Recall that the area of an equilateral triangle with side lengths [tex]s[/tex] is [tex]\dfrac{\sqrt3}4s^2[/tex]. If cross sections are to be taken perpendicular to the y-axis, then each section's side length will be determined by the horizontal distance between the right and left sides of the circle.
Since [tex]x^2+y^2=16[/tex], you have [tex]x=\pm\sqrt{16-y^2}[/tex], where the positive root corresponds to the right half and the negative roots corresponds to the left half.
The volume is given by
[tex]\displaystyle\int_{-4}^4\frac{\sqrt3}4\left(\sqrt{16-y^2}-(-\sqrt{16-y^2})\right)^2\,\mathrm dy[/tex]
[tex]=\displaystyle\sqrt3\int_{-4}^4(16-y^2)\,\mathrm dy=2\sqrt3\int_0^4(16-y^2)\,\mathrm dy[/tex]
where the last equality follows from the fact that the integrand is symmetric about [tex]y=0[/tex]. The volume is then
[tex]2\sqrt3\left(16y-\dfrac13y^3\right)\bigg|_{y=0}^{y=4}=2\sqrt3\left(64-\dfrac{4^3}3\right)=\dfrac{256}{\sqrt3}[/tex]
