## Calculus 2: Fluid Force

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= $$2\sqrt{4-x^2}$$ dy

Pressure = pg(5+x)

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

2. Relevant equations

3. The attempt at a solution
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 Quote by Nyasha F= pg∫ $$2\sqrt{4-x^2}$$ (5+x) dy
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
 Recognitions: Gold Member Science Advisor Staff Emeritus 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. $$pg\int_0^2 2\sqrt{4- x^2}(5+x)dx$$ is exactly the same as $$pg\int_0^2 2\sqrt{4- y^2}(5+y)dy$$

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## Calculus 2: Fluid Force

 Quote by 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. $$pg\int_0^2 2\sqrt{4- x^2}(5+x)dx$$ is exactly the same as $$pg\int_0^2 2\sqrt{4- y^2}(5+y)dy$$
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)

 Quote by 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. $$pg\int_0^2 2\sqrt{4- x^2}(5+x)dx$$ is exactly the same as $$pg\int_0^2 2\sqrt{4- y^2}(5+y)dy$$
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

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 Quote by 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
Make the substitution and see!

What do you get?

 Quote by tiny-tim Make the substitution and see! What do you get?
I get :

$$pg\int_0^2 2\sqrt{4- R^2sin\theta^2}(5+Rsin\theta)Rcosd\theta$$

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

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 Quote by Nyasha $$pg\int_0^2 2\sqrt{4- R^2sin\theta^2}(5+Rsin\theta)Rcosd\theta$$
… and now put R = 2 …

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

 Quote by tiny-tim … and now put R = 2 … (oh, and you haven't substituted the limits)

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

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

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 Quote by Nyasha $$pg\int_0^2 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta$$ I am not supposed to multiple this integral by two using the symmetry argument
Nyasha, you still haven't changed the limits of integration …

and what is √(4 - 4sin2θ) ?

 Quote by tiny-tim Nyasha, you still haven't changed the limits of integration … and what is √(4 - 4sin2θ) ?

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

What am l supposed to do with √(4 - 4sin2θ) ? I'm l supposed to use some trig identity
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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?

 Quote by 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?

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

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.

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

Is $$8\cos\theta\sin\theta}$$ = $$sin(8\theta) ?$$

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 Quote by nyasha1 $$pg\int_0^2 {4\cos\theta}(5+2sin\theta)2cosd\theta$$ 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. $$pg\int_0^2 ({20\cos\theta +8\cos\theta\sin\theta})2cosd\theta$$ Is $$8\cos\theta\sin\theta}$$ = $$sin(8\theta) ?$$
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?

 Quote by 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?
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 ?

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

Is this integral correct ?
 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.
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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.
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