Have you tried getting a custom-made all-steel driveshaft, so then you can find out how strong the rear end is, and then the half-shafts...
I think if one were to calculate the thrust vectors in the stock rear end, it would not be hard to weld some aluminum reinforcements to the pumpkin in the right places to keep the thing together. The gear teeth themselves don't seem to be a source of trouble, more the pumpkin itself failing to retain the force of the various thrust vectors.
Assuming the torque at the driveshaft is 1000 ft-lbs, which it is not, but for the sake of easy math, the torque out the back end if it is a 2.62 is 2,620 ft-lbs. Just holding the pumpkin in place requires 2,620 ft-lbs in this case, as it is trying to rotate about the pitch axis, backwards, since it is trying to rotate the half-shafts forward.
Simultaneously, it is trying to rotate about its roll axis in the same direction as engine rotation with 1000 ft-lbs.
Then, there are the forces of the ring gear interacting with the pinion, which are mainly two vectors:
A) a vector perpendicular to the angle of the gear teeth on the ring gear relative to radial, because it is not a straight-cut gear, in which case the force would be perpendicular to radial from the ring gear. The gear teeth are cut in such a way that this force is trying to push the pinion forward and upward, which is a reason why pinion gears need conical bearings, not just plain ol' roller or ball bearings.
B) a vector that is proportional to the angle of the actual gear teeth faces relative to the center curve between gear teeth. If, for an extreme and unrealistic example, the gear teeth are cut at a slope of 45 degrees per side, so each gear tooth has faces that are 90 degrees from each other, then the force trying to separate the gears from each other that is perpendicular to the plane of the ring gear, roughly, would be exactly the same as the force of the pinion rotating the ring gear. 1000 ft-lbs, assuming a convenient 2.0 inch diameter, aka 1 inch radius of the ring gear, is 12,000 pounds of force, which, if the gear teeth were a horrendous 45 degree cut as mentioned, would result in the ring and pinion needing 12,000 pounds of force to keep them together.
The Ford Nine Inch rear end is a favorite of hot rodders since it supports the pinion gear so well, so it can take a lot of torque for its size. By including ring and pinion in one assembly in the famous "third member" (not an Austin Powers movie)(yet.) the assembly is able to be set up in advance of ever bolting the third member to the front of the axle, a method that seems to have escaped Chrysler except in its 8 3/4" rear end, if I recall correctly.
There are a few forces at work that influence design of the rear end housing. As the pinion rotates the ring gear, the resistance to said movement via inertia, drag, etc. is causing a counter-force acting upon the pinion gear, trying to rotate it in its entirety, over the top of the pumpkin, backwards, then, in addition to that, the gear teeth themselves are constantly trying to reject each other, also.
Merely piling more thickness on the housing everywhere is not the most cost- or weight-efficient method for reinforcing the rear end. Witness the light, graceful Ford nine inch versus the clunky, heavy Mopar Dana 60. All big rig axles are of the third member type. If they tried to make them in the same manner as the Dana 60/70/80, they would end up with axles that weighed far too much.
Now, a SOLID axle has more vectors to contend with, namely, braking torque being restrained by the axle ends and transmitted to the vehicle by that means, in the opposite direction as the rotation from acceleration.
But, it is not the same with an IRS. The braking torque from friction brakes is transmitted to the suspension uprights and is handled by the chassis-to-upright suspension linkage.
With some judicious measurements in a test rig, one could estimate the forces involved, and their directions, and suitably beef up the stock rear end, without adding too much weight.