## Calculus (3rd Edition)

$$16\sqrt 2 \ \pi.$$
Since $y=x$ then $y'=1$. Hence the surface area is given by $$2\pi\int_0^4 y\sqrt{1+(y')^2}dx =2\pi\int_0^4 x\sqrt{1+1}dx\\ =2\sqrt 2\pi\int_0^4 x dx=\sqrt 2 \pi x^2|_0^4=16\sqrt 2 \ \pi.$$