A conical​ container, oriented such that its vertex is at the​ bottom, has radius 5 ft and height 20 ft. It is filled to a height of 17 ft of a liquid weighing 60.6 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 3 ft above the​ cone's rim?

To solve this problem, we will calculate the work required in two parts:

1. **Work to pump the liquid to the rim of the cone.**
2. **Work to pump the liquid to a level 3 feet above the cone's rim.**

### Step 1: Understanding the Problem

We have:
- **Radius of the cone (R):** 5 ft
- **Height of the cone (H):** 20 ft
- **Height of the liquid (h):** 17 ft
- **Weight density of the liquid (w):** 60.6 lb/ft³

The cone is oriented with its vertex at the bottom, meaning the top surface of the liquid is horizontal.

### Step 2: Work to Pump the Liquid to the Rim

The formula for work is given by:

\[
\text{Work} = \int_{a}^{b} F(y) \, dy
\]

Where:
- \( y \) is the variable height from the bottom of the cone.
- \( F(y) \) is the force required to lift a thin slice of the liquid at height \( y \) to the top.

#### Force \( F(y) \)
To find \( F(y) \), consider a thin slice of the liquid at height \( y \) with thickness \( dy \).

The volume of this slice \( dV \) is the area of the circular cross-section at height \( y \) times \( dy \). The radius of the circular cross-section at height \( y \) is given by similar triangles:

\[
\frac{r(y)}{y} = \frac{R}{H} \implies r(y) = \frac{R}{H} \cdot y = \frac{5}{20} \cdot y = \frac{y}{4}
\]

The area of the slice at height \( y \):

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

Thus, the volume of the slice \( dV \) is:

\[
dV = \frac{\pi y^2}{16} \, dy
\]

The weight of the slice:

\[
\text{Weight} = w \cdot dV = 60.6 \cdot \frac{\pi y^2}{16} \, dy
\]

#### Work to Pump to the Rim
The distance that each slice needs to be lifted to reach the rim is \( 20 - y \). So the work required for each slice is:

\[
dW = \text{Weight} \times \text{Distance} = 60.6 \cdot \frac{\pi y^2}{16} \cdot (20 - y) \, dy
\]

Now, integrate this expression from \( y = 0 \) to \( y = 17 \):

\[
\text{Work}_1 = \int_0^{17} 60.6 \cdot \frac{\pi y^2}{16} \cdot (20 - y) \, dy
\]

### Step 3: Work to Pump the Liquid 3 Feet Above the Rim

The distance to lift the liquid to 3 feet above the rim is \( 23 - y \). The work required for this is:

\[
dW = 60.6 \cdot \frac{\pi y^2}{16} \cdot (23 - y) \, dy
\]

Now, integrate this expression from \( y = 0 \) to \( y = 17 \):

\[
\text{Work}_2 = \int_0^{17} 60.6 \cdot \frac{\pi y^2}{16} \cdot (23 - y) \, dy
\]

### Step 4: Calculation

Let’s compute the work:

\[
\text{Work}_1 = \frac{60.6 \pi}{16} \int_0^{17} y^2(20 - y) \, dy
\]

\[
\text{Work}_2 = \frac{60.6 \pi}{16} \int_0^{17} y^2(23 - y) \, dy
\]

We can compute these integrals to find the total work required.

I'll calculate these integrals now.The work required to pump the liquid:

1. **To the rim of the cone** is approximately **141,275.23 ft·lb**.
2. **To a level 3 feet above the rim** is approximately **199,733.95 ft·lb**.

Would you like a detailed breakdown of the integration steps or any further explanations?

Here are some related questions you might find helpful:
1. How would the work change if the cone were filled to the top?
2. What if the weight density of the liquid was different?
3. How does the work required change if the cone's dimensions are altered?
4. What if the liquid height was 10 feet instead of 17 feet?
5. How would the work differ if the cone was inverted?

**Tip:** When dealing with work in lifting fluids, remember that the shape of the container and the distance each slice of fluid is lifted both contribute significantly to the total work required.

A conical container, oriented such that its vertex is at the bottom, has radius 5 ft and height 20 ft. It is filled to a height of 17 ft of a liquid weighing 60.6 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 3 ft above the cone's rim?

Calculate the work required to pump liquid from a conical tank with a 5 ft radius and 20 ft height, filled to 17 ft, to both the rim and 3 feet above the rim. This problem uses integration to solve real-world physics problems involving force and work in lifting fluids.

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