Slant cone volume problem I was given a a problem to solve, I thought I solved it but my answers don't look like the ones provided.
The Problem
A cone with radius of base r and height h, is stretched to the left and right by length a and b respectively such that height of the stretched cones equals to the height of the original cone and apexes of all the three cones are colinear.​
Devise an algorithm to calculate the volume of the union of the two stretched cones (colored blue in the image).

My first question was: what is the volume of a slant cone? What I've found so far is that the formula for the volume of a slant cone is the same as that of a regular cone (which made sense to me).
So I though the answer would be 2 times the area of a cone given by radius r and height h minus the intersection of the two slant cones, which is also a cone with radius r. I just need to find out the height of that smaller cone.
My thought was that I could just take the cross section of of the bodies and handle the problem in 2D. Where the base of the cone will lie on the x axis and the center for the circle lies in point (0,0).
What I did was calculate the equation of the line that goes from (-a,h) to (r,0). And I would also calculate the equation of the line going from (-r,0) to (b,h).
I proceeded to calculate the line equations and then to calculate where they intersect. I would then take y value at intersection and use it as the height of the smaller cone.
I really thought I nailed it since when I had a = 0 and b = 0 I got the volume of the original cone. However, when the answers where provided mine didn't match the ones given. Can someone point out the hole in my logic?
 A: For the principles of projection onto parallel planes, on a plane $z=c$, the section of the three cones are circles of  same radius $r(h-c)/h$,
and centers shifted in the same ratio $c/h$ wrt $a,b$.
So the intersection is a " conic wedge"  structure: the superposition of  the intersections of the shifted circles.
Then by Cavalieri principle, you can slant the cones symmetrically by  $(a+b)/2$ wrt the right cone,
compute the intersection of the two circles $\cdot dz$, and integrate.
A: The volume of the intersection is given by the following integral based on a general equation for the area of the intersection of two same-sized  circles in this scenario, dependent on the value of y: 
$$V_i =\int_0^{h_i}\big(2\cos^{-1}(\frac{y(a+b)}{2r(h-y)})\cdot(r-\frac{ry}{h})^2 - \sqrt{(r-\frac{ry}{h})^2-(\frac{y(a+b)}{2h})^2}\cdot(\frac{y(a+b)}{h})\bigr) dy$$

Example: $a=3, b=4, h=5\ \text{and}\ r=3$
$h_i = \frac{30}{13}$ which is the height of the conic wedge
$$V_i =\int_0^{30/13} 2\cos^{-1}(\frac{7y}{6(5-y)})\cdot(3-\frac{3y}{5})^2 - \sqrt{(3-\frac{3y}{5})^2-(\frac{7y}{10})^2}\cdot(\frac{7y}{5}) dy$$
An integral solver couldn't come up with an anti-derivative but used an approximation method and came up with $25.25762\ \#^3$
