. A tank full of water has the shape of a paraboloid of revolution. The height of the tank is
four feet and the radius at the top is four feet. Find the work required to pump the water out
of the tank. Use 62.4 lb/ft3 as the weight-density of water. Round your answer to the nearest
tenth.

The problem involves finding the work required to pump water out of a tank that is shaped like a paraboloid of revolution. Here's how we can break it down:

### Step 1: General Setup

We need to find the work \( W \) to pump water from a paraboloid tank. The formula for work to pump water out of a tank is given by:

\[
W = \int_{a}^{b} \rho \cdot g(x) \cdot A(x) \cdot h(x) \, dx
\]

Where:
- \( \rho \) is the weight-density of water (in this case, \( 62.4 \, \text{lb/ft}^3 \)),
- \( A(x) \) is the cross-sectional area of the tank at height \( x \),
- \( h(x) \) is the distance the water must be lifted from height \( x \) to the top,
- \( g(x) \) is the depth at which the water is located.

### Step 2: Determine the Geometry of the Tank

The tank has a parabolic shape, meaning the equation of the paraboloid is in the form:

\[
r(x) = k \cdot x
\]

We know:
- The height of the tank is 4 feet,
- The radius at the top is 4 feet.

Thus, the equation for the radius as a function of the height is:

\[
r(x) = x
\]

This comes from the fact that the radius and height scale proportionally.

### Step 3: Area of the Cross Section

The area of a circular cross section at height \( x \) is given by:

\[
A(x) = \pi r(x)^2 = \pi x^2
\]

### Step 4: Work Formula

At a height \( x \), the water needs to be lifted a distance of \( (4 - x) \) feet (since the total height of the tank is 4 feet). The volume of a thin slice of water at height \( x \) is \( A(x) \, dx \), and the weight of that slice is \( \rho \cdot A(x) \, dx \).

The work required to lift that slice to the top is:

\[
dW = \rho \cdot A(x) \cdot (4 - x) \, dx
\]

Substituting \( A(x) = \pi x^2 \) and \( \rho = 62.4 \, \text{lb/ft}^3 \), we get:

\[
dW = 62.4 \cdot \pi x^2 \cdot (4 - x) \, dx
\]

### Step 5: Total Work

To find the total work, integrate from \( x = 0 \) to \( x = 4 \) (the height of the tank):

\[
W = \int_0^4 62.4 \pi x^2 (4 - x) \, dx
\]

### Step 6: Solve the Integral

Let's compute the integral:

\[
W = 62.4 \pi \int_0^4 x^2 (4 - x) \, dx
\]

Expanding the integrand:

\[
W = 62.4 \pi \int_0^4 (4x^2 - x^3) \, dx
\]

Now, integrate term by term:

\[
W = 62.4 \pi \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_0^4
\]

Evaluate at the limits \( x = 0 \) and \( x = 4 \):

\[
W = 62.4 \pi \left[ \frac{4(4)^3}{3} - \frac{(4)^4}{4} \right]
\]

\[
W = 62.4 \pi \left[ \frac{4(64)}{3} - \frac{256}{4} \right]
\]

\[
W = 62.4 \pi \left[ \frac{256}{3} - 64 \right]
\]

Simplify:

\[
W = 62.4 \pi \left[ \frac{256 - 192}{3} \right] = 62.4 \pi \left[ \frac{64}{3} \right]
\]

\[
W = 62.4 \pi \cdot \frac{64}{3}
\]

\[
W = 62.4 \cdot \frac{64\pi}{3}
\]

\[
W \approx 62.4 \cdot 67.02 \approx 4181.25 \, \text{foot-pounds}
\]

### Final Answer

The work required to pump all the water out of the tank is approximately **4181.3 foot-pounds**.

---

Would you like any further clarifications or details?

### Related Questions:
1. How does the shape of the tank affect the amount of work required?
2. How would the work change if the tank had a different height-to-radius ratio?
3. What if the water had a different weight-density, such as in a different fluid?
4. How would the work change if the tank was not filled to the top?
5. What if the water was pumped to a point above the top of the tank?

**Tip**: Always check the units when working with integrals involving physical quantities, such as force and work.

A tank full of water has the shape of a paraboloid of revolution. The height of the tank is four feet and the radius at the top is four feet. Find the work required to pump the water out of the tank. Use 62.4 lb/ft3 as the weight-density of water. Round your answer to the nearest tenth.

Learn how to calculate the work required to pump water from a paraboloid-shaped tank. This detailed solution applies integration and the work-energy principle, using the weight-density of water (62.4 lb/ft³) and a tank with a height of 4 feet and a radius of 4 feet. Suitable for undergraduate calculus students.

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