To calculate the amount of work required to pump the liquid from a conical tank, we can break down the task into smaller steps. The formula for work in this context is:

$W = \int_{a}^{b} F(x) \, dx$

Where $F(x)$ is the force needed to lift a thin slice of liquid at a height $x$, and $[a, b]$ represents the bounds of integration (the range over which the liquid is lifted).

Step 1: Setting up the Problem

The cone has a height of 32 ft, and it is filled with liquid up to a height of 29 ft. The liquid has a weight density of 64.4 lb/ft³.

Key Details:

• Radius of cone = 8 ft
• Height of cone = 32 ft
• Liquid is filled to a height of 29 ft
• Weight density = 64.4 lb/ft³

We want to calculate how much work is needed to pump the liquid from a height of 29 ft to the rim of the cone, which is at 32 ft.

Step 2: Understanding the Geometry of the Cone

For a cone, the radius at any height $h$ is proportional to $h$. The radius at height $h$ can be given by:

$r(h) = \frac{8}{32} \times h = \frac{h}{4}$

Thus, the radius of the liquid at a height $h$ (where $h$ ranges from 0 to 29 ft) will be $\frac{h}{4}$.

Step 3: Calculating the Work

Work is the force required to lift a thin layer of liquid over the distance it needs to be pumped. We can calculate the volume of a thin slice of liquid at height $h$ (with thickness $dh$) and then determine the force required to lift that slice.

The volume $dV$ of a thin slice of liquid at height $h$ is the volume of a disk, which is:

$dV = \pi r^2 dh = \pi \left(\frac{h}{4}\right)^2 dh = \frac{\pi h^2}{16} dh$

The weight of this slice is given by:

$dW = \text{Weight density} \times dV = 64.4 \times \frac{\pi h^2}{16} dh = \frac{64.4 \pi h^2}{16} dh$

To find the total work, we need to multiply the weight of the slice by the distance it has to be lifted. The liquid is being pumped from height $h$ to the top of the cone, which is at 32 ft. Therefore, the distance to lift the slice is $32 - h$.

The work to lift a thin slice is:

$dW = \left( \frac{64.4 \pi h^2}{16} \right) \times (32 - h) dh$

Now, integrate this expression from $h = 3$ ft (since the liquid is filled to a height of 29 ft) to $h = 29$ ft to find the total work:

$W = \int_3^{29} \frac{64.4 \pi h^2 (32 - h)}{16} dh$

Step 4: Solving the Integral

Let's calculate the work step by step:

1. Simplify the expression:

$W = \frac{64.4 \pi}{16} \int_3^{29} h^2 (32 - h) dh$

2. Distribute $h^2$ inside the parentheses:

$W = \frac{64.4 \pi}{16} \int_3^{29} (32h^2 - h^3) dh$

3. Integrate each term separately:

$\int 32h^2 dh = \frac{32h^3}{3}, \quad \int h^3 dh = \frac{h^4}{4}$

Thus,

$W = \frac{64.4 \pi}{16} \left[ \frac{32h^3}{3} - \frac{h^4}{4} \right]_3^{29}$

4. Evaluate the integral at $h = 29$ and $h = 3$:

$W = \frac{64.4 \pi}{16} \left( \left( \frac{32(29)^3}{3} - \frac{(29)^4}{4} \right) - \left( \frac{32(3)^3}{3} - \frac{(3)^4}{4} \right) \right)$

You can compute the exact value using a calculator or further simplify. After evaluating this expression, you will find the total work required to pump the liquid to the rim of the cone.

Let me know if you'd like help with the exact numerical evaluation or any further breakdown of steps!