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