Find the force (in Newtons) exerted on the dam

[beanryu]

The Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.1x^2 and below the line y=180 . (Here, distances are measured in meters.) The water level can be assumed to be at the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure. Water has a density of 1000kg/m^3. Since this is a metric problem, you must multiply the mass to be lifted by 9.8m/s^2 to convert to a weight.

First give the integrand expressed in terms of y (the width of the dam must be expressed as a function of y).

area * pressure = force
pressure = 1000 * 9.8 * 180

area = integral of (width * dy) from 180 to 0
width = sqrt(y*4/0.1)

force = (1000 * 9.8 * 180) * integral of (sqrt(y*4/0.1)*dy) from 180 to 0

this is wrong... why?

okay i got it...

(1000 * 9.8 * 180) * integral of (sqrt(y*4/0.1)(180-y)*dy) from 180 to 0

 

[HallsofIvy]

The Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.1x^2 and below the line y=180 . (Here, distances are measured in meters.) The water level can be assumed to be at the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure. Water has a density of 1000kg/m^3. Since this is a metric problem, you must multiply the mass to be lifted by 9.8m/s^2 to convert to a weight.

First give the integrand expressed in terms of y (the width of the dam must be expressed as a function of y).

area * pressure = force
pressure = 1000 * 9.8 * 180

No, the height of a column of water above that particular point on the dam is not "180", it is 180- y, the height of the top of the dam above the particular y value.

area = integral of (width * dy) from 180 to 0
width = sqrt(y*4/0.1)

force = (1000 * 9.8 * 180) * integral of (sqrt(y*4/0.1)*dy) from 180 to 0

this is wrong... why?

okay i got it...

(1000 * 9.8 * 180) * integral of (sqrt(y*4/0.1)(180-y)*dy) from 180 to 0