#### To determine

**To find:**

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

#### Answer

4.43×106 J

#### Explanation

**1) Concept:**

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

weight=gravity·volume

Area of circular disk is πr2m3

**2) Given:**

**3) Calculation:**

From the given figure observe that the given tank is spherical in shape.

The radius of the tank below the center is -3m and from the tap to the center, the height is -4m.Let centre be orgin and let the vertical down the centre and passing through the north and south pole of sphere be x-axis. The x-axis is positive downwards.

Consider circular disk slice of water having thickness ∆x m lying at the coordinate x. Its’ radius r=9-x2

Volume of the disk of water =πr2 ∆x m3=π9-x22∆x m3=π9-x2∆x m3

Therefore, the weight of the slice is =g.density·volume=9.8×103π9-x2∆x N

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

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

=9.8×103x+4π9-x2∆x J

Therefore, the total work reqiured is

W=∫abf(x)dx

W≈∫-339.8×103πx+49-x2dx

=9.8×103·π∫-33x9-x2+4(9-x2)dx

In the integral ∫-33x9-x2dx the function x9-x2 is odd since the degree of x is 3 which is an odd number, therefore, the value of ∫-33x9-x2dx=0

And in the integral ∫-334(9-x2)dx the function 4(9-x2) is even since the degree of x is 2 which is even therefore,

∫-334(9-x2)dx=2∫034(9-x2)dx

=9.8×103·π· 2∫034(9-x2)dx

=9.8×103·π·2·49x-13x303

=9.8×103·π· 89(3)-1333

=9.8×103·π· 818

=4431.168×103

=4.43×106 J

**Conclusion:**

The work required to pump the water out of the spout is ≈4.43×106 J