Solutions - 0120-01

(a) (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) Setup:

Coordinate system: set $y=0$ at the top of the tank, increasing downwards.

Horizontal slice of the tank: disk of radius $r$ at depth $y$, satisfies:

$$r^2+y^2={10}^2,A(y)=\pi r^2=\pi ({10}^2-y^2)$$

Distance pumped up (add $2$ for the spigot): $h(y)=y+2$

Thus:

$$W\gg \gg {\int }_0^{10}\rho g\pi (y+2)({10}^2-y^2)dy$$

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(b) (1) Change upper bound, top of water at $y=4$:

$$W\gg \gg {\int }_4^{10}\rho g\pi (y+2)({10}^2-y^2)dy$$