The ends of a water trough have the shape of the region bounded by the graphs of y = x2 and y = 4 with x and y both measured in feet. To what depth must the trough be filled with water so that the force exerted by the water on either end of the trough is 779.423 lb? (Water has a density of 62.5 lb per cubic foot).

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To find the depth at which the trough must be filled with water, we need to determine the volume of water in the trough and then use the formula for force exerted by water to find the depth.

Step 1: Find the endpoints of the region bounded by the graphs of y = x^2 and y = 4.
To determine the endpoints, we need to find the x-values where the two graphs intersect. Set the two equations equal to each other:

x^2 = 4

Taking the square root of both sides:

x = ±2

So, the endpoints are x = -2 and x = 2.

Step 2: Determine the area of the region between the graphs.
To find the area, we need to integrate the difference between the two functions within the given bounds (-2 and 2):

Area = ∫[from -2 to 2] (4 - x^2) dx

Step 3: Calculate the volume of water.
Since water fills up the region between the graphs, we can use the area calculated in Step 2 to find the volume. Since the trough has symmetry, the total volume is twice the calculated area:

Volume = 2 * Area

Step 4: Find the depth of the water.
Using the formula for the force exerted by water:

Force = Density * Volume * Gravity

We can rearrange the formula to solve for the depth:

Depth = Force / (Density * Area * Gravity)

Step 5: Substitute the given values and calculate.

Density of water = 62.5 lb/ft^3 (given)
Gravity = 32.2 ft/sec^2 (acceleration due to gravity)

Depth = 779.423 lb / (62.5 lb/ft^3 * Area * 32.2 ft/sec^2)

Substituting the calculated Area from Step 2:

Depth = 779.423 lb / (62.5 lb/ft^3 * ∫[from -2 to 2] (4 - x^2) dx * 32.2 ft/sec^2)

To determine the depth to which the trough must be filled with water, we need to calculate the volume of water in the trough.

Let's first find the x-values where the graphs of y = x^2 and y = 4 intersect.

Setting x^2 = 4, we have:
x^2 - 4 = 0

This equation can be factored as:
(x - 2)(x + 2) = 0

Therefore, the x-values where the graphs intersect are x = -2 and x = 2.

Next, we need to find the area of the cross-section of the trough at a given x-value.

The area of the cross-section can be determined by finding the difference in the y-values between the graphs of y = x^2 and y = 4.

At x = -2, y = (-2)^2 = 4.
At x = 2, y = 2^2 = 4.

Therefore, at x = -2 and x = 2, the depth of water in the trough is 4 - 0 = 4 ft.

To find the total volume of water in the trough, we need to integrate the area of the cross-section over the interval [-2, 2].

Let's define the function A(x) as the area of the cross-section at a given x-value.

A(x) = 4 - x^2

Now we can find the total volume V by integrating A(x) from -2 to 2.

V = ∫[from -2 to 2] A(x) dx
V = ∫[from -2 to 2] (4 - x^2) dx

Evaluating this integral will give us the total volume of water in the trough.

After finding V, we can calculate the mass of the water using its density of 62.5 lb per cubic foot:
mass = density * volume

Finally, we can find the depth of water needed to exert a force of 779.423 lb on either end of the trough using the equation:
force = pressure * area
pressure = force / area
Depth = pressure / (density * gravity)

Where gravity is the acceleration due to gravity (32.2 ft/s^2).