Question

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases linearly from 1 meter at the south endpoint to 9 meters at the north endpoint. Find the pool volume.

Answer 1

Volume is
`2000\pi\ m^(3)`

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

`c = (1)/(5)`

Therefore,

`h(y) = (1)/(5)y + 5`

Now, the Volume of the pool is given by:

`V = \int h(y)dA`

where

A =
`r\theta`

`A = rdr\ d\theta`

Thus

`V = \int ((1)/(5)y + 5)dA`

`V = \int_(0)^(2\pi)\int_(0)^(20) ((1)/(5)rsin\theta + 5) rdr\ d\theta`

`V = \int_(0)^(2\pi)\int_(0)^(20) ((1)/(5)r^(2)sin\theta + 5r}) dr\ d\theta`

`V = \int_(0)^(2\pi) ((1)/(15)20^(3)sin\theta + 1000) d\theta`

`V = [- 533.33cos\theta + 1000\theta]_(0)^(2\pi)`

`V = 0 + 2\pi * 1000 = 2000\pi\ m^(3)`