CalcRef

Applications of Integrals · Work & Force

Work (Pumping)

W=∫abρgA(y)(h−y) dyW = \int_a^b \rho g A(y)(h - y)\, dy

Work to pump fluid out of a tank. A(y) is the cross-sectional area at height y, h is the height to which fluid is pumped.

Worked examples

Find the work to pump water out of a full cylindrical tank of radius 2m and height 5m.
  1. A(y) = π(2²) = 4π. Pump to height h = 5.
  2. W = ∫₀⁵ 1000(9.8)(4π)(5-y) dy = 39200π ∫₀⁵ (5-y) dy
  3. = 39200π[5y - y²/2]₀⁵ = 39200π(25-12.5) = 39200π(12.5)

Answer: 490,000π ≈ 1,539,380 J

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