Recall that the area of an equilateral triangle with side lengths \(s\) is \(\dfrac{\sqrt3}4s^2\). 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 \(x^2+y^2=16\), you have \(x=\pm\sqrt{16-y^2}\), where the positive root corresponds to the right half and the negative roots corresponds to the left half.
The volume is given by
\(\displaystyle\int_{-4}^4\frac{\sqrt3}4\left(\sqrt{16-y^2}-(-\sqrt{16-y^2})\right)^2\,\mathrm dy\)
\(=\displaystyle\sqrt3\int_{-4}^4(16-y^2)\,\mathrm dy=2\sqrt3\int_0^4(16-y^2)\,\mathrm dy\)
where the last equality follows from the fact that the integrand is symmetric about \(y=0\). The volume is then
\(2\sqrt3\left(16y-\dfrac13y^3\right)\bigg|_{y=0}^{y=4}=2\sqrt3\left(64-\dfrac{4^3}3\right)=\dfrac{256}{\sqrt3}\)
