How can you explain a paraboloid in a rotating liquid without centrifugal force or gravity? How does a rotating container with liquid in it cause a paraboloid to form? My physics teacher says that it is incorrect to use Centrifugal force and Gravity, but after reading though many forums on here and elsewhere, I've become convinced that you can use it as a correction term. Is this accurate?
Also, is it okay if I do use these terms as correction terms if my rubrics say to use'accurate and up-to-date scientific terms', and to talk about 'the plane perpendicular to force field gravity direction and other forces'?
 A: Proceed by choosing a point O on the symmetry axis Z. Choose a differential fluid element having dimensions $\mathrm{d} r$, $\mathrm{d}x$ ,$\mathrm{d}y$ where x is the polar angle in your horizontal plane of observation and y is the azimuthal angle perpendicular to this plane. An F.B.D of this element should easily give the following 2 equations:


*

*$P_z = p - \rho\,z\,g$ ($\rho$ is density)

*$P_r = \rho\,\omega^2\, r^2/2$ ($\omega$ is angular velocity).


Since the free surface has no shear stress, put the total differential $\mathrm{d}P =0$  which should give you a differential equation in z and r. Integrating, you will obtain:
$$z = \frac{\omega^2\,r^2}{2\,g}$$
Where z is measured from the lowest point of the free surface. I hope you get the idea that we simply used FBD and solved using the forces on the element.
A: There is nothing wrong with your proposed approach, but I think maybe you may have misunderstood your physics teacher; most likely he/ she would like you to be completely comfortable and proficient with the analysis of dynamical systems from an inertial frame (i.e. unaccelerated frame) before shifting on to analysis from accelerated frames, which involve inertial forces such as centrifugal force. So he / she simply wants you to look at the bucket, draw a free body diagram for an element of fluid and write down Newton's second law for it. 
As in Lelouch's answer we draw a free body diagram for a triangular element of fluid at the fluid's free surface, as below:

The thing about a fluid is that forces act normal to boundaries; there is never a shear component along a boundary. Thus the only forces on the element are the element's weight $\rho\,g\,\times \frac{1}{2} r\,\delta r\,\delta \phi\,\delta z$ (we consider a small sector subtending an azimuthal angle $\delta\phi$ so that the element's volume is $\frac{1}{2} r\,\delta r\,\delta \phi\,\delta z$) and the pressure $p$ times the areas of the two faces normal to the two faces. All these forces must add up to the necessary centripetal force towards the center of rotation. So the weight is balanced by the pressure force on the base, this gives you an equation for $p$, which you then plug into the expression $p\,\delta z\,r\,\delta \phi$ for the nett force on the element towards the center of rotation. This nett inwards force must be the density times the volume $\frac{1}{2} r\,\delta r\,\delta \phi\,\delta z$ times the centripetal acceleration $\omega^2\,r$, by Newton's second law.  These thoughts allow you to get the equation  for the free surface.

Question From OP

Is there a possibility to help explain this at a high-school (i suppose, I'm 15) level? We haven't been introduced to shear force yet and have just finished Newton's Laws and Inertia

I'm guessing that the notion of pressure and shear force is the most mysterious to you. The way we measure the way dollops of fluid interact with their neighbors is through the stress tensor. This is a slightly frightening word, but bear with me. You are to imagine a small polyhedron of fluid and we wish to calculate the forces on the faces from the neighboring fluid. You specify the face by a vector pointing in a direction normal to the face and whose length is proportional to the face's area. Then the tensor is a matrix - a linear function - that takes this vector and returns another vector - the force on the face. You multiply the face vector by the matrix to get the force. 
In general, you probably know that matrices change the directions of the vectors they multiply. The component of force in the direction of the face vector, i.e. normal to the face, is called the pressure; that along the face's plane is called the shear. 
The definition of a fluid is that it cannot withstand shear unless the neighbors are in motion relative to one another. Any shear shifts the fluid until it finds an equilibrium. At equilibrium the stress tensor matrix multiplies the any face vector, no matter what its direction may be, and returns another vector in the same direction, i.e. normal to the face. The only matrix that has this property is a constant times the identity matrix. It is for this reason that we know the pressure p acting towards the center of rotation on the little triangle is equal to the pressure thrusting up to support the triangle's weight. So the weight gives the pressure, which can then be used to find the nett force towards the center of rotation.
Alternatively, one can imagine, instead of a fluid, a tight packing of identical steel cubes with fantastic lubrication between them; pressure is then the force normal to the cubes' faces, shear is the friction between the faces. Perfectly lubricated cubes can only push normal to each others' faces, there is no force transmission associated with any tendency to relative sideways motion - no friction. 
