I think i have solved the exercise 15 of Apostol Vol.1 exercises 2.13 in which he says " A solid has a circular base of radius 2. Each cross section cut by a plane perpendicular to a fixed diameter is an equilateral triangle. Compute the volume of the solid."
So my solution involves realizing that this solid is a cone with circular base trough which perpendicular planes, that pass trought the center of the circle, intersect the cone forming a equilateral triangle
And so i calculated the line that passed by points $(2 \,,\, 0)$ and $(0\, ,\, 2\sqrt{3})$ and got $$y=2\sqrt{3}-x\sqrt{3} \,\iff \, x= \frac{-y\sqrt{3}}{3}+2$$
Therefore i simply calculated the integral that gives the volume of a solid of revolution
$$V=\pi\int_{0}^{2\sqrt{3}}\left(2-\frac{y\sqrt{3}}{3}\right)^2dy = \frac{8\pi\sqrt{3}}{3}$$
but that is only $\frac{1}{2}$ of the anser that Apostol gives and in other solutions people give things like $\frac{32\sqrt{3}}{3}$ as the anser , that is 4 times my anser and two times Apostol's anser. Which is right ?
The answer that you calculated is incorrect in that you have assumed that the solid is a cone. In fact, it is not. A cone does not have similar triangles in its cross sections; they are hyperbolas. We can find the volume of this solid by noting that it can be expressed as the integral of the cross-sectional areas, for example
$$V=\int_{-2}^2 A_{\text{triangle}}(x)~dx$$
Now, the triangles have a base and sides of $2y$, and the area is then is then $1/2$ the base times the height, or $A_{\text{triangle}}=\sqrt{3}y^2=\sqrt{3}(4-x^2)$. Then, using symmetry for the integral,
$$ \begin{align} V &=2\int_0^2 \sqrt{3}(4-x^2)~dx\\ &=2\sqrt{3}\left(4x-\frac{x^3}{3}\right)\biggr|_0^2\\ &=2\sqrt{3}\left(8-\frac{8}{3}\right)\\ &=\frac{32\sqrt{3}}{3} \end{align} $$
EDIT: I have added a quick and dirty 3D image of this volume below.
EDIT: I've added another image showing the triangle that are being integrated