A trough is 7 feet long and 11 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x4�=�4 from x=−1�=−1 to x=1�=1. The trough is full of water. Find the amount of work required to empty the trough by pumping the water over the top. 

This problem involves calculating the work required to pump water out of a trough. We can approach it by integrating the work done to lift infinitesimally small slices of water to the top of the trough.

### Step 1: Define the situation

The trough has a length of 7 feet, a height of 11 feet, and the cross-sectional shape follows the curve \( y = x^4 \) from \( x = -1 \) to \( x = 1 \).

Let’s break down the steps:

1. **Volume of a slice of water**: 
   Consider a small horizontal slice of water at height \( y \), where the width of the slice is determined by the equation \( y = x^4 \). For a particular height \( y \), the horizontal distance between the two sides of the trough is the width, which is:
   
   \[
   \text{Width} = 2 \cdot (y^{1/4})
   \]

   The area of a cross-section at height \( y \) is then:
   
   \[
   A(y) = \text{Width} \times 1 = 2 \cdot (y^{1/4})
   \]

2. **Volume of the small slice**: 
   The volume of a small slice of water with thickness \( dy \) is:
   
   \[
   dV = A(y) \times 7 \times dy = 14 \cdot (y^{1/4}) \cdot dy
   \]

3. **Work done to lift the slice**: 
   The weight of a slice of water is \( \rho \cdot g \cdot dV \), where \( \rho \) is the density of water (62.4 lb/ft³), and \( g \) is the gravitational constant (in this case, the standard weight per volume is already given by \( \rho g = 62.4 \)).

   The height it must be lifted is \( 11 - y \), since the trough is 11 feet tall. Therefore, the work to lift a slice is:

   \[
   dW = \rho g \cdot dV \cdot (11 - y) = 62.4 \cdot 14 \cdot (y^{1/4}) \cdot (11 - y) \cdot dy
   \]

4. **Integrating the total work**: 
   To find the total work, we integrate from \( y = 0 \) (the bottom of the trough) to \( y = 1 \) (the top of the water):

   \[
   W = \int_0^1 62.4 \cdot 14 \cdot (y^{1/4}) \cdot (11 - y) \, dy
   \]

Let me compute this integral.import sympy as sp

# Define the variable
y = sp.symbols('y')

# Define the integrand: 62.4 * 14 * (y**(1/4)) * (11 - y)
integrand = 62.4 * 14 * (y**(1/4)) * (11 - y)

# Compute the integral from y = 0 to y = 1
work = sp.integrate(integrand, (y, 0, 1))
workIt appears that the integral involves complex terms that SymPy cannot handle directly. Let me evaluate this integral numerically to get the result for the work done.import scipy.integrate as integrate
import numpy as np

# Define the function for the integrand
def integrand(y):
    return 62.4 * 14 * (y**(1/4)) * (11 - y)

# Perform the numerical integration from y=0 to y=1
work, _ = integrate.quad(integrand, 0, 1)
workIt seems like I can’t do more advanced data analysis right now. Please try again later. However, if you want, I can walk you through setting up the calculations manually or using a different method. Let me know how you'd like to proceed!

Discover how to calculate the work required to empty a trough filled with water using calculus and integration. This problem involves determining the volume of water and the distance it needs to be lifted based on the trough's dimensions and cross-sectional area. Suitable for advanced high school students studying calculus and physics.

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