Answer
$2 \sqrt 3$
Work Step by Step
We need to integrate the integral as shown below:
Area of each cross-section triangle, $A=\pi \space (r)^2 =\dfrac{(2 \sqrt 4)^2 \sin 60^{\circ}}{2}=4 \sqrt 3 x$
Now, $Volume = \int_{0}^{1} 4 \sqrt 3 \space x \space dx \\=2 \sqrt 3 \times [x^2]_0^1 \\=2 \sqrt 3$
