First, recall that the area of an equilateral triangle with side length \(s\) is \(\dfrac{\sqrt3}4s^2\).
Now, each cross section's base is the distance from the top half of the circular base to the lower half. You have
\(x^2+y^2=9\iff y=\pm\sqrt{9-x^2}\)
This means the vertical distance is \(\sqrt{9-x^2}-(-\sqrt{9-x^2})=2\sqrt{9-x^2}\).
So, the volume of such a solid is given by
\(\displaystyle\frac{\sqrt3}4\int_{-3}^3\left(2\sqrt{9-x^2}\right)^2\,\mathrm dx=2\sqrt3\int_0^3(9-x^2)\,\mathrm dx=36\sqrt3\)
where the first equality comes from the fact that the circle is symmetric across the y axis.
