A conical​ container, oriented such that its vertex is at the​ bottom, has radius 8 ft and height 32 ft. It is filled to a height of 30 ft of a liquid weighing 62.4 lb divided by ft cubed. How much work will it take to pump the contents to the​ rim? How much work will it take to pump the liquid to a level of 2 ft above the​ cone's rim? not rounded


### Given Information:

- The container is a **cone** with:
  - **Radius (R) = 8 ft**
  - **Height (H) = 32 ft**
- The liquid is filled to a height of **30 ft**.
- The weight density of the liquid is **62.4 lb/ft³**.
- We need to find the work to:
  1. Pump the liquid to the **rim of the cone**.
  2. Pump the liquid to a level **2 ft above the cone's rim**.

### Work to Pump Liquid to the Rim:

1. **Volume element** at height \( y \):

   The cone is similar at any horizontal cross-section. The radius of the cone at any height \( y \) is proportional to \( y \), so we use similar triangles:

   \[
   r(y) = \frac{8}{32} y = \frac{y}{4}
   \]

   The area of a cross-section at height \( y \) is:

   \[
   A(y) = \pi r(y)^2 = \pi \left( \frac{y}{4} \right)^2 = \frac{\pi y^2}{16}
   \]

2. **Volume of a thin slice** of liquid at height \( y \) with thickness \( dy \) is:

   \[
   dV = A(y) dy = \frac{\pi y^2}{16} dy
   \]

3. **Mass of the slice** (since mass = density × volume):

   \[
   dM = \rho dV = 62.4 \cdot \frac{\pi y^2}{16} dy
   \]

4. **Work to pump the liquid** at height \( y \) to the rim (height = 32 ft):

   The work to pump this slice is the weight of the slice times the distance it needs to be moved (32 − y):

   \[
   dW = \left( 62.4 \cdot \frac{\pi y^2}{16} \right) \cdot (32 - y) dy
   \]

5. **Total work** to pump the liquid to the rim (integrating from \( y = 0 \) to \( y = 30 \)):

   \[
   W = \int_0^{30} 62.4 \cdot \frac{\pi y^2}{16} \cdot (32 - y) \, dy
   \]

### Work to Pump Liquid 2 ft Above the Rim:

To pump the liquid to **34 ft** (2 ft above the rim), the only difference is the distance to pump the liquid, which is now \( 34 - y \).

6. **Total work** to pump the liquid to 34 ft:

   \[
   W_{\text{above}} = \int_0^{30} 62.4 \cdot \frac{\pi y^2}{16} \cdot (34 - y) \, dy
   \]

I'll now compute both integrals to get the exact values.The exact work required to pump the liquid:

1. **To the rim of the cone (32 ft)** is \( 333450 \pi \) ft-lb.
2. **To 2 ft above the rim (34 ft)** is \( 403650 \pi \) ft-lb.

If you prefer, I can also express these values numerically.

Would you like to see more details, or do you have any questions?

Here are some related questions for further exploration:

1. How does the shape of the container affect the work needed to pump the liquid?
2. What would the work be if the liquid was filled to a different height?
3. How would the work change if the cone had a different radius or height?
4. How does the weight density of the liquid influence the work done?
5. How does pumping the liquid to different heights above the cone impact the work required?

**Tip**: For objects with varying cross-sections like cones, the work calculation requires integrating both volume and distance.

A conical container, oriented such that its vertex is at the bottom, has radius 8 ft and height 32 ft. It is filled to a height of 30 ft of a liquid weighing 62.4 lb divided by ft cubed. How much work will it take to pump the contents to the rim? How much work will it take to pump the liquid to a level of 2 ft above the cone's rim?

This problem demonstrates how to calculate the work required to pump liquid from a conical container (radius 8 ft, height 32 ft) filled to a height of 30 ft. It covers concepts of calculus, including volume integration and the work-energy theorem. The solution explores two scenarios: pumping the liquid to the rim and to a height 2 ft above the rim.

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