## Calculus (3rd Edition)

$$10\sqrt {17} \ \pi.$$
Since $y=4x+3$, then $y'=4$. Hence, the surface area is given by $$2\pi\int_0^1 y\sqrt{1+(y')^2}dx =2\pi\int_0^1(4x+3)\sqrt{1+4^2}dx\\ =2\sqrt {17} \pi\int_0^14 x+3 dx=2\sqrt {17} \pi( 2x^2+3x)|_0^1=10\sqrt {17} \ \pi.$$