Calculus 2: Fluid Force

Nyasha

1. The problem statement, all variables and given/known data
A circular "pressure release gate" is located in the vertical face of a dam. The radius of the gate is 2 metres and the gate is positioned so that its top lies 5 metres below the surface of the water held in the reservoir behind the dam.Find the total fluid force acting on the door of the gate, when it is in a fully closed position.


2. Relevant equations
Area= [tex]2\sqrt{4-x^2}[/tex] dy

Pressure = pg(5+x)

3. The attempt at a solution
F= pg∫ [tex]2\sqrt{4-x^2}[/tex] (5+x) dy
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

tiny-tim

Hi Nyasha!

(have a square-root: √ and a rho: ρ )

almost right , but the area of the slice is 2√(4 - y²) dy, at depth 5+y

HallsofIvy

I don't see your point, tiny-tim. There is no coordinate system given with the problem. It perfectly legitimate to set up a coordinate system in which x is vertical.

[tex]pg\int_0^2 2\sqrt{4- x^2}(5+x)dx[/tex] is exactly the same as [tex]pg\int_0^2 2\sqrt{4- y^2}(5+y)dy[/tex]

tiny-tim

erm … the point is you can't use x and y at the same time!

(i assume that's why Nyasha was puzzled about how to integrate)

Nyasha

If my integration is correct then how do l further simplify this integration using the area of a disk of radius "R" where x=Rsinθ and dx=Rcosθ and a simple symmetry argument

tiny-tim

Make the substitution and see!

What do you get?

Nyasha

I get :


[tex]
pg\int_0^2 2\sqrt{4- R^2sin\theta^2}(5+Rsin\theta)Rcosd\theta
[/tex]


From here l don't know how to further simplify this

tiny-tim

… and now put R = 2 …

(oh, and you haven't substituted the limits)

Nyasha

[tex]

pg\int_0^2 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta

[/tex]


I am not supposed to multiple this integral by two using the symmetry argument

tiny-tim

Nyasha, you still haven't changed the limits of integration …

and what is √(4 - 4sin²θ) ?

Nyasha

[tex]


pg\int_0^4 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta


[/tex]


What am l supposed to do with √(4 - 4sin²θ) ? I'm l supposed to use some trig identity

tiny-tim

If x = 2sinθ, and x goes from 0 to 2, what does θ go from?

And you must learn your trignonometric identities … what is 1 - sin²θ?

Nyasha

[tex]


pg\int_0^2 {4\cos\theta}(5+2sin\theta)2cosd\theta


[/tex]


The limit of the integral is from 0 to π/2. I don't know how to put it into the integral using Latex source code.


[tex]

pg\int_0^2 ({20\cos\theta +8\cos\theta\sin\theta})2cosd\theta



[/tex]

Is [tex] 8\cos\theta\sin\theta} [/tex] = [tex] sin(8\theta) ?
[/tex]

tiny-tim

Hi nyasha!

(you can put π/2 into the integral with ^{\pi /2} )

You don't like undoing those brackets, do you?

It's a combination of cos²θ and cos²θsinθ.

No, sin2θ = 2sinθcosθ, but it doesn't work for sin8θ.

btw, going back to the original integral, it would have been simpler to just integrate x√(4 - x²) and √(4 - x²) separately, without substituting …

and shouldn't the limits have been -2 to +2?

Nyasha

If we integrate x√(4 - x²) and √(4 - x²) separately then do we need x=Rsinθ. The question says in order to evaluate the integral l will need the use the formula of disk of radius "R" where x=Rsinθ.

How did you get your limits to be -2 to +2 ?

[tex] pg\int_0^{\pi /2} {4\cos\theta}(5+2sin\theta)2cosd\theta



[/tex]


Is this integral correct ?

Nyasha

According to the solutions manual when l evaluate one of my integrals l should get zero. What am l doing wrong on this question ? Please help.

tiny-tim

Hi Nyasha!

Your integral (from 0 to 2) is only over the top half of the gate …

you need to integrate from -2 to 2.