A spherical tank is given. The tank has radius 6 m and spot coming out of the top with height 2 m. A tank is half full of oil that has a density of 900 kg/m3. Find the work W (in J) required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Round your answer to the nearest whole number.)

To find the work \( W \) required to pump the oil out of the spout, we can use the following method based on calculus and the physics of work. The work done to move a small volume of oil \( dV \) a distance \( y \) is given by:

\[
dW = \rho g y dV
\]

where:
- \( \rho \) is the density of the oil (\( \rho = 900 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \)),
- \( y \) is the distance the oil must be lifted,
- \( dV \) is a small volume of oil.

### Step 1: Geometry of the tank

The tank is spherical with a radius \( R = 6 \, \text{m} \). It is half full, so the oil fills the lower hemisphere of the tank.

We place the center of the sphere at the origin of a coordinate system, so the equation of the sphere is:

\[
x^2 + y^2 = 36
\]

where \( y \) is measured from the center of the sphere.

### Step 2: Volume of a thin slice of oil

Consider a thin horizontal slice of oil at height \( y \) with thickness \( dy \). The volume of this slice, \( dV \), is the area of a circular cross-section times the thickness \( dy \). The radius of the circular cross-section at height \( y \) is \( \sqrt{36 - y^2} \), so the area is \( \pi (36 - y^2) \). Thus, the volume of the slice is:

\[
dV = \pi (36 - y^2) \, dy
\]

### Step 3: Distance the oil is lifted

Since the spout is 2 m above the top of the tank (which is at \( y = 6 \)), the distance the oil must be lifted to reach the spout from height \( y \) is:

\[
\text{Distance lifted} = 8 - y
\]

### Step 4: Work done to lift a slice of oil

The work done to lift a small slice of oil is:

\[
dW = \rho g (8 - y) \, dV = \rho g (8 - y) \, \pi (36 - y^2) \, dy
\]

Substituting \( \rho = 900 \, \text{kg/m}^3 \) and \( g = 9.8 \, \text{m/s}^2 \), we get:

\[
dW = 900 \cdot 9.8 \cdot \pi (8 - y)(36 - y^2) \, dy
\]

### Step 5: Integrating the work over the range of the oil

Since the oil fills the bottom half of the tank, the oil ranges from \( y = -6 \) to \( y = 0 \) (the midpoint of the tank). Therefore, the total work is the integral of \( dW \) from \( y = -6 \) to \( y = 0 \):

\[
W = 900 \cdot 9.8 \cdot \pi \int_{-6}^{0} (8 - y)(36 - y^2) \, dy
\]

### Step 6: Solving the integral

Now we compute the integral:

\[
W = 900 \cdot 9.8 \cdot \pi \left[ \int_{-6}^{0} (8(36 - y^2) - y(36 - y^2)) \, dy \right]
\]

Breaking it into two parts:

\[
W = 900 \cdot 9.8 \cdot \pi \left[ \int_{-6}^{0} 8(36 - y^2) \, dy - \int_{-6}^{0} y(36 - y^2) \, dy \right]
\]

We can now calculate these integrals.

It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can guide you through manually solving the integral step-by-step, or we can discuss alternative ways to approach this problem.

Would you like to proceed with any of these options? Here are some related questions you may also be interested in:

1. How is the work-energy theorem applied in physics problems?
2. What is the role of density in fluid mechanics?
3. How do integrals apply to calculating work in variable force systems?
4. What is the formula for the volume of a spherical segment?
5. How can calculus be used in real-world engineering problems?

**Tip**: When calculating work in physics, always make sure that the force, displacement, and direction of the force are clearly defined for each small segment.

Learn how to calculate the work required to pump oil from a spherical tank with a radius of 6 m. This problem involves calculus concepts such as integration, the work-energy principle, and the geometry of spherical tanks. Suitable for college-level students studying physics and calculus.

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