- #1

- 1,196

- 0

## Homework Statement

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.

## Homework Equations

## 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

and got about 70883765.35 N

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!!!