A water tank in the shape of a hemisphere of radius $2\textrm{ m}$ is full of water. How much work is required to pump all the water to a level $3\textrm{ m}$ above the tank?

Solution We position an $x$-axis so the bottom of the tank is at $x=0$ and the top of the tank is at $x=2$, as illustrated in Figure 59(a).*

Figure 59

The work required to pump the water to a level $3\textrm{ m}$ above the top of the tank depends on the weight of the water and its distance from a level $3\textrm{ m}$ above the tank. The water fills the container from $x=0$ to $x=2.$

Partition the interval $[ 0, 2]$ into $n$ subintervals, each of width $\Delta x=\dfrac{2}{n}$, and choose a number $u_{i}$ in each subinterval. Now think of the water in the tank as $n$ circular layers, each of thickness $\Delta x$. As Figure 59(b) illustrates, the radius of the circular layer $u_{i}$ meters from the bottom of the tank is $\sqrt{4u_{i}-u_{i}^{2}}$. Then $$\begin{align*} \hbox{Volume }V_{i}\hbox{ of }i{\hbox{th }}\hbox{layer}&=\pi (\hbox{Radius} )^{2}\hbox{(Thickness)}\\[5pt] &=\pi \Big( \sqrt{4u_{i}-u_{i}^{2}}\Big) ^{2}\Delta x =\pi \big( 4u_{i}-u_{i}^{2}\big) \Delta x \end{align*}$$

The density of water is $\rho =1000\textrm{ kg} \textrm{/m}^{3},$ so $$\begin{align*} \hbox{Weight of }i{\hbox{th }}\hbox{layer} & =\rho gV_{i}=(1000) ( 9.8) \pi \big(4u_{i}-u_{i}^{2}\big)\Delta x\, \\ \hbox{Distance }i{\hbox{th}}\hbox{ layer is lifted} & =5-u_{i} \\ \hbox{Work done in lifting }i{\hbox{th}}\hbox{ layer} & =9800\,\pi \big(4u_{i}-u_{i}^{2}\big)(5-u_{i})~\Delta x \end{align*}$$

The work $W$ required to lift all the water from the tank to a level 3 m above the top of the tank is given by $$\begin{eqnarray*} W &=&\int_{0}^{2}9800~\pi (4x-x^{2})(5-x)~{\it dx}=9800\pi \displaystyle \int_{0}^{2}(x^{3}-9x^{2}+20x) ~{\it dx} \notag \\[5pt] &=&9800~\pi \left[ \dfrac{x^{4}}{4}-3x^{3}+10x^{2}\right] _{0}^{2}=196{,}000\pi \approx 615{,}752~\textrm{J} \end{eqnarray*}$$