Why perpendicular axis theorem is applicable only for laminar (2-D)objects? I was taught that perpendicular axis theorem is valid only for laminar objects and not for 3-D objects.I have difficulty in understanding this intuitively.
I mean why would such a condition even exist?
We can consider a 3-D object to be a collection of infinitely divided elements which are stacked one upon on the other to form a 3-D object.Now,we know that for a 2-D object the theorem is valid and we can find moment of inertia of each element about an axis passing through its geometrical centre and perpendicular to its plane and now we can add up moment of inertia of each element using simple addition to get moment of inertia of the 3-D body(because all elements are symmetrical).
This is the same as using perpendicular axis theorem for 3-D object right?    
 A: For a 3D object although the slices are all laminar, and the Parallel Axis Theorem applies independently for each lamina, it does not apply for all of them together, ie for the 3D object. The reason is that although the laminas have the same z axis, they do not have the same x and y axes. So you cannot simply add the moments of inertia (MOI) for the x and y axes for each lamina. The x and y axes for each lamina are parallel but they are off-set from each other and from the x, y axes you are using for the 3D object. As @Triatticus points out, you need to apply the Parallel Axis Theorem to each lamina.
You can add the values $I_z$ for each lamina to get the value of $I_z$ for the 3D body, provided that all the z axes coincide. The lamina do not have to be symmetrical to do this, and the z axis does not even have to pass through the centre of mass of each lamina. However, this is not an application of the Perpendicular Axis Theorem - there are no perpendicular axes here.

Suppose your 3D object consists of two co-axial disks each of mass $m=\frac12 M$ and radius $a$ separated by a distance $2b$. 
The MOI for this 3D object about the common z axis is $I_z=2\times \frac12 ma^2=\frac12 Ma^2$. It equals the sum of the $I_z$ for the 2 disks because the z axes coincide. If the 3D object satisfied the Perpendicular Axis Theorem the MOIs about perpendicular axes would be $I_x=I_y=\frac14 Ma^2$. However, where is the origin for these x, y axes located? When you apply the Perpendicular Axis Theorem for each disk you are using a different origin for each, usually the centre of each disk. When you check the Perpendicular Axis Theorem for the 3D object you are using the geometrical centre of the 2 disks as origin, which is the midpoint of the centres of the 2 disks.
The MOIs for each disk about x and y axes through their own centres of mass are $I_x=I_y=\frac14 ma^2$ which satisfies the Perpendicular Axis Theorem. But these axes do not coincide with the x and y axes through the COM for the 3D object. The disk x, y axes are each offset by a distance $b$ from the 3D object x, y axes. Using the Parallel Axis Theorem the MOIs about x and y axes through the COMs of the 3D object are $$I_x=I_y=2(\frac14 ma^2+mb^2)=\frac14 Ma^2+Mb^2 \ne \frac12 I_z$$ Equality only applies when $b=0$ - ie when the disks coincide.
Generally the moments of inertia for a 3D object are defined by distance from the relevant axis : $$I_x=\int (y^2+z^2)dm$$ $$I_y=\int (x^2+z^2)dm$$ $$I_z=\int (x^2+y^2)dm$$ Note that these definitons use the same co-ordinate system throughout, including the same origin. Then $$I_x+I_y=\int (x^2+y^2+2z^2)dm=I_z+2\int z^2 dm$$ The Perpendicular Axis Theorem $I_x+I_y=I_z$ only applies if $z=0$ for all points of the object - ie if the object is confined to the $xy$ plane.
A: Your method is correct, but I wouldn't call it the same thing as the perpendicular axis theorem. This is because of the points others have already brought up. For a 2D object in the x-y plane we know that $I_z=I_x+I_y$. Objects in 3D can also be rotated about these same axis, and it is not true in general that $I_z=I_x+I_y$ (as has already been pointed out). I would instead say you are just applying the perpendicular axis theorem to each slice so that you can add up each $I_z$ to find $I_z$ of the composite body. But I guess at this point it all just depends on what you mean when you "define" the "3D perpendicular axis theorem".
Addition:
As pointed out, you also have to keep in mind while the composite $I_z$ is the sum over each $I_z$, the same cannot be said of the moments of inertia about the other axes. $\Sigma I_x$ is not equal to $I_x$ for example. 
