Solutions - 0120-05

(1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg \int h(y)\rho gdV\\ \\ \gg \gg \int \rho gh(y)A(y)dy\end{array}$$

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(2) Integrand components:

Option 1: Set $y=0$ at the base, going up.

Take a cross-sectional slice with a vertical plane. This intersects the surface of the pyramid in a triangle whose width $w(y)$ is the side length of the square (the horizontal cross section) at height $y$.

$$w(y)=230+\frac{0-230}{146}y$$

Note that $A(y)=w{(y)}^2$. So we have:

$$W={\int }_0^{146}\rho gy{(230-\frac{230}{146}y)}^{2}dy$$

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Option 2:

Set $y=0$ at the vertex, going down.

$$w(y)=0+\frac{230-0}{146}y$$

And:

$$W={\int }_0^{146}\rho g(146-y){\left(\frac{230}{146}y\right)}^{2}dy$$