A right circular conical tank has radius $3$ m at the top and height $6$ m. It is filled with water up to the brim. Find the total work (in kJ) required to pump all the water out through a pipe at the top of the tank, assuming $\rho =1000$ kg/m${}^3$ and $g=9.8$ m/s${}^2$.

Step-by-step solution

1. Express the cone's radius as a function of depth $y$: at a distance $y$ from the vertex, radius $r=\frac{1}{2}y$ (since radius doubles over height).
2. The volume of a thin slice at depth $y$ and thickness $dy$ is $dV=\pi r^2,dy=\pi {\left(\frac{y}{2}\right)}^{2}dy=\frac{\pi y^2}{4},dy$.
3. The weight of this slice is $dW=\rho g,dV=1000\times 9.8\times \frac{\pi y^2}{4},dy=2450\pi y^2,dy$.
4. The slice must be lifted a distance $(6-y)$ m to reach the top. Thus the work to lift this slice is $dmathcalW=2450pi{y}^2(6-y),dy$.
5. Integrate $y$ from $0$ to $6$: $\mathcal{W}=2450piin{t}_0^6{y}^2(6-y),dy$.
6. Expand inside the integral: $y^2(6-y)=6y^2-y^3$.
7. Compute the integral: ${\int }_0^6(6y^2-y^3),dy=6\frac{y^3}{3}{|}_0^6-\frac{y^4}{4}{|}_0^6=12\times 216-\frac{1296}{4}=2592-324=2268$.
8. Multiply by $2450\pi$: $\mathcal{W}=2450\pi \times 2268=2450\times 2268\pi$.
9. Compute numeric value: $2450\times 2268=5,556,600$. So $\mathcal{W}=5,556,600\pi$ joules.
10. Using $piapprox3.1416$, $\mathcal{W}approx17,454,000$ J.
11. Convert to kJ: $\mathcal{W}approx17,454$ kJ.
12. Round to one decimal: $\boxed{117.3}$ kJ.

Concept tested: Work to pump liquid from a tank

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked