A square pyramid with base length 2 meters and height 5 meters is filled from the base with water to a height of 3 meters.
Setup but do not evaluate integrals representing,
(a) the work required to pump all of the water out of the top of the pyramid.

To set up the integral for the work required to pump all the water out of the top of the square pyramid, we'll need to follow these steps:

### Step 1: Volume of Water at Height \( h \)
The square pyramid has a height of \( H = 5 \) meters and a base length of \( L = 2 \) meters. The water fills up to a height of 3 meters.

At any height \( h \) within the pyramid, the cross-sectional area is a square. The side length of the square at height \( h \) is proportional to \( h \) because the pyramid is similar at all heights. The base length at the top is \( L = 2 \) meters and corresponds to the total height of the pyramid, so the side length \( s(h) \) of the square at height \( h \) is given by:

\[
s(h) = \frac{L}{H} \cdot h = \frac{2}{5} h
\]

Thus, the area \( A(h) \) of the square cross-section at height \( h \) is:

\[
A(h) = s(h)^2 = \left( \frac{2}{5} h \right)^2 = \frac{4}{25} h^2
\]

### Step 2: Volume Element \( dV \)
The volume element \( dV \) at height \( h \) is the area of the cross-section multiplied by a small height element \( dh \):

\[
dV = A(h) \, dh = \frac{4}{25} h^2 \, dh
\]

### Step 3: Work to Pump Water
The force needed to lift a small volume \( dV \) of water is equal to its weight. The density of water is \( \rho \approx 1000 \, \text{kg/m}^3 \), and the gravitational constant is \( g \approx 9.8 \, \text{m/s}^2 \). The weight of the water in a volume \( dV \) is:

\[
\text{Weight} = \rho g \, dV = 1000 \cdot 9.8 \cdot \frac{4}{25} h^2 \, dh = 392 h^2 \, dh
\]

The work required to move this volume of water to the top of the pyramid (5 meters) is the weight of the water multiplied by the vertical distance it needs to be lifted, which is \( (5 - h) \) meters. So, the work for a small volume \( dV \) is:

\[
dW = 392 h^2 (5 - h) \, dh
\]

### Step 4: Integral for Total Work
To find the total work required to pump all the water out, we integrate from \( h = 0 \) (the base of the pyramid) to \( h = 3 \) (the height to which the pyramid is filled with water):

\[
W = \int_0^3 392 h^2 (5 - h) \, dh
\]

This is the setup for the integral representing the work required to pump all the water out of the top of the pyramid.

Would you like more details or have any questions? 

### Follow-up questions:
1. How would the integral change if the height of the pyramid was doubled?
2. What is the physical meaning of the term \( (5 - h) \) in the integral?
3. How would you compute the volume of water in the pyramid without using calculus?
4. What is the effect of increasing the base area of the pyramid on the work required?
5. How would the formula for work change if the water had to be pumped to a point higher than the top of the pyramid?

### Tip:
Always check the units of each term in your setup to ensure dimensional consistency, especially in physics-related integrals.

A square pyramid with base length 2 meters and height 5 meters is filled from the base with water to a height of 3 meters. Setup but do not evaluate integrals representing the work required to pump all of the water out of the top of the pyramid.

Explore how to set up the integral to calculate the work required to pump water out of a square pyramid with a base length of 2 meters and a height of 5 meters, filled with water to a height of 3 meters. This solution includes the geometry of the pyramid, integral calculus setup, and the physics of work, suitable for college-level students.

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