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).
The valve is at (y = 3). A slice at position (y) must be lifted vertically from (y) up to 3. Distance = (3 - y). 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]. ] He placed the center of the circular cross-section at (0,0)
[ W = 196000 \int_-3^0 (3 - y)\sqrt9-y^2 , dy. ] The valve was at the top of the circle, at (y = 3)
The water filled from the bottom ((y = -3)) up to the center line ((y = 0)), so half-full.
Numerically: (27\pi/4 \approx 21.20575), plus 9 = 30.20575. Multiply by 196000: (W \approx 5,920,327) Joules, or about (5.92) MJ.