00:01
Hello students, given the radius of the cylinder r is equal to 15 feet and height of the cylinder h is equal to 6 feet and height of the water filled that is hw is equal to 2 by 3 h that is equal to 2 by 3 into 6 gives the value 4 feet is the 2 3 of the filled water and we need to find the work required to pump all the water over up the rim.
00:35
Using the density of the water, we know that density of the water is equal to 62 .4 lb for per feet square.
00:44
Now, work done is given by w is equal to integral a to b row pi r square.
00:55
This implies we have w is equal to integral 4 to 6, 62 .4 into pi into radius is 50 .4 .5.
01:06
Square h let be x into d x this implies we have 62 .4 pi 15 square into integral 4 to 6 x into d x this implies we have 1 440 pi into x square by 2 within the limits 4 to 6 applying the limits and simplifying therefore we get work done is equal to 440 856 feet lb.
01:48
Next a cone -shaped water reservoir is 28 feet in diameter across the top given and depth 21 feet.
02:01
If the reservoir is filled to a depth of 14 feet we need to find the work done to require to pump all the water to top of the reservoir.
02:09
Let consider a symmetry of the reservoir of 14 feet depth 21 feet and x is the radius and y is the dy is the thickness of the reservoir and y is the distance from the apex of conical reservoir now therefore volume is given by dv is equal to pi x square d .y.
02:38
Let be equation 1...