Integral Calculus Reviewer By Ricardo Asin Pdf 54 · Instant Download

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).

Thus: [ \int_-3^0 y\sqrt9-y^2,dy = -9. ] So minus that term: ( -\int_-3^0 y\sqrt9-y^2 , dy = -(-9) = +9). Integral Calculus Reviewer By Ricardo Asin Pdf 54

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). Each slice’s thickness = (dy)

The water filled from the bottom ((y = -3)) up to the center line ((y = 0)), so half-full. Thus: [ \int_-3^0 y\sqrt9-y^2,dy = -9

His foreman yelled, “Rico, how much work will the pump do? We need to budget for fuel!”