## Calculus with Applications (10th Edition)

$14\pi$
If y=0, then x=1, x=3, so that a=1, b=3. The volume is $V=\int^{3}_{1}\pi(\sqrt x +5)^{2}dx=\pi[\frac{(x+5)^{2}}{2}]^{3}_{1}=14\pi$