Posted by Soroban on August 15, 2002 at 23:42:41:
In Reply to: Application of Integration posted by Irene on August 13, 2002 at 15:17:03:
: Please can anyone help me? I'm stuck at this problem. I've tried it for hours and still can't find the solution. please help!!!
: Find the fluid force on the vertical side of the tank, where the dimensions are given in feet. Assume that the tank is full of water:
: A parabola 4 for height, and 4 for length. Eq: y = x^2
Irene,
I assume the tank has that parabola for a vertical end-plate. This is hard to explain
without a graph, but I'll try. (Keep the graph of y = x^2 in mind.)
We have the Integral (from y = 0 to y = 4) of: (density-weight)x(depth)x(width) dy
The density of water is taken to be 62.5
The depth is y
The width is 2x or 2y^(1/2)
The integral is: INT (62.5)(y)2y^(1/2) dy
= 125 INT y^(3/2) dy
= 50y^(5/2) [from 0 to 4]
and I get a result of 1600.
[Or am I way off base? Anyone?]