Hydrostatic Force Problem - Calculus

HanRam
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Homework Statement


Find hydrostatic force on the vertical side of the tank that has the shape of the region bounded by the curves y=2x2, y=8. Assume that the tank is full of water.

Homework Equations


F=pgAd (force = density*gravity*area*depth)

The Attempt at a Solution


I know I need to set up and evaluate an integral, and I believe it must be evaluated from 0 to 8, since the top of the tank is at y=8. Other than this though, I have been unable to set up the integral that I need. Any help as to how the integral would be set up would be appreciated.
 
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After hours working at this problem, I think I may have figured it out, but I'm not at all sure that I did it correctly. Anyway, here is what I came up with.

\int(1000)(9.8)(8-2x^2)(2x)dx

After calculating this integral from 0 to 2 (I did this because x=0 corresponds to the bottom, and x=2 corresponds to the top), I multiplied my result by 2 to compensate for the fact that I was only integrating for the right side (0 to 2), and by multiplying by 2, that would cover the left side (-2 to 0).

My final result was 313600 N. Can anybody tell me if I went wrong somewhere? (Which I have a feeling I did).
 
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I'll give you 10 dollars for a typed response. 10 dollars. Anybody want to make 10 dollars and respond virtually? No?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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