Can the lift generated by a helicopter be justified using Bernoulli theorem? When the shaft of the helicopter rotates, it creates a low pressure. Because of the low pressure, the helicopter lifts. Is my understanding that this is just an application of Bernoulli's theorem?
 A: The blades of a helicopter are contoured much like the wing of an airplane. The physics behind both are, basically, the same.
Bernoulli's Effect is usually quoted as the reason behind flight in so many physics textbooks. While this isn't wrong, Bernoulli's Effect isn't actually the main reason that blades/wings can cause flight. 
If you've noticed, the wing of an airplane is tilted a bit. This is so that the air molecules hit the bottom surface at an angle. If you have a ceiling fan, you can observe the slight angle in their blades too. This air hits the blade and is rebounded downwards. From the wing/blade point of view, it's being pushed upwards. This is what causes lift. 
You can try this, by holding out a piece of cardboard while you're traveling in a fast car. Keep it horizontal and you won't experience lift. Tilt it a bit and you'll feel it being pushed up.
The Bernoulli effect simply adds to this.
A: You are basically correct. 
The principle making a helicopter fly is basically the same as for a fixed wing plane. The "wings" of a helicopter are the rotor blades. They rotate at high speed and so have apparent velocity relative to the air, just as plane wings do when moving forward.
As the air moves over the blade it generates lift by deflecting the air and by the low pressure on top of the wing due to Bernoulli's principle.
There are some technical complexities to stop the helicopter spinning round or flipping over due to the torque generated but these are not really to do with lift generation.
