# Homework Help: Find the total hydrostatic force on the face of a semi-circular dam

1. Aug 13, 2011

### GreenPrint

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

Find the total hydrostatic force on the face of a semi-circular dam with radius of 20 m, when its reservoir is full of water. The diameter of the semicircle is the top of the dam.

2. Relevant equations

3. The attempt at a solution

the width at any point would be 2sqrt(400-y^2) from x^2+y^2 = r^2
the density would be 1000 kg/m^3, this is the rough estimate for the density of water I am told to use in the class
gravity is 9.8 m/s^2, for some reason we don't use 9.81 in the class

I got F= 19600*integral[0,20] (20-y)sqrt(400-y^2)dy

were 19600 is just density times gravity times 2 = 1000*9.8*2
20-y is the depth at any given point y

I have no idea how to evaluate the indefinite integral
integral (20-y)sqrt(400-y^2)dy
when I tried doing so I spent a good amount of time trying to evaluate it using a trig sub but ended up doing something terribly wrong for some reason I got
integral (20-y)sqrt(400-y^2)dy = (ysqrt(400-y^2))/40 - 10 cos^(-1)(y/20) + sqrt(400-y^2)/3 - (y^2 sqrt(400-y^2))/1200 + c
I got frustrated with it and just plugged into my calculator
19600*integral[0,20] (20-y)sqrt(400-y^2)dy

This question however has reappeared on a practice test in which the problem was solved differently but I think it may have been solved wrong, the answer key my professor made had
F = integral[0,20] 9800sqrt(400-h^2)2h dh
as the integral to be evaluated were they just used h for the depth instead of (20-h). I thought you had to include this though in the integral and not just h because you want the depth at any point of any of the horizontal slices and not just h

So I thought that the problem was solved wrong on the answer key but sense my integral I had a really tough time evaluating by hand 19600*integral[0,20] (20-y)sqrt(400-y^2)dy I think I may have done something wrong and I was just suppose to use y instead of (20-y).

What's the correct integral to solve this problem? Thanks for any help!!!

2. Aug 13, 2011

### upsidedowntop

I get the same integral as your professor.

I worked it out in terms of depth, so if P(h) is the pressure as a function of depth:
P(h) = gdh [g=gravity, d=density, h=depth].

If w(h) is the width of the dam as a function of depth, then
w(h) = 2(r^2 -h^2)^(1/2) [r=radius of the semi-circle]

The force on a region of the dam, dh is pressure times area, so
dF = P(h)w(d)dh
so
F = $\int_0^r P(h)w(d)dh$
or
F = $\int_0^r (9.8)(1000)(h)(2)(400-h^2)^{1/2}dh$