## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{ 8 \sqrt 3}{15}$
We need to integrate the integral as shown below: Area of each cross-section triangle, $Area=\dfrac{(2 \sqrt x-x)^2 \sin 60^{\circ}}{y}=\dfrac{\sqrt 3 }{4} (2 \sqrt x-x)^2$ $I= \int_{0}^{4} [\dfrac{\sqrt 3 }{4} (2 \sqrt x-x)^2] dx \\= \dfrac{\sqrt 3}{4}\int_{0}^{4} [4x-4x \sqrt x+x^2] dx \\=\dfrac{ 8 \sqrt 3}{15}$