Answer
$\dfrac{ 8 \sqrt 3}{15}$
Work Step by Step
The area of each cross-section triangle is given as: $\dfrac{(2 \sqrt x-x)^2 \sin 60^{\circ}}{y}=\dfrac{\sqrt 3 (2 \sqrt x-x)^2 }{4}$
We integrate the integral to calculate the area as follows:
$A= \int_{0}^{4} \dfrac{\sqrt 3 (2 \sqrt x-x)^2 }{4} dx$
or, $= \dfrac{\sqrt 3}{4}\int_{0}^{4} [4x-4x \sqrt x+x^2] dx$
or, $=\dfrac{ 8 \sqrt 3}{15}$
