Ricardo Asin Pdf 54 | Integral Calculus Reviewer By

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

[ dW = \textforce \times \textdistance = 196000\sqrt9-y^2 \cdot (3 - y) , dy. ] Integral Calculus Reviewer By Ricardo Asin Pdf 54

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

His foreman yelled, “Rico, how much work will the pump do? We need to budget for fuel!” ] [ W = 196000 \int_-3^0 (3 - y)\sqrt9-y^2 , 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).

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). ]