#### To determine

**(a)**

**To find:**

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

#### Answer

80003π.

#### Explanation

**1) Concept:**

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

**2) Given:**

The height of the paraboloid is 4 ft.

The radius of the paraboloid is 4 ft.

**3) Calculations:**

The vertex of the parabola is at origin and passing through point 4,4.

Therefore, the equation of the parabola is y=ax2.

Substitute values of (x, y) from the given point

4=a·42=16a

Solving for a,

a=14

Hence,y=14x2

That is,x2=4y

Therefore, x=2y.

Each circular disk has radius 2y and is moved 4-y ft.

And water weighs 62.5 lb/ft3

Therefore, the work required to pump the water out of the tank is

W=∫0462.5πr2h dy=∫0462.5π2y24-y dy

W=250π∫044-yy dy=250π∫044y-y2dy

After integrating,

W=250π2y2-y3340

Applying the Fundamental Theorem of Calculus,

W=250π32-643=250π323=80003π

**Conclusion:**

The work required to pump the water out of the tank is 80003π ft.lb.

#### To determine

(b)

**To find:**

The depth of water remaining in the tank after 4000 ft·lb of work has been done.

#### Answer

2.06 ft.

#### Explanation

**1) Given:**

W=4000 ft·lb

**2) Calculations:**

Let the depth of remaining water be h.

Therefore, from part (a),

W=250π∫h44-yy dy

After integrating,

W=250π2y2-y334h

Applying the Fundamental Theorem of Calculus,

W=250π2(4)2-433-250π2h2-h33

Work done is given to be 4000.

Therefore,

4000=250π32-643-250π2h2-h33

4000250π=323-2h2+h33

16π=323-2h2+h33

h3-6h2+32=48π

By using a graphing calculator,h is ≈2.06.

Therefore, the depth of water remaining in the tank after 4000ft·lb of work has been done is ≈2.06 ft.

**Conclusion:**

The depth of water remaining in the tank after 4000 ft·lb of work has been done is ≈2.06 ft.