A vertical plate is submerged in water. What is the hydrostatic force by riemann sum?

A semi circle has equation x^2+y^2=R^2    where axis X and Y are on the diameter x an y
At y from the bottom  x^2 = R^2-y^2
x= sqrt (R^2-y^2 ) , so the lenght is L= 2x
L= 2sqrt (R^2-y^2)
An element of area at y will be dA= Ldy
dA= 2 sqrt (R^2-y^2) dy

Hidrostatic pressure will be P = G h  , h measured from the water level , where
 h= (R-2) -y  from the X Y axis ,. thus , when y=0  ( bottom)  ,h=R-2

dforce = P dA
dForce = G ( (R-2)- y) (2 sqrt (R^2-y^2) )dy

dF= 2G (R-2) sqrt (R^2-y^2) - 2G y sqrt (R^2-y^2) dy

F= 2G(R-2) INT sqrt (R^2-y^2) dy - 2G INT y sqrt (R^2-y^2) dy

This is the method .-

But the best way to do this is in the same way , but using the central angle
Doing the same , T is the central angle from the origin ,between a vertical radius and any radius ,
L= 2RsinT
y= RcosT
dy = -RsinT dT
dA= 2RsinT I dyI
dA= 2R^2 sin^2T dT

Pressure = P = G h  , h= (R-2) -y = 4-y
h=4-RcosT
dForce = P dA
dF= G (4-RcosT ) (2R^2 sin^2T) dT
When y = R-2= 4  ,
4= RcosT
4=6cosT
cos T = 4/6= 2/3 ~ 48.2°~ 0.8425 rad
When y=0 , cos T = 0  , T= pi/2 , so
0.8425
F= INT G (4-RcosT ) (2R^2 sin^2T) dT

F= 8R^2 G INT sin^2 T dT - 2R^3 G INT cosT sin^2 T dT

F= 8R^2G INT ( 1-cos2T)/2 dT -2R^3G (sin^3T) /3

F= 8R^2G ( T/2 - (1/4) sin2T ) - 2R^3G/3 sin^3T

F= 8R^2G (( pi/4- (1/4) sin pi ) - ( 0.8425/2 -(1/4) sin 1.685 )) - 2R^3G/3 ( 1-sin^3 0.8425)

G= 1 Ton/m3
F= 288 ( (0.7854 -0)  -(0.4212 -0.9935/4 ))  - 144 (1-0.7463)

F= 288 (0.6126)  - 144*0.2537

F= 176.42-36.53 ~ 140 ton