#### To determine

**To find:**

The work required to pump the water out of the spout.

#### Answer

1.04×105ft-lb

#### Explanation

**1) Concept:**

The work is calculated by using the formula W=∫abf(x)dx

**2) Given:**

**3) Calculation:**

Consider the horizontal slice (shaped like a disk) of water ∆x ft thick and lying at coordinate x.Calculate the radius of the slice by using a similar triangle method

In triangle ∆PQR & ∆TRM

Let x be the depth below the spout

d8-x=38

d=38(8-x)

Therefore,

r=3+d=3+38(8-x)

Simplifying this,

The radius of the disk is r=3816-x ft The volume the disk of water is πr2∆x=π·96416-x2∆x ft3. It weighs about 62.59π6416-x2∆x lb and must be lifted x ft by the pump

Therefore, the work needed to pump the disk water out of the spout is 62.5x9π6416-x2∆x ft-lb The total work required is

W=∫abf(x)dx

W≈∫0862.5x9π6416-x2dx

W≈62.59π64∫08x16-x2dx

=62.59π64∫08(256x-32x2+x3) dx

=62.59π64256x22-32x33+x44 08

=62.59π64128x2-32x33+x44 08

=62.59π6412882-32833+844

=33,000π= 33,000*3.14

≈1.04×105 ft-lb

**Conclusion:**

The work required to pump the water out of the spout is ≈1.04×105 ft-lb