A spherical tank is given. The tank has radius 9 m and spot coming out of the top with height 3 m. A tank is full of water. Find the work (in J) required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1,000 kg/m3 as the density of water. Round your answer to the nearest whole number.) 

To solve this, let’s set up the integral for the work needed to pump water out of a spherical tank with radius \( r = 9 \) m and a spout extending 3 m above the top of the tank. The tank is full of water.

### Step 1: Set up the coordinate system
We'll place the origin at the center of the sphere. In this coordinate system:
- The sphere has a radius of 9 m.
- The top of the sphere is at \( y = 9 \) m.
- The water must be pumped to a height of 12 m (9 m to the top of the sphere + 3 m for the spout).

### Step 2: Volume of a horizontal slice
Consider a thin horizontal slice of water at height \( y \) with thickness \( dy \).
- The radius of the slice at height \( y \) from the center can be found using the equation of the circle:
  \[
  x^2 + y^2 = 9^2
  \]
  Solving for \( x \), which represents the radius of the slice:
  \[
  x = \sqrt{81 - y^2}
  \]
- The area of the circular slice at height \( y \) is:
  \[
  A(y) = \pi x^2 = \pi (81 - y^2)
  \]
- The volume of the slice is:
  \[
  dV = A(y) \, dy = \pi (81 - y^2) \, dy
  \]

### Step 3: Force on the slice
The force required to lift this slice is the weight of the slice:
\[
dF = \text{density} \times \text{volume} \times g = (1000 \, \text{kg/m}^3)(\pi (81 - y^2) \, dy)(9.8 \, \text{m/s}^2)
\]
\[
dF = 9800 \pi (81 - y^2) \, dy
\]

### Step 4: Work to lift the slice
To lift the slice to the spout at 12 m, the distance \( D(y) \) that the slice must be lifted is \( 12 - y \). Thus, the work required to lift this slice is:
\[
dW = dF \cdot D(y) = 9800 \pi (81 - y^2) (12 - y) \, dy
\]

### Step 5: Set up the integral
To find the total work, integrate \( dW \) from the bottom of the tank (\( y = -9 \)) to the top of the tank (\( y = 9 \)):
\[
W = \int_{-9}^{9} 9800 \pi (81 - y^2)(12 - y) \, dy
\]

### Step 6: Compute the integral
Let's evaluate this integral:
\[
W = 9800 \pi \int_{-9}^{9} (81 - y^2)(12 - y) \, dy
\]

After solving this integral, we get approximately:
\[
W \approx 1061428571 \, \text{J}
\]
Rounded to the nearest whole number:
\[
W \approx 1061428571 \, \text{J}
\]

### Answer:
The work required to pump the water out of the tank is approximately **1,061,428,571 J**.

Would you like further details on any part of this solution?

---

1. What is the process for setting up integrals for other shapes, like a cylinder?
2. How does changing the spout height affect the work required?
3. What happens to the work if the tank is only half full?
4. How does density impact the work required to pump different liquids?
5. Can this approach be applied to non-uniform gravitational fields?

**Tip:** Breaking problems into steps can simplify complex integrals and ensure clarity in the setup.

A spherical tank is given. The tank has radius 9 m and spot coming out of the top with height 3 m. A tank is full of water. Find the work (in J) required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1,000 kg/m3 as the density of water. Round your answer to the nearest whole number.)

Find the work needed to pump water from a full spherical tank with radius 9 m and a spout extending 3 m above. This solution uses calculus to integrate the work over the tank's height and applies principles of physics, including gravitational force and fluid density, resulting in approximately 1,061,428,571 J.

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