Question related two different ways to do solution

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Homework Statement
An upside-down square pyramid with a height of 3 m and side lengths of 8m is filled to the brim with water. How much work is needed to lift all the water out of the container through the top. (ρ = 1000 kg/m3, g = 9.8 m/s2)
Relevant Equations
Work = Force * displacement
My work:
Handwritten work on a work-done integral for water in an inverted pyramid tank: cross-section side 2a, a/h = 4/3, mass 1000 V, weight F = 9.8m, integral of 64/9 h^2 times 1000 times 9.8 times (3-h) from 0 to 3, giving about 470,400 J (470 kJ).


Answer Key (red font one) *may not be right.

My question is why is sidelength of cross section written as 8-8x/3, while my way of doing it doesn't involve that (involves suspicious similar triangle relationships)? Which answer (if any) is right here? 470 kj or 117 kj? Thank you!

Handwritten work on a work-done integral for water in an inverted pyramid tank: cross-section side 2a, a/h = 4/3, mass 1000 V, weight F = 9.8m, integral of 64/9 h^2 times 1000 times 9.8 times (3-h) from 0 to 3, giving about 470,400 J (470 kJ).


Red text on grey boxes giving a work calculation: side length 8 - 8x/3 m, cross-section volume (8 - 8x/3)^2 m^2 dx, work 9.8 m/s^2 times 1000 kg/m^3 times x times (8 - 8x/3)^2 m^2 dx, integral from 0 to 3 of that, and total W = 117 kJ.
 
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