An inverted cone (base above the vertex) is 2 m high and has a base radius of 1/2 m. If the tank is full, how much work is required to pump the water to a level 1 m above the top of the tank?

Step 1
Given, an inverted cone is 2 m high and has a base radius of ${\displaystyle \frac{1}{2}}$ m. If the tank is full, then how much work is required to pump the water to a level 1 m above the top of the tank.
Step 2: Concept Used
If force is a function F(x) of position x then in moving from x = a to x = b work done is
${\displaystyle W={\int }_a^{b}F\left(x\right)dx}$
Step 3: Calculation
Consider the water tank conical in shape. We will make a small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w.
From similarity, we will get
${\displaystyle \frac{w}{\frac{1}{2}}=\frac{2-h}{2}}$
${\displaystyle 2w=\frac{2-h}{2}}$
${\displaystyle w=\frac{2-h}{4}}$
The weight of the slice under consideration is
${\displaystyle Weight=volume\times density\times gravitationalcons\tan t}$
${\displaystyle Weight=\pi \times w^2\times dh\times 62.4(Densityofwater=62.4\frac{lb}{f{t}^3})}$
${\displaystyle Weight=62.4\pi w^{2}dh}$
Now by definition the work done is
${\displaystyle W={\int }_1^{2}dW}$
${\displaystyle ={\int }_1^2\left(62.4\pi w^{2}dh\right)h}$
${\displaystyle =62.4\pi \times \frac{1}{16}{\int }_1^2{(2-h)}^{2}hdh}$
${\displaystyle =62.4\pi \times \frac{1}{16}{\int }_1^2(4+h^2-4h)hdh}$
${\displaystyle =62.4\pi \times \frac{1}{16}{\int }_1^2(4h+h^3-4h^2)dh}$
${\displaystyle =62.4\pi \times \frac{1}{16}{\{\frac{4h^2}{2}+\frac{h^4}{4}-\frac{4h^3}{3}\}}_1^2}$
${\displaystyle =62.4\pi \times \frac{1}{16}\{\frac{16}{2}+\frac{16}{4}-\frac{32}{3}-\frac{4}{2}-\frac{1}{4}+\frac{4}{3}\}}$
${\displaystyle =62.4\pi \times \frac{1}{16}\{8+4-\frac{32}{3}-2-\frac{1}{4}+\frac{4}{3}\}}$
${\displaystyle =1.625\pi lb}$
Hence work done is ${\displaystyle 1.635\pi lb}$.