Thoughts on how to come up with the answer for this question? Do I plug it into the formula Integral from $a$ to $b$, $2\pi(x)(fx)$
Thanks!
By definition (in fact, it follows from definition), the wanted surface area is
$$2\pi\int\limits_0^5 y\sqrt{1+y'^2}dx=2\pi\int\limits_0^5\sqrt{3x}\sqrt{1+\frac3{4x}}\;dx=\sqrt3\,\pi\int\limits_0^5\sqrt{4x+3}\,dx=$$
$$\left.\sqrt3\,\pi\frac14\frac23(4x+3)^{3/2}\right|_0^5=\frac\pi{2\sqrt3}\left[23^{3/2}-3^{3/2}\right]$$