A circular swimming pool has a diameter of 24 ft, the sides...

Name: A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? Use the fact that water weighs 62.5 lb/ft³ and g=1. Note: Just set up the integral. Do not evaluate it.
Uploaded: 2022-05-24T23:33:00-08:00
Duration: 5 min 49 s
Channel: Mary Wakumoto
Description: A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? Use the fact that water weighs 62.5 lb/ft³ and g=1. Note: Just set up the integral. Do not evaluate it.

A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? Use the fact that water weighs 62.5 lb/ft³ and g=1. Note: Just set up the integral. Do not evaluate it.

Transcript

00:01 All right, so here we have a water tank.

00:04 It's not drawn to scale, but the radius is 12 feet.

00:09 The height, it's actually, i think a swimming pool.

00:11 The height is five feet, but we have four feet of water in the tank.

00:15 And we're told that the weight of water is 62 .5 pounds per cubic foot.

00:22 And our goal is to find the work it would take to basically drain the swimming pool, to drain the water tank.

00:28 So the idea is that what we're going to do is, um, imagine if we slice the water into a lot of little flat cylinder, little like flat discs, and each of these discs have a weight associated with it, and we're going to lift them up a certain distance.

00:49 So basically we need to find the volume of this disk.

00:54 We have a little height, d -y.

00:57 We have the volume of our little disc, d -volume.

01:01 Is going to be pi r square d y.

01:04 So that gives me the little volume piece.

01:07 And then it needs to be lifted up against gravity.

01:11 The top one only has to go up one foot, but the bottom one has to go up five feet.

01:18 So we need to take that into account.

01:21 Okay, so basically, in general, the work that's done is equal to our weight, our mg, our weight, times the height we need to go.

01:33 And notice that the top one only goes.

01:38 So if my, so what i'm going to do is kind of make a system here where y will be going this way.

01:46 So as y gets bigger, we have further to go.

01:49 So if y is zero, we only go one meter.

01:53 And but if y is four, then we're going to go up.

02:02 So five, so not meters, but feet.

02:04 So with y is zero, we only go up one feet.

02:07 But if y is down at the four meter mark, then we're going to go up five feet in terms of the work that we do.

02:15 So this w is work.

02:18 So this is work.

02:21 Okay.

02:22 So, but for us, we're going to add a basic, our total work is going to be made up of adding up all the work contributions to our little disk...

Question

A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? Use the fact that water weighs 62.5 lb/ft³ and g=1. Note: Just set up the integral. Do not evaluate it.

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