Integral Calculus Reviewer By Ricardo Asin Pdf 54 Apr 2026
His foreman yelled, “Rico, how much work will the pump do? We need to budget for fuel!”
[ dW = \textforce \times \textdistance = 196000\sqrt9-y^2 \cdot (3 - y) , 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.
Rico remembered Ricardo Asin’s golden rule: “For work problems, slice it, find the force on each slice, multiply by the distance that slice travels, then integrate.” Integral Calculus Reviewer By Ricardo Asin Pdf 54
Engineer Rico, a young civil engineer fresh out of review, stared at a cylindrical water tank on a construction site. The tank lay on its side—a common setup for fuel or water storage. Its radius was 3 meters, and its length was 10 meters. The tank was half-full of water, and he needed to pump all the water out through a valve at the very top of the tank.
He grabbed a notebook. Page 54 of his old reviewer flashed in his mind—a similar problem with a horizontal cylinder.
First integral: (\int \sqrt9-y^2, dy) is a standard semicircle area formula. From (y=-3) to (0), it’s a quarter circle of radius 3. Area of quarter circle = (\frac14\pi (3^2) = \frac9\pi4). So (3 \times \frac9\pi4 = \frac27\pi4). His foreman yelled, “Rico, how much work will the pump do
The valve is at (y = 3). A slice at position (y) must be lifted vertically from (y) up to 3. Distance = (3 - y).
So bracket = (\frac27\pi4 + 9).
Thus: [ \int_-3^0 y\sqrt9-y^2,dy = -9. ] So minus that term: ( -\int_-3^0 y\sqrt9-y^2 , dy = -(-9) = +9). Rico remembered Ricardo Asin’s golden rule: “For work
[ W = 196000 \int_-3^0 (3 - y)\sqrt9-y^2 , dy. ]
The water filled from the bottom ((y = -3)) up to the center line ((y = 0)), so half-full.
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.
Therefore: [ W = 196000 \left( \frac27\pi4 + 9 \right) \quad \textJoules. ]