Hydrostatic Force Problem In Calc. 2

[coco87]

Hey,
I've done an even problem from my book, and am not 100% sure if it's correct. It asks to find the Hydrostatic Force acting against an Area.

1. The problem statement, all variables and given/known data

A vertical dam has a semicircular gate as shown in the figure. Find the Hydrostatic force against the gate.

http://sites.google.com/site/lcphr3ak/Home/prbm14.png

w=2 \mbox{ m}

i=4 \mbox{ m}

l=12 \mbox{ m}

This figure represents a Dam. The top rectangle represents... I guess air? The w is for height, at 2 \mbox{ m}. The lower rectangle is the water. The half circle at the bottom is the Object will act as the area. The i in the circle is the diameter of the half circle, which is 4 \mbox{ m}. And last, but not least, l represents the entire length of the Dam, which is 12 \mbox{ m}.

2. Relevant equations

\int\sqrt{a^2-u^2}du \Rightarrow \frac{u}{2} \sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}{\frac{u}{a}}+C

F = pgAd

p = 1000 \mbox{ }kg/m^3

g = 9.8 \mbox{ }m/s^2

Now, I am not 100% sure I am applying these formulas correctly. F is what I'm assuming to be the Hydrostatic Force since the book states that it is, "The force exerted by the fluid on an area". And the book gives p as the density of water, and of course g as gravity.

3. The attempt at a solution

d = 12-2 = 10

x^2+y^2=(2)^2 \Rightarrow y=\sqrt{4-x^2}

A=\int_{-2}^{2}\sqrt{4-x^2}dx

\Rightarrow \frac{x}{2} \sqrt{4-x^2} + 2\sin^{-1}{\frac{x}{2}}=I

A=I(2)-I(-2)=6.283185

F=(1000)(9.8)(6.283185)(10)=615752.13 \mbox{ N}

Did I correctly apply the concept? If so, is my Arithmetic correct also?

Thanks!