Hydrostatic force, half circle

[kevtimc]

Homework Statement

Find the hydrostatic force on any side of a bottom half-circle with a 6 m diameter with the top 1 m above water level.

http://img377.imageshack.us/my.php?image=45095885io3.png"

Homework Equations

P = 1000 * 9.8 (mass density of water) * displacement
A = l * w
F = P * A

The Attempt at a Solution

Pressure = density * gravity * displacement
displacement = (2 - y)

x = 2\sqrt{9 - y^2}
A = x * delta y (for any rectangular strip)

I'm concerned with my displacement; would it be 2 - y?
Would I be integrating from -1 to -3?

EDIT: I looked over the problem again, I think what we really have is a semi circle with a radius of 2 at sea level.
The area would be 2 * sqrt(4 - y^2) and the displacement just y, I think :)

 

[HallsofIvy]

No, you do not have a semi-circle with a radius of 2: the problem specifically says a "half-circle with a 6 m diameter" so you have a semi-circle with a radius of 3 m, not two. The fact that the water only comes up to 1 m below the top means that the depth for any y (with x= \sqrt{9- y^2}) is y-1. You will want to integrate from y= 1 to y= 3.
(Note: I've taken my positive y-axis downward to avoid negatives.)

 

[kevtimc]

No, you do not have a semi-circle with a radius of 2: the problem specifically says a "half-circle with a 6 m diameter" so you have a semi-circle with a radius of 3 m, not two. The fact that the water only comes up to 1 m below the top means that the depth for any y (with x= \sqrt{9- y^2}) is y-1. You will want to integrate from y= 1 to y= 3.
(Note: I've taken my positive y-axis downward to avoid negatives.)

Thanks, but I'm having trouble with the second integral in the problem:

2pg \int (y-1) * \sqrt{9-y^2}

For: 2pg\int \sqrt{9-y^2} from 1 - 3 (when integrals are separated)

doesn't that evaluate to 2pg \int9*cos^2\theta ?

I'm not getting the correct answer when I evaluate this integral.

 

[lakeccrunner]

I'm having a problem with the exact same problem. The answer should be 6.7x10⁴N, but I have yet to get this answer.

I evaluate 2pg\int(y-1)\sqrt{9-y^{2}}dy from 1 to 3

And get: 2pg\inty\sqrt{9-y^{2}dy - 2pg\int\sqrt{9-y^{2}}dy from 1 to 3.

I then do u substitution for the first integral using u=9-y^{2}, du=-2y, -\frac{1}{2}du=ydy

For the second integral, since the are is a semicircle, I use 2pg\pi3^{2}

So I have: -pg\inty\sqrt{u}du - 2pg\pi3^{2} from 1 to 3.

Then I take the antiderive of the first integral to get

-pg\frac{2}{3}(9-y^{2})^{\frac{3}{2}} - 2pg\pi3^{2}

From here I plug in the numbers and get:

-147832-277088 which give me -424920 I then divide it by 2, since it says to find the Hydrostatic Force on one side of the semicircle to get 2.1x10^{5}

I think I'm making a little mistake somewhere, but I can't seem to figure out where. Any help would be appreciated.