Total thrust on the face of a vertical dam "A vertical dam is a parabolic segment of width $12m$ and maximum depth $4m$ at the center. If the water reaches the top of the dam, find the total thrust on the face."
Is it possible to answer this question without knowing the height of the dam. I've copied the question exactly from the book, the answer is $51.2w$ (where w is the weight of unit volume of water).
I've attempted to model it as part of a circle with radius 4 m giving a height for the dam of $4\sin\theta$ m then trying to find the horizontal force on the vertical projection of the curved surface:
$$F_h=w\times A \times \bar{z}=w\times12\times4\sin\theta\times2\sin\theta\\=96w\sin^2\theta$$ where $\bar{z}$ is the depth of the centroid of the vertical projection of the curved surface (centroid of a rectangle).
Then for the vertical component $F_v$ which would be the weight times the volume of water on the face of the dam. I modeled the volume as $12\times$(area of circular sector minus the area of a triangle) which gives
$$F_v=12w(\frac{1}{2}4^2\theta-\frac{1}{2}4\cos\theta\times4\sin\theta)\\=96w(\theta -\sin\theta\cos\theta)$$
The resultant force would then be $\sqrt{F_h^2+F_v^2}$ which gives something that i cannot simplify without knowing $\theta$. 
Perhaps I'm over complicating this question. I think that there must be a deceptively simple method that I'm not seeing. Maybe a fresh pair of eyes will see it. Any help will be appreciated!
 A: A parabolic segment looks like the shaded region of this graph:

The height of the dam is what the problem says is the "depth", that is, $4$ meters.
The face of the dam is the region that is under the line $y=4$
and above the line $y=ax^2$ for some suitable $a$
(so that the bottom edge meets the top edge at $x=6$).
If the problem is the least bit realistic,
it is not enough (or one might say it is doing too much)
to "weigh" the water behind the dam.
What you want is the pressure at each point on the face of the dam.
This is zero at $y=4$ and (since the water comes up to the top)
increases linearly as you go down, with maximum value at $y=0$.
I do hope your book explained how to compute hydrostatic pressure
before springing this problem on you.
The usual formula for the pressure at depth $h$ below the surface
(as given in this reference, for example), is
$$P = \rho g h$$
where $\rho$ is the density of the fluid and $g$ is the acceleration of gravity.
Density is just the mass per unit volume, and $g$ is weight per unit mass,
so $\rho g = w$ (the given weight per unit volume).
(I'm ignoring the air pressure on the surface of the water here, on the assumption
that we want the net force on the dam caused by the water alone;
the air exerts a pressure on the other side of the dam that cancels out
its contribution to the water pressure. 
But this may be more detail than your book cares about.)
Pressure is force per unit area.
Since the depth varies, the total force (aka thrust) is an integral over the dam face,
where the ratio of force on a small square region to the area of that region
approaches $P$ as the area of the region goes to zero.
Armed with that information, I get the same answer as the book does,
$51.2w$ units of force.
