Hydrostatic pressure on a square Vertically inserted into the water I have a rectangle  6 feet wide and 4 feet high that is submerged under the water with 2 feet of water above it.
Using a riemann sum how do I find the pressure? I am not able to use the triangle ratio trick so what do I do now?
I tried to do the ratio and I got
$$\frac{a}{4-x}$$
$$a = 6 - \frac{3x}{2}$$
$$\int_2^6 19620 (6x - \frac{3x^2}{2})dx$$
This is wrong but I do not know why.
 A: The geometry here is simpler than in your previous triangle question.  We give a highly informal description that should lead you to the answer.
Divide the rectangle into a large number of horizontal strips of width $\Delta x$. Consider such a strip, which is more or less at depth $x$ feet beneath the water.  The area of the strip is $6\,\Delta x$. We want to compute the force against it, and add up over all the strips. Then if we wish we can divide by the area to find the pressure (force per unit area) against the rectangular plate.
The force against the strip is approximately $6\,\Delta x$ multiplied by a certain constant $K$ times $x$. The constant is $g$ times the mass of a cubic foot of water. I recall that the acceleration due to gravity is (in United States units) about $32$ feet per second$^2$ but cannot recall the mass of a cubic foot of water. (Life is easier in metric!) 
We will "add up" the pressures on the strips, from $x=2$ (the depth of the top of the rectangular plate) to $x=2+4=6$ (the depth of the bottom of the rectangle). More precisely, we let $\Delta x \to 0$, and the limit of the sums is an integral. So the force is 
$$\int_2^6 6Kx\,dx.$$
A: The pressure at a certain depth is given by $P = \rho g d$, where $\rho$ is the denstity of the fluid, $g$ is gravitational acceleration, and $d$ is depth.
Hydrostatic force $F$ can be calculated from the equation $F = PA$, where $P$ is the pressure and $A$ is the area.
The idea here is that water pressure changes based on depth, and so the force changes at different depths. So you want to sum across infinitesimally small changes in depth. The idea is that for an infinitesimally small band, you can assume the whole band is at the same depth. The area will be this infinitesimally small change in height times the width of the plate. And you sum across the depths.
This is, in effect, a word problem describing your situation. For more, including two worked examples, I refer you to Paul's Online Math Notes again. I also note that I have taught from Stewart, and it's covered in there as well.
As a final note, I see that your constant $19620 = 2 \cdot 9810$ which is the constant in the metric system. But the problem is presented in feet, not meters.
A: The pressure difference from sea level at depth $x$ is given by $P(x) = \rho g x$, where $\rho$ is the density, and $g$ is the acceleration due to gravity. The force on a element '$dx$' tall at depth $x$ is $P(x) w \, dx$, where $w$ is the width.
Hence the total force in the plate is:
$$ F = w\int_2^6 P(x) \, dx = \rho g w \int_2^6 x \, dx = 96 \rho g.$$
Substitute your value of $\rho g$ (salt or fresh water?) for a final answer; I have substituted linear dimensions in feet already.
