A tank has the shape of an inverted cone with height 10m and base radius 5m . The tank is filled with water to a depth 7m . How much work, in Joules, is required to pump all water over the top of the tank? (If necessary, round to two decimal places.)


To determine the work required to pump all the water over the top of the tank, we must integrate the force needed to move small slices of water from their respective heights up to the top of the tank. Here’s a step-by-step breakdown of how to solve the problem.

### Step 1: Key Formula for Work
The work required to pump a small amount of water \( \delta W \) a height \( h \) is given by:

\[
\delta W = \text{force} \times \text{distance}
\]

Since the force is due to the weight of water, and the weight is \( \delta W = \rho \cdot g \cdot \delta V \), where:
- \( \rho \) is the density of water (\(1000 \, \text{kg/m}^3\)),
- \( g \) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)),
- \( \delta V \) is the volume of a small slice of water at a height \( h \).

The distance the water needs to be moved is the height from the slice of water at \( y \) to the top of the tank, which is \( 10 - y \) meters (since the total height of the tank is 10 meters).

### Step 2: Finding the Volume of a Slice
The tank has a shape of an inverted cone. For any height \( y \) from the bottom of the cone, the radius \( r(y) \) of the cross-section at that height can be found using similar triangles. The radius increases linearly from 0 at the bottom to 5 meters at the top (height 10 meters). Thus:

\[
\frac{r(y)}{y} = \frac{5}{10} = \frac{1}{2}
\]

So,

\[
r(y) = \frac{y}{2}
\]

The volume of a thin slice of water \( \delta V \) at height \( y \) is the area of the cross-section times the thickness \( \delta y \), where the area of the cross-section is \( \pi r^2 \). Therefore:

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

### Step 3: Work to Move Each Slice
The work to move a small slice of water is:

\[
\delta W = \rho g \delta V (10 - y)
\]

Substituting \( \delta V \):

\[
\delta W = 1000 \times 9.8 \times \frac{\pi y^2}{4} \times (10 - y) \delta y
\]

\[
\delta W = 2450 \pi y^2 (10 - y) \delta y
\]

### Step 4: Total Work
To find the total work, we integrate \( \delta W \) from the bottom of the tank (\( y = 0 \)) to the water level (\( y = 7 \)):

\[
W = \int_0^7 2450 \pi y^2 (10 - y) \, dy
\]

Let’s compute this integral.The total work required to pump all the water over the top of the tank is approximately \( 4,180,059.20 \) Joules.

Would you like further details or have any other questions?

Here are 5 related questions to explore:
1. How would the work change if the tank were filled to a different depth?
2. What if the tank had a different radius while keeping the same height?
3. How does the shape of the tank (e.g., cylinder vs cone) affect the required work?
4. How would the work change if the acceleration due to gravity differed (e.g., on the Moon)?
5. How would the problem change if the water needed to be pumped to a height above the tank?

**Tip:** Understanding the geometric relationship (similar triangles) in these problems is crucial when dealing with cones or other changing-radius shapes.

A tank has the shape of an inverted cone with height 10m and base radius 5m. The tank is filled with water to a depth of 7m. How much work, in Joules, is required to pump all the water over the top of the tank? (If necessary, round to two decimal places.)

This detailed solution calculates the work required to pump water from a cone-shaped tank with a height of 10 meters and base radius of 5 meters, filled to a depth of 7 meters. Using integration and the work-energy principle, this example is ideal for university-level calculus students.

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