#### To determine

**To find:**

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

#### Answer

2.55*106 J

#### Explanation

**1) Concept:**

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

And weight=gravity·volume

**2) Given:**

**3) Calculation:**

Let the circular disk slice of water have thickness ∆x m .

It is lying at the coordinate x with radius r=9-x2 m

The given tank is half full of oil, therefore, the lower limit changes from -3 to 0 and the density of oil is 900 kg/m3

Volume of the slice of oil =πr2 ∆x m3=π9-x22∆x m3=π9-x2∆x m3

Therefore, the weight of the slice is =g·density·volume=9.8×900π9-x2∆x N

This weight is lifted by the pump about x+4m.

Therefore, the work needed to pump the water out of the spout is

=9.8×900x+4π9-x2∆x J

Therefore, the total work reqiured is

W=∫abf(x)dx

W≈∫039.8×900πx+49-x2dx

=9.8×900·π∫03x9-x2+4(9-x2)dx

=9.8×900·π∫039x-x3+36-4x2dx

=9.8×900·π92x2-14x4+36x-43x303

=9.8×900·π9232-1434+36(3)-43(3)3

=9.8×900·π·(92.25)

=2556141 J

**Conclusion:**

The work required to pump the water out of the spout is ≈2.55*106 J