A spherical tank with radius 3 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 2 m spout at the top. Find the work W required to pump the liquid out of the spout. (Use 9.8 m/s2 for g.)

To find the work \( W \) required to pump the liquid out of the spout, we need to apply the concept of work in a fluid system. Here's the step-by-step breakdown of the solution.

### Step 1: Concept of Work
The formula for work \( W \) to pump a small volume of liquid \( dV \) from a depth \( y \) is:

\[
dW = \text{force} \times \text{distance} = (\rho \cdot g \cdot dV) \times \text{distance}
\]

where:
- \( \rho = 900 \, \text{kg/m}^3 \) is the density of the liquid,
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( dV \) is the small volume element of liquid,
- The distance is the height the liquid needs to be pumped, which depends on the depth from the spout.

### Step 2: Volume Element \( dV \)
We will slice the sphere horizontally and consider a thin slice of thickness \( dy \) at a depth \( y \) from the center of the sphere. The radius of this slice is \( r(y) \), and the volume element of this slice is:

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

From the equation of a circle, \( r(y) \) is given by:

\[
r(y) = \sqrt{9 - y^2}
\]

The limits for \( y \) will range from \( -3 \) to \( 0 \) since the tank is half full and the radius of the tank is 3 m.

### Step 3: Work Done to Pump Each Slice
The distance each slice must be lifted to the spout is \( 5 - y \), since the spout is 2 m above the top of the tank (3 m + 2 m = 5 m total).

Thus, the work to pump a slice is:

\[
dW = \rho g \pi r(y)^2 (5 - y) \, dy = 900 \times 9.8 \times \pi (9 - y^2)(5 - y) \, dy
\]

### Step 4: Total Work
Now we integrate this expression over the range \( y = -3 \) to \( y = 0 \):

\[
W = 900 \times 9.8 \times \pi \int_{-3}^{0} (9 - y^2)(5 - y) \, dy
\]

### Step 5: Solve the Integral
Let's now solve the integral:

\[
I = \int_{-3}^{0} (9 - y^2)(5 - y) \, dy
\]

First, expand the integrand:

\[
(9 - y^2)(5 - y) = 45 - 9y - 5y^2 + y^3
\]

So, the integral becomes:

\[
I = \int_{-3}^{0} (45 - 9y - 5y^2 + y^3) \, dy
\]

Integrating term by term:

\[
\int (45) \, dy = 45y
\]
\[
\int (-9y) \, dy = -\frac{9y^2}{2}
\]
\[
\int (-5y^2) \, dy = -\frac{5y^3}{3}
\]
\[
\int (y^3) \, dy = \frac{y^4}{4}
\]

Evaluating these from \( y = -3 \) to \( y = 0 \):

\[
I = \left[ 45y - \frac{9y^2}{2} - \frac{5y^3}{3} + \frac{y^4}{4} \right]_{-3}^{0}
\]

At \( y = 0 \), all terms are zero, so we only need to evaluate at \( y = -3 \):

\[
I = 45(-3) - \frac{9(-3)^2}{2} - \frac{5(-3)^3}{3} + \frac{(-3)^4}{4}
\]
\[
= -135 - \frac{81}{2} - \frac{-135}{3} + \frac{81}{4}
\]
\[
= -135 - 40.5 + 45 + 20.25
\]
\[
= -110.25
\]

### Step 6: Final Work Calculation
Now, substituting this into the expression for \( W \):

\[
W = 900 \times 9.8 \times \pi \times (-110.25)
\]

Let's calculate this.The work required to pump the liquid out of the spout is approximately \( W = -3,054,900 \, \text{J} \). 

However, since work can't be negative in this context, the magnitude of the work is \( 3,054,900 \, \text{J} \).

Would you like more details on this solution, or do you have any further questions?

Here are 5 related questions:
1. How would the work change if the tank were full instead of half-full?
2. How does the shape of the tank affect the work required to pump out the liquid?
3. How would the required work change if the density of the liquid were different?
4. What happens to the work calculation if the spout height is changed?
5. How do different gravitational environments (like on the Moon) impact the work needed?

**Tip:** When solving problems involving work and fluid, always consider the shape of the container and how the distance each fluid element must be moved changes with depth.

This problem explores the work required to pump liquid out of a half-full spherical tank with radius 3 m and liquid density 900 kg/m3. Using physics and calculus concepts, we calculate the total work needed with detailed integration steps. Suitable for undergraduate students.

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