A hemispherical tank of radius $5\text{m}$ is filled with water. Find the total work (in kJ) required to pump all the water out through a pipe located at the top of the tank. Take $\rho =1000{\text{kg/m}}^3$ and $g=9.8{\text{m/s}}^2$.

Step-by-step solution

1. Place the origin at the centre of the hemisphere; the top opening is at $y=5$ metres.
2. Consider a thin horizontal slice of water at height $y$ (measured from the centre) with thickness $dy$.
3. Radius of the slice is $r=\sqrt{5^2-y^2}$ by the equation of the circle $x^2+y^2=5^2$.
4. Volume of the slice: $dV=\pi r^{2}dy=\pi (5^2-y^2)dy$.
5. Mass of the slice: $dm=\rho dV=1000\pi (25-y^2)dy$.
6. Distance each slice must be lifted to the top opening is $d=5-y$ (since the top is at $y=5$).
7. Work done on the slice: $dW=dmgd=1000\pi (25-y^2)g(5-y)dy$.
8. Substitute $g=9.8$: $dW=9800\pi (25-y^2)(5-y)dy$.
9. Integrate $y$ from $-5$ (bottom of hemisphere) to $5$ (top): $W=9800\pi {\int }_{-5}^5(25-y^2)(5-y)dy$.
10. Expand the integrand: $(25-y^2)(5-y)=125-25y-5y^2+y^3$.
11. Integrate term‑by‑term: ${\int }_{-5}^{5}125dy=125\times 10=1250$, ${\int }_{-5}^5(-25y)dy=0$ (odd function), ${\int }_{-5}^5(-5y^2)dy=-5\times \frac{2\cdot 5^3}{3}=-\frac{1250}{3}$, ${\int }_{-5}^5{y}^{3}dy=0$.
12. Thus the integral equals $1250-\frac{1250}{3}=\frac{2500}{3}$.
13. Multiply by the constant: $W=9800\pi \times \frac{2500}{3}=\frac{24,500,000\pi }{3}$ joules.
14. Compute numerically: $\frac{24.5\times 10^6\pi }{3}\approx 8.1667\times 10^6\pi \approx 2.565\times 10^7$ J.
15. Convert to kilojoules: $W\approx 2.565\times 10^4$ kJ.
16. Round to three significant figures: $W\approx 2.57\times 10^4\text{kJ}$.
17. State the final answer.

Concept tested: Work done in pumping fluid out of a tank

Published 10 September 2026 · How this solution was prepared and checked