#### To determine

**To compute:**

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

#### Answer

486864π J≈1.53×106 J.

#### Explanation

**1) Given:**

Diameter of cylinder is 4m.

Height of cylinder is 5m.

Density of oil is 920kg/m3.

**2) Calculations:**

The volume of a horizontal slice of cooking oil having ∆x meter thickness is πr2∆x.

That is,π22∆x m3=4π∆x m3

Density of oil is 920kg/m3.

Therefore,

Mass of slice is 920×4π∆x=3680 π∆x

Weight is 9.8×3680π∆x=36064π∆x N

Now, the height of the cylinder is 5m, and the height of the spout is 1m, so the total height is 6m.

Also, the cylinder is filled to a level of 3m.

Thus, to calculate the total work needed to empty the tank, integrate the weight from 3m to 6m.

Therefore,

W=limn→∞∑i=1n36064πxi*∆x=∫3636064πx dx

After integrating,

W=36064 πx2263

Applying the Fundamental Theorem of Calculus,

W=36064π262-32

=18032π36-9

=486864π=486864×3.14=1528752.96J

≈1.53×106J.

**Conclusion:**

The work required to pump the oil out of the tank is 486864π J≈1.53×106 J.