# Homework Help: Calculus 2: Fluid Force

1. Feb 18, 2009

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

2. Feb 18, 2009

### 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

3. Feb 18, 2009

### 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 $$pg\int_0^2 2\sqrt{4- y^2}(5+y)dy$$

4. Feb 18, 2009

### 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)

5. Feb 18, 2009

### 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

6. Feb 18, 2009

### tiny-tim

Make the substitution and see!

What do you get?

7. Feb 18, 2009

### Nyasha

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

8. Feb 18, 2009

### tiny-tim

… and now put R = 2 …

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

9. Feb 18, 2009

### 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

10. Feb 18, 2009

### tiny-tim

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

and what is √(4 - 4sin2θ) ?

11. Feb 18, 2009

### Nyasha

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

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

12. Feb 18, 2009

### 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?

13. Feb 18, 2009

### Nyasha

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

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

14. Feb 18, 2009

### 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?

15. Feb 18, 2009

### 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 ?

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

Is this integral correct ?

Last edited: Feb 18, 2009
16. Feb 23, 2009

### 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.

17. Feb 23, 2009

### 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.

18. Feb 23, 2009

### Nyasha

I used the symmetry argument so that l could integrate from 0 to 2 and then multiply it by two. Is that correct ?

19. Feb 23, 2009

### tiny-tim

Well, it would be … if it wasn't for the (5 + y).

(in physical terms: the pressure on the bottom half of the gate is obviously more than on the top half)

20. Feb 23, 2009

### Nyasha

Thanks very much tiny-tim it now makes perfect sense.

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