# How to pump water out of a trough to ground level?

Here is the question: A trough shaped tank is located 3 meters underground (The top of the tank is 3 meters away from ground level.) The dimensions of the tank are: height h = 5 meters, length c = 10 meters, base width a = 4 meters and top width b = 8 meters. If the tank is completely full, find the work done in pumping all the water out of the tank to ground level.

I think I have the integral set up to pump the water from out of the top. Could someone please explain how to set up the integral to pump the water out to the ground level? The weight of water is 9800 N/m^3.

• Sorry, typo. I meant a trough, basically a trapezoidal prism. – Nick Salling Mar 28 '19 at 16:57

Given some height $$x$$ over the bottom of the tank, take a layer which is $$\Delta x$$ thick. The volume of that layer is $$\Delta x\times A(x)$$ where $$A(x)$$ is the area of the top side of the layer. It is 10 meters long and $$4+\frac{4x}5$$ wide. So the the volume is $$10\left(4+\frac{4x}{5}\right)\Delta x$$ The total weight of the layer is $$9800$$ times this, and the work needed to lift the layer up to the ground is equal to the weight times the distance up from the layer to the ground, which is $$8-x$$.
Putting this all together, we get that the total work to lift the layer is $$9800\cdot 10(8-x)\left(4+\frac{4x}{5}\right)\Delta x$$ Adding together all these layers, and making the layers thinner and thinner results in the integral $$\int_0^59800\cdot 10(8-x)\left(4+\frac{4x}{5}\right)\,dx$$