Numerically: (27\pi/4 \approx 21.20575), plus 9 = 30.20575. Multiply by 196000: (W \approx 5,920,327) Joules, or about (5.92) MJ.
He grabbed a notebook. Page 54 of his old reviewer flashed in his mind—a similar problem with a horizontal cylinder. Integral Calculus Reviewer By Ricardo Asin Pdf 54
Weight of the slice = volume × density of water (1000 kg/m³ × 9.8 m/s² = 9800 N/m³): [ dF = 9800 \cdot 20\sqrt9-y^2 , dy = 196000\sqrt9-y^2 , dy \quad \text(Newtons). ] Numerically: (27\pi/4 \approx 21
His foreman yelled, “Rico, how much work will the pump do? We need to budget for fuel!” Page 54 of his old reviewer flashed in
Split it: [ W = 196000 \left[ 3\int_-3^0 \sqrt9-y^2 , dy ;-; \int_-3^0 y\sqrt9-y^2 , dy \right]. ]
Second integral: Let (u = 9-y^2), (du = -2y,dy), so (y,dy = -\frac12du). [ \int_-3^0 y\sqrt9-y^2,dy = \int_y=-3^0 \sqrtu \left(-\frac12 du\right) = -\frac12 \int_u=0^9 u^1/2 du = -\frac12 \cdot \frac23 u^3/2 \Big| 0^9 = -\frac13 (27) = -9. ] But careful with limits: actually (y=-3 \to u=0), (y=0 \to u=9), so (\int 0^9 \sqrtu (-\frac12 du) = -\frac12 \cdot \frac23 [27-0] = -9). Yes.
Therefore: [ W = 196000 \left( \frac27\pi4 + 9 \right) \quad \textJoules. ]