# Chapter 6 - Applications of the Integral - 6.5 Work and Energy - Exercises - Page 317: 22

$9800 c (\dfrac{ah^2}{3}+\dfrac{bh^2}{6}) \ J$

#### Work Step by Step

The force of one layer is equal to: $\ Force= \ Mass \times \ gravity = 9800 c (a+\dfrac{b-a}{h})^2 \Delta y \ N$ Therefore, the work done can be computed as: $W=\int_{0}^{h} 9800 c (a+\dfrac{b-a}{h})^2 \Delta y \ N \\=\int_{0}^{h} 9800 c (ah+(b-a) y \ dy -9800 c \int_{0}^{h} (ay+\dfrac{(b-a)}{h} y^2) \ dy \\= 9800 c [ah^2+\dfrac{h^2(b-a)}{2}-\dfrac{ah^2}{2}-\dfrac{(b-a) h^2}{3} ] \\= 9800 c (\dfrac{ah^2}{3}+\dfrac{bh^2}{6}) \ J$

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