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hoa [83]
3 years ago
5

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

arly from 1 meter at the south endpoint to 9 meters at the north endpoint. Find the pool volume.
Mathematics
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

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 = \frac{1}{5}

Therefore,

h(y) = \frac{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 (\frac{1}{5}y + 5)dA

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta

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

V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta

V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}

V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}

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