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Me again
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More correctly they describe the height the fluid must be lifted- the fluid might be directed horizontally for a while but that would not require any work. In any integral, the limits of integration give the minimum and maximum values of the variable of integration. Here x represents the height any "piece of oil" or "layer of oil" must be lifted.I am still not understanding what the difference is between the limits of integration for a problem like this and the x inside the integral. Aren't they both describing the distance the fluid must flow?
Not quite. The force (weight) of each thin layer of water is actually [itex]4\pi \rho dx[/itex], the volume of a thin disk times the density.Also the [itex]4\pi \rho[/itex] makes sense here as being the Force that needs to be applied.
Again, not quite. You seem to be thinking "force times distance" as "constant in front of integral times integral". That's not the case. In both problems, x is the "distance". It just happens that the force in the second problem is a constant and so can be taken out of the integral. Of course, [itex]x\sqrt{9- (x-7)^2}[/itex], is not a "distance"- look at the units. If x is measured in feet then that is feet squared. It's actually the area of a "layer of oil". multiplying it by density and "thickness", dx, gives the weight or force needed to lift it.So taking this force times the distance (integral) gives us the work. Right? So does this mean the expression I had for work in the first equation, [itex]x\sqrt{9- (x-7)^2}[/itex], is just distance and has nothing to do with Force?
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