Expert: Scotto - 2/17/2009

Question
QUESTION: 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.

(HINT: In order to evaluate the integral(s) you obtain, it may be useful to use the formula for the area of a disk of radius "R" where x=Rsinθ  and a simple symmetry argument. )

ANSWER: WE need to integrate the width * depth from one side of the
cylinder to the other.  This means we need to do the following:

Itegrate 2(5+x)RsinΘ from x = 0 to x = 4.
Note that sinΘ is a function of x.
Notes
It is sinΘ = x/R and substitute this in.
The variable R is a constant - it is 2.
Both sides were taken care of since we multiplied by 2 in front.
The problem comes down to integrating x to a power.
The x+5 is used since that is the depth of the water.


---------- FOLLOW-UP ----------

QUESTION:  l am not supposed to integrate 4∫(5+x)(cosΘ) since x=RsinΘ  then dx is going to be RcosΘ. R=2 so l pull it out of the integral and then multiply it by two using the symmetry argument.

Why are you integrating from x=o to x=4 ? How did you get that limit.

Answer
The depth of the opening has radius 2 meters, which means it's diameter is 4 meters.

Since x is our variable in question, we want sin(Θ) expressed as
an expression of x.  There is no dΘ involved, so we don't have
to worry about changing variable in terms of dx.  We just need
to convert all of the variables to x first before we can integrate.