## Calculus 8th Edition

$80.6095$
$y = x^{1/2}$ then $y' = \frac{1}{2} x^{-1/2}$ and $1+(y')^{2} = 1+1/4x$ So $S = \int^{4}_{0} 2\pi (4-\sqrt{x}) \sqrt{1_1/(4x)} dx$ by using CAS $S = 2\pi \ln{(\sqrt{17} +4)} + \frac{\pi}{6}(31\sqrt{17} + 1) \approx 80.6095$