Integral Calculus Reviewer By Ricardo Asin Pdf 54 | Recent × HONEST REVIEW |

The water filled from the bottom ((y = -3)) up to the center line ((y = 0)), so half-full.

Rico told the foreman, “About 5.9 megajoules.” The foreman nodded, and the pump worked perfectly—thanks to a slice, a distance, and an integral from page 54 of Ricardo Asin’s reviewer.

He placed the center of the circular cross-section at (0,0). The circle’s equation: (x^2 + y^2 = 9). The tank’s length (into the page) was 10 m. The valve was at the top of the circle, at (y = 3).

Thus: [ \int_-3^0 y\sqrt9-y^2,dy = -9. ] So minus that term: ( -\int_-3^0 y\sqrt9-y^2 , dy = -(-9) = +9). Integral Calculus Reviewer By Ricardo Asin Pdf 54

[ W = 196000 \int_-3^0 (3 - y)\sqrt9-y^2 , dy. ]

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.

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. The water filled from the bottom ((y =

Therefore: [ W = 196000 \left( \frac27\pi4 + 9 \right) \quad \textJoules. ]

The valve is at (y = 3). A slice at position (y) must be lifted vertically from (y) up to 3. Distance = (3 - y).

His foreman yelled, “Rico, how much work will the pump do? We need to budget for fuel!” The circle’s equation: (x^2 + y^2 = 9)

[ dW = \textforce \times \textdistance = 196000\sqrt9-y^2 \cdot (3 - y) , dy. ]

So bracket = (\frac27\pi4 + 9).