Integral Calculus Reviewer By Ricardo Asin Pdf: 54

Split it: [ W = 196000 \left[ 3\int_-3^0 \sqrt9-y^2 , dy ;-; \int_-3^0 y\sqrt9-y^2 , dy \right]. ]

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.”

Thus: [ \int_-3^0 y\sqrt9-y^2,dy = -9. ] So minus that term: ( -\int_-3^0 y\sqrt9-y^2 , dy = -(-9) = +9).

Therefore: [ W = 196000 \left( \frac27\pi4 + 9 \right) \quad \textJoules. ] Integral Calculus Reviewer By Ricardo Asin Pdf 54

He grabbed a notebook. Page 54 of his old reviewer flashed in his mind—a similar problem with a horizontal cylinder.

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

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. Split it: [ W = 196000 \left[ 3\int_-3^0

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). ]

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

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). Therefore: [ W = 196000 \left( \frac27\pi4 +

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.

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

Each slice’s thickness = (dy). Width of the slice = (2x = 2\sqrt9 - y^2). Volume of the slice = length × width × thickness = (10 \cdot 2\sqrt9 - y^2 \cdot dy = 20\sqrt9-y^2 , dy).