Elongation of a wire of non-uniform cross-section under it's own weight (solid mechanics) Recently finished a chapter on Solid Mechanics at school.
Among the numerous scenarios/examples our teacher provided, was a case of a wire of non-uniform cross sectional area allowed to stretch under its own weight by suspending it by one end, from the ceiling.
The wire looks like a truncated cone, with the smaller end having a radius $r$ and the larger end having a radius $R$. The length/height (prior to suspension) of the cone is $h$. It has a mass $M$ and a Young's modulus $Y$.

^ Best image I could find online 
And our teacher proceeded to provide us the "formula" to be used to find the change in length $x$:
$$x =  \frac {Mgh}{πRrY}$$
Where $g$ is the acceleration due to gravity.
The issue? We don't know how it's derived, much less...if it's even correct.
Staring at this formula jotted down on a piece of paper made me realize that it bore a striking similarity to the general equation relating Young's modulus with force applied, length of object, cross sectional area of object and the increase in length.
$$Y =  \frac {F.L}{A.x}$$
Which implies;
$$x =  \frac {F.L}{A.Y}$$
And this is analogous to the formula provided by my teacher.
In this case, however, I take a strong exception to the use of $πRr$ as $A$. Since we're dealing with a wire of non-uniform cross section, simply substituting $A$ with $πRr$ seems ridiculous. 
However, since the variation/gradient in radius with respect to length is steady (for a truncated cone), it seems prudent to replace $A$ with the mean/average cross sectional (since the radius varies smoothly across the wire, simply taking the average of the extreme radii values would give the same result as using integration to obtain the same...the latter being more convenient) area of the wire:
$$A =  \frac {πR^2 + πr^2}{2}$$
Which, in the general formula, should yield;
$$x =  \frac {F.L}{A.Y}$$
$$=> x =  \frac {(Mg).L}{(\frac {πR^2 + πr^2}{2})Y}$$
Which is different from the one my teacher provided.
So my question;
1) Is the formula provided by my teacher correct? If so, how was it derived?
2) If the formula I came up with is wrong, then where did I err?
 A: 
Where did I err?

The equation $x=\frac{F}{A}\frac{L}{Y}$ applies to each infinitesimally-thin horizontal layer of the wire. 
If the cross-sectional area $A$ of the wire were constant then $F$ (= weight of wire below any layer) would increase linearly with $y$, the height from bottom to top. Because the only variation (that in  $F(y)$) is linear, the total extension $x$ for the whole wire can be found by using the average value of $F$ at the ends. This is just the same as finding the area under a straight-line graph from the range and the average height at the ends.
However, for non-uniform wire area $A(y)$ decreases (linearly) from bottom to top while $F(y)$ increases but no longer linearly. The overall variation is not likely to be linear. So taking the average between the end values no longer works. This is the same as the reason why the area under any non-linear curve is not necessarily equal to the range times the average height at the two ends.
See also the worked solution to this problem on the Mathalino website.
A: You need to do an integration rather than averaging. and use the formula for the volume of a truncated cone.
Relate the density of the cone $\rho$ to its mass$M$ and its dimensions $h,\, r$ and $R$.
Let $z$ be the distance of a disc of thickness $dz$ from the bottom of the cone and the radius of this disc be $x$.
You need to relate the radius of the disc $x$ to the distance of the disc from the bottom $z$ and to find the weight of the truncated cone of height $z$ below the disc.
This will enable you to work out the extension of the disc and then integrate for all the discs that make up the truncated cone to find the total extension.
A: If you do a differential force balance on the section of wire between x and x + dx (where x is measured upwards), you obtain:
$$\sigma(x+dx)A(x+dx)-\sigma(x)A(x)=\rho g A(x)dx$$where $\sigma$ is the tensile stress and A is the local cross sectional area.  This leads to the differential equation:$$\frac{d(\sigma A)}{dx}=\rho g A\tag{1}$$The local strain in the wire is $$\epsilon(x)=\frac{\sigma(x)}{Y}\tag{2}$$So the change in length of the wire is $$\Delta L=\int_0^h{\epsilon dx}\tag{3}$$So you need to solve Eqn. 1 for $\sigma$ as a function of x, subject to the initial condition that $\sigma=0$ at x = 0 (the bottom of the wire).  You then substitute this into Eqns 2 and 3 to get the change in length.
The solution obtained by implementing this methodology is:
$$\Delta L=\frac{Mgh(R+r/2)}{\pi Yr(R^2+Rr+r^2)}\tag{4}$$
Your teacher's result matches this only if R>>r.
