Expert: Scotto - 2/24/2009

Question
QUESTION: Scotto can you please tell me what l am doing wrong on this question is has been giving me headaches for some time. According to the solutions manual one when l evaluate my integrals one of them should get zero.

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.



ANSWER: What I have used int() for is the integral (not integer less than).

2pg*int(0 to 2)(2√(9-y²)(5+y)dy, breaks into
2pg*int(0 to 2)(10√(9-y²)dy + 2pg*int(0 to 2)(2y√(9-y²)dy.

For the first integral, what is u = sin(é) used for?

Take a right triangle, Θ as the angle, 3 as the hypoteneuse,
y as the far side, and √(9-y²) as the near side.
sin(Θ) is known to be y/3 and cos(é) is √(9-y²)/3.
cos(Θ)d(é) = dy/3.

When y is 0, Θ is 0.  Wheyn y is 2, cos(Θ) is (√5)/3.

It becomes
2pg*int(0 to arcos(√5/3))10*3cos(Θ)*3cos(Θ)dΘ =
180pg*int(0 to arcos(√5/3))cos²(Θ)dΘ.
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By the way, where you have 9-y², shouldn't it be 4-y²?
That way, Θ would go from 0 to π/2.  I need to see the
original question again.  This applies to the second
intgral as well.
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On the second integral, let u=9-y², and du = -2y dy.
This quickly becomes -2pg*int(0 to ??)√u du, which is
-2pg*(√u)³2/3 = -(4/3)√u at -u=?? + u=0


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

QUESTION: Here is the original 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.

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I have attached another image which is more clear.U= 2sinΘ is what l should substitute for the first integral in order to evaluate it.

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The problem l am having is that the solutions manual says one of the integral when evaluated will be zero but when l evaluate my integrals none of them is zero  

Answer
I would appreciate it if the writing non the paper
was a little bit darker.  It's almost all readable,
but it could be better.

For example, the 9's and 4's look to similar.

If the radius is 2, then it should be x²+y²=4,
and that might be what's written there.

On the next line, you have x² = 9 - y² - is that where your error is?
Where does the x² = 9 - y² come from?
The radius is 2, so the equation for radius gives us x² = 4 - y².

Try putting 4 into both integrals in place of the 9.

Also, the integrals should go from -2 to 2 since we are taking the
midpoint as the center of the graph.  This integral should still be
times 2, since it is only the right half of the circle if you take
the circle minus the far side, which is the negative circle.  That's
like subtracting a negative of the same thing, so the result is
multiplying by 2.

I believe that one of the 2nd integral should be 0 if this is done.
Try graphing the function and you will see that there is the same
amount of area above and below the axis.