a tank filled with oil is in the shape of an inverted right rectangular pyramid with the base at the top with a width of 3 feet, a depth of 3 feet, and a height of 8 feet. Oil inside the tank weighs 40 pounds per cubic foot. Setup an integral which would calculate how much work it would take to pump oil from the tank to a level 2 feet above the tank if the tank were completely full.

To calculate the work required to pump the oil from the tank to a level 2 feet above the tank, we can follow these steps.

### Step 1: Understanding the Geometry
The tank is an inverted right rectangular pyramid. The base of the pyramid (at the top) has dimensions of 3 feet by 3 feet, and the height of the tank is 8 feet. The vertex of the pyramid is at the bottom.

### Step 2: Volume Element
Since the tank is pyramidal, the width and depth of horizontal slices of the tank decrease as we move downward. Let’s take a thin slice of the tank at a height \( y \) (measured from the bottom of the tank), where \( y \) ranges from 0 to 8 feet (since the total height of the tank is 8 feet).

The dimensions of the slice at height \( y \) will be proportional to its distance from the bottom. The scaling factor is based on the similar triangles formed by the slice and the full base of the tank.

At the top (height \( y = 8 \)), the width is 3 feet, and at the bottom (height \( y = 0 \)), the width is 0. Therefore, the width and depth at any height \( y \) are given by:
\[
w(y) = d(y) = \frac{3}{8}y
\]
The area of a horizontal slice at height \( y \) is:
\[
A(y) = w(y) \cdot d(y) = \left(\frac{3}{8}y\right) \cdot \left(\frac{3}{8}y\right) = \frac{9}{64} y^2
\]

### Step 3: Volume of a Slice
The volume of a thin slice of thickness \( dy \) at height \( y \) is:
\[
dV = A(y) \, dy = \frac{9}{64} y^2 \, dy
\]

### Step 4: Weight of a Slice
Since oil weighs 40 pounds per cubic foot, the weight of a slice is:
\[
dW = 40 \cdot dV = 40 \cdot \frac{9}{64} y^2 \, dy = \frac{360}{64} y^2 \, dy = \frac{45}{8} y^2 \, dy
\]

### Step 5: Distance to Pump the Oil
The work to pump a slice of oil is the weight of the slice times the distance it must be lifted. Since we are pumping the oil to a level 2 feet above the top of the tank, the distance from height \( y \) to this level is:
\[
\text{distance} = 10 - y
\]
(10 feet is the total height to which the oil must be lifted—8 feet for the height of the tank plus 2 feet above it).

### Step 6: Work for Each Slice
The work to pump a thin slice of oil is the weight of the slice times the distance it must be lifted:
\[
dW_{\text{pump}} = \frac{45}{8} y^2 (10 - y) \, dy
\]

### Step 7: Total Work
To find the total work, we integrate from \( y = 0 \) to \( y = 8 \):
\[
W = \int_0^8 \frac{45}{8} y^2 (10 - y) \, dy
\]

This is the setup for the integral to calculate the total work.

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### Final Setup:
\[
W = \frac{45}{8} \int_0^8 y^2 (10 - y) \, dy
\]

Would you like me to solve the integral as well?

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### Related Questions:
1. How would the result change if the oil weighed 50 pounds per cubic foot instead of 40?
2. How would the work change if the pyramid's height were increased to 10 feet?
3. What if the oil were pumped to a level 3 feet above the tank instead of 2 feet?
4. How would the shape of the tank affect the work if it were a cylinder instead of a pyramid?
5. How much work would be required if the tank were only half full of oil?

### Tip:
In work problems, always pay attention to how the distance an object is lifted changes with its position in the system. This affects the integral setup significantly.

A tank filled with oil is in the shape of an inverted right rectangular pyramid with the base at the top with a width of 3 feet, a depth of 3 feet, and a height of 8 feet. Oil inside the tank weighs 40 pounds per cubic foot. Set up an integral which would calculate how much work it would take to pump oil from the tank to a level 2 feet above the tank if the tank were completely full.

Learn how to set up an integral to calculate the work required to pump oil from an inverted right rectangular pyramid tank, with a base of 3 feet by 3 feet, height of 8 feet, and oil weighing 40 pounds per cubic foot. This problem involves applying calculus concepts to calculate work in physics using integration techniques.

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