Hydrostatic Force Problem!!!

• Jun 23rd 2008, 04:14 PM
elocin
Hydrostatic Force Problem!!!
A large tank is designed with vertical ends in the shape of the region between the curves y = x^2/2 and y = 13, measured in feet. Find the hydrostatic force on one end of the tank if it is filled to a depth of 8 ft with gasoline. (Assume the gasoline's density is 42.0 lb/ft^3

Sooo, I am not sure what I'm dealing with, but I think we are integrating from 0 to 8 and pull out the 42 of the integral, of something, I think the curves are confusing me. I know that hydrostatic force is area*pressure, and I think the pressure is 42 times x. But I don't know what the area is. Please help!
• Jun 23rd 2008, 04:45 PM
galactus
The sides are just a parabola with depth 13. Did you graph it. That helps to see.

$\displaystyle depth = h(y)=13-y$

$\displaystyle length = L(y)=2\sqrt{2y}$

$\displaystyle 42\int_{0}^{8}2(13-y)(\sqrt{2y})dy$

$\displaystyle 84\int_{0}^{8}\left(13\sqrt{2y}-\sqrt{2}y^{\frac{3}{2}}\right)dy$
• Jun 23rd 2008, 05:01 PM
elocin
Oh I see, is the square root over the 2y for both terms?
• Jun 23rd 2008, 05:02 PM
galactus
No, just the first one.
• Jun 23rd 2008, 05:26 PM
elocin
Okay, so I got 7239 and it said it was wrong, I probably did something wrong, once again. I integrated, from 0 to 13 (and multiplying the whole thing by 84) the integral of (13*sq rt of 2y)-((sq rt of 2)*y^3/2) dy and that got me to, after integrating, 13sqrt of 2 over 2)*y^3/2 (-) 2*sqrt of 2 over 5)*y^5/2). That's probably really hard to read! I'm sorry!
• Jun 23rd 2008, 05:35 PM
galactus
My mistake. That is a typo. You integrate from 0 to 8.
• Jun 23rd 2008, 05:41 PM
elocin
Oh that's okay!! I got 8870 integrating from 0 to 8, but it's still wrong! Hmm I will double check my work again.
• Jun 23rd 2008, 06:01 PM
elocin
Oh and you did write to integrate from 0 to 8, I the one who made the mistake!! But I still have 8870.