I am trying to derive the volume of a cylindrical hoof in cylindrical coordinates given the radius $a$ and angle the tangent plane makes through the diameter, $\alpha$
My plan is to evaluate $\int dV$ over the curve where $dV = A (ad\phi)$, $A=\frac{1}{2}aZ(\phi)$ is the area of the triangle, and $ad\phi$ is a very small arc length.
$$\int dV=\int^\pi_0 A(\phi) ad\phi=\int^\pi_0 \frac{a^3 \sin\phi \tan\alpha}{2}d\phi=a^3\tan\alpha$$
Given in the book, I know the answer is $\frac{2}{3} a^3 \tan\alpha$. Unless I have my arc length approach is fundamentally wrong, I really don't know how this is wrong. Looking at the integrand $f(\phi)$ we have some really nice properties: unbounded when $\alpha=\pi/2$, $f(0)=0$, $f(\pi/2)=\frac{1}{2}(a)(a\tan\alpha)[ad\phi]$
In your method the element of volume is a pyramid whose thickness varies from $ad\phi$ at the curved surface to $0$ at the origin. However in your calculation you have assumed that this element is a slab of constant thickness $ad\phi$. Instead of $dV=(\frac12 Z a)(a d\phi)$ you should have $dV=\frac13 (Z a d\phi)a$. Hence a correction factor of $\frac23$ is required.