What is the force due to hydrostatic pressure on one flat parabolic side in this problem?

multicube

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6f936fc7-3086-30b7-8b02-b74b76bc665d___ab37bb53-9014-3b43-8a18-f070a870c558.gif
As you can see in this picture, a trough has a parabolic cross section CodeCogsEqn.gif, where a is a constant. The height is H meters and the length is L meters. The trough is filled with water. Approximate the weight density of water to be ρg=10^4 Newtons per cubic meter.
The problem is to calculate the force due to hydrostatic pressure on one flat parabolic side.
And here's the calculation I've come up with for this problem.
CodeCogsEqn (1).gif
But this does not seem to be the answer, although I find it reasonable. Could anyone please shed some light on the problem, please?
 
View attachment 19004
As you can see in this picture, a trough has a parabolic cross section View attachment 19005, where a is a constant. The height is H meters and the length is L meters. The trough is filled with water. Approximate the weight density of water to be ρg=10^4 Newtons per cubic meter.
The problem is to calculate the force due to hydrostatic pressure on one flat parabolic side.
And here's the calculation I've come up with for this problem.
View attachment 19006
But this does not seem to be the answer, although I find it reasonable. Could anyone please shed some light on the problem, please?
Please share with us - your method of deriving the expression:

1589910853291.png

By the way, you do not tell use "what" does your expression calculate!
 
Please share with us - your method of deriving the expression:

View attachment 19007

By the way, you do not tell use "what" does your expression calculate!
Sorry for my poor explanation.
To explain, I drew the given graph(I hope that tells where the y came from)
and that equation is an integration of all the pressures multiplied by the surface area from height 0 to H.
Since the weight density is 10^4 Newtons per meter.
I put that on the front of the integration.
 
Sorry for my poor explanation.
To explain, I drew the given graph(I hope that tells where the y came from) ................................................ where is the graph
and that equation is an integration of all the pressures multiplied by the surface area from height 0 to H.
Since the weight density is 10^4 Newtons per meter. .................................... no that should be Newtons per meter cubed
I put that on the front of the integration.
Please show all the steps through mathematics - instead of words
 
Please show all the steps through mathematics - instead of words
I've tried to derive the Newton(force) to this question by multiplying the (Area at side of the tank)*(height from top of tank)*10000
since this is the same as (N/m^3)*(m^2)*m

CodeCogsEqn.gif can be shown as this graph
다운로드.png
and I wanted to derive the area of the side of the tank by slicing the area on the side like this drawing.
6f936fc7-3086-30b7-8b02-b74b76bc665d___ab37bb53-9014-3b43-8a18-f070a870c558.gif
but I am stuck-not only with deriving dy, but also the height from the top of the tank.
 
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