step explanation:
The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula
[tex]SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx[/tex]
If [tex]f(x)=x^2,[/tex] then
[tex]f'(x)=2x[/tex]
and
[tex]b=0\\ \\a=2[/tex]
Therefore,
[tex]SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx[/tex]
Apply substitution
[tex]x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du[/tex]
Then
[tex]SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du[/tex]
Now
[tex]\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))[/tex]
Hence,
[tex]SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}[/tex]
