Thus: [ \int_-3^0 y\sqrt9-y^2,dy = -9. ] So minus that term: ( -\int_-3^0 y\sqrt9-y^2 , dy = -(-9) = +9).
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).
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). Integral Calculus Reviewer By Ricardo Asin Pdf 54
Therefore: [ W = 196000 \left( \frac27\pi4 + 9 \right) \quad \textJoules. ]
His foreman yelled, “Rico, how much work will the pump do? We need to budget for fuel!” Thus: [ \int_-3^0 y\sqrt9-y^2,dy = -9
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.
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). Width of the slice = (2x = 2\sqrt9 - y^2)
He grabbed a notebook. Page 54 of his old reviewer flashed in his mind—a similar problem with a horizontal cylinder.
I’m unable to provide a direct PDF file or a specific page (like “page 54”) from Ricardo Asin’s Integral Calculus Reviewer , as that would likely violate copyright laws. However, I can offer you an original, illustrative story inspired by the kind of integral calculus problem you might find on such a page—complete with a worked-out solution in the spirit of Asin’s teaching style. Inspired by typical problems on page 54 of many integral calculus reviewers—specifically, “Applications: Work Done in Pumping Liquid.”
So bracket = (\frac27\pi4 + 9).