Integral Calculus Reviewer By Ricardo Asin Pdf 54 ★ Direct & Direct

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.”

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

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). Integral Calculus Reviewer By Ricardo Asin Pdf 54

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

[ W = 196000 \int_-3^0 (3 - y)\sqrt9-y^2 , dy. ] I’m unable to provide a direct PDF file

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.

His foreman yelled, “Rico, how much work will the pump do? We need to budget for fuel!” A slice at position (y) must be lifted

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