Answer
$\frac{8\pi}{3} *(5^{\frac{3}{2}} - 1)$
Work Step by Step
Starting with the formula for surface area under rotation around the x-axis:
$S = \int_{a}^{b} 2 \pi y* \sqrt{1 + (\frac{dy}{dx})^2} dx$
In our case, $a = 0, b=4, y = 2\sqrt{x}$. Thus, by the power rule for differentiation, $\frac{dy}{dx} = (2)*(\frac{1}{2})*(x^{-\frac{1}{2}}) = x^{-\frac{1}{2}}$
$\implies S = \int_{0}^{4} 2 \pi *2\sqrt{x}* \sqrt{1 + (x^{-\frac{1}{2}})^2} dx$ $= \int_{0}^{4} 4 \pi \sqrt{x + 1}$ $ dx $
Doing a u-substitution where $u = 1+x$, we have $du =dx$ and:
$S = \int_{1}^{5} 4 \pi \sqrt{u}$ $ du $ = $4\pi *(\frac{2}{3})*u^{\frac{2}{3}} \Bigr\rvert_{u = 1}^{u=5} = (8\pi/3)*(5^{\frac{3}{2}} -1)$
