A tank is is half full of oil that has a density of 900kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 for g and 3.14 for . Round your answer to three significant digits.)

Image of the tank has been attached.

From online sources, we will use radius r = 9 meters with a spout of height h = 3 meters.

W = 75800 KJ

Explanation:

Let's assume that we inscribe a cylinder which is x units from the center of the sphere and also has a height denoted by ∆x.

Now, At the center of the sphere with radius r, the equation of the sphere is given as:

x² + y² + z² = r².

Now, at the point location where the cylinder is x units from the center point, the width(y) of the sphere will be gotten as:

0² + y² + x² = 9²

y² + x² = 81

y² = 81 - x²

Thus;

y = √(81 - x²)

So y = r is the radius of the cylinder.

Thus; volume of the cylinder is given as:

V = πr²h

∆V = π(81 - x²)∆x

Integrating, we have;

V = π(81 - x²)

The weight of the water is given as:

F = mg

We know that density is;

ρ = m/V

So, m = ρV

Thus;

F = ρVg

F = (900)[π(81 - x²)](9.8)

F = 8820π(81 - x²)

Work needed to move this water up is;

W = Fd

The water will move up x to units to the center which is 9 units up to the top of the sphere, and 3 units to get to the top of the spout. Thus, d = 9 + 3 + x = 12 + x

Thus;

W = 8820π(81 - x²)(12 + x).

Due to the fact that x varies from -9 to 0 at the bottom half of the sphere, the required work will be;:

W = ∫8820π(81 - x²)(9 + x)dx (from x = -9 to 0)

Using online integral calculator, we have;

W = 8820π × 2733.75

W = 75749.061 KJ

To 3 significant figures = 75800 KJ

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