Answer
$\dfrac{4 \pi}{3}$
Work Step by Step
The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$
Now, $V=2\pi \int_{0}^{\sqrt 3} (x) (2-\sqrt {1+x^2}) \ dx \\= 2\pi (x^2-\dfrac{(1+x^2)^{3/2}}{3}]_{0}^{\sqrt 3} \\=2 \pi [3-\dfrac{(1+(3)^2)^{3/2}}{3}) \\=2 \pi (3-\dfrac{8}{3}+\dfrac{1}{3}) \\=\dfrac{4 \pi}{3}$
