## Thomas' Calculus 13th Edition

$\dfrac{ 72 \pi }{35}$
The area of each cross-section triangle is given as: $A=\pi r^2 =\pi (\sqrt x- \dfrac{x^2}{8}) ^2$ We integrate the integral to calculate the volume as follows: $V= \int_{0}^{4} \pi (\sqrt x- \dfrac{x^2}{8}) ^2 dx$ or, $=\pi [\dfrac{x^2}{2}-\dfrac{x^{7/2}}{14} +\dfrac{x^5}{320}]_0^4$ or, $=\pi (8-\dfrac{64}{7} +\dfrac{16}{5})$ or, $=\dfrac{ 72 \pi }{35}$