Find the work. Application to physics

Question

A trough in a shape of a semicircular cylinder , is filled with water whose mass density is 1000 $kg/m^3$ . Suppose the water was initially filled to a depth of 3 m. Set up an integral for the work required to pump water to 10 m above the top of the trough, until water level drops 1 foot. Use y to label your slice, where y is measured from the initial water surface.

What I am confusing about this is I am not sure what does " until water level drops 1 foot" mean? What I am trying is considering about W=Fd

I sliced it horizontally, rewrite

y is initial water surface, which means y starts 1m below the top of the shape. such that work for slice = weight force of slice * distance it travelled = $[(2\sqrt{16- (1+y)^2})( 10)(density)dy][(y+11)]$

When I am triying to find total work, the upper and lower bounds because it backs to the original question that I don't understand what does " until water level drops 1 foot" mean? Does it mean I should leave 1 foot water at the last? If the previous thought is correct, what is the bounds should be?

Answer 1

Personally I would have set up $y$ differently (the top of the tank seems like a convenient reference that would simplify the area calculation, and I like my positive $y$ axis to point upward)--but the problem statement requires $y$ to be defined a certain way. So we're stuck with that.

I think this means the interval to integrate over starts at $y=0$ (the surface of the water and ends at $y=0.3048$ (that is, it "drops" from the original depth to something one foot lower; one foot is $0.3048$ meters, and a downward distance is considered positive). If the problem statement was misprinted and was really supposed to drop one meter then the integral ends at $y=1$.

Given the required definition of $y$, your formula for the work done one one slice looks OK. You can check the answer afterward by comparing it to some simpler problems, for example the work required to lift a $10$ m by $8$ m by $1$ foot slab of water to a height $11$ meters above the top of the slab (raising the entire mass an average of $11$ m plus $1/2$ foot)--your answer should be less than that, but not less than half that. That should catch errors such as missing factors of $2$ in case we both missed one.

Answer 2

"until water level drops one foot" means until the water level drops from 3 feet to 2 feet. Imagine a thin "layer" of water. Each "layer" is a rectangle with length 10 m. To get the width, write the circular end as $x^2+ y^2= 16$ so that for a given "x", $y= \sqrt{16- x^2}$. The area of that rectangle is $10\sqrt{16- x^2}$ square meters. Taking the thinkness to be "dx", its volume is $10\sqrt{16- x^2}dx$. Multiply that by the density of water (I am going to call it "$\delta$") to get the mass of each "layer", $10\delta\sqrt{16- x^2}dx$. Finally, since "x" is measured from the top of the trough (a diameter of the circle) to pump the water out of the trough you must lift it a distance "x" m. Work is "force times distance" so the work to lift that "layer" is $10\delta x\sqrt{16- x^2}dx$. Initially, the water has "height" 3 feet, 1 foot below the top of the trough and "drops one foot" going down to height 2 feet so to 2 feet below the top. Integrate from x= 1 to x= 2: $10\delta\int_1^2 x\sqrt{16- x^2}dx$