Question

A new auditorium is built with a foundation in the shape of one-fourth of a circle of radius 50 feet. So, it forms a region bounded by the graph of x^2 + y^2 = 50^2 with x ≥ 0 and y ≥ 0. The following equations are models for the floor and ceiling. Floor: z = x+y/ 5 Ceiling: z = 20 + xy /100 Calculate the volume of the room, which is needed to determine the heating and cooling requirements.

Answer 1

Explanation:

To calculate the volume of the room, we need to integrate the difference between the ceiling and floor functions over the region bounded by the circle. We can use double integrals to do this.

First, let's rewrite the equations for the floor and ceiling in terms of x and y:

Floor: z = (x + y) / 5

Ceiling: z = 20 + xy / 100

The volume of the room can be calculated using the following double integral:

V = ∫∫(20 + xy/100 - (x + y)/5) dA

where the limits of integration are x = 0 to x = 50, and y = 0 to y = √(50^2 - x^2).

We can simplify the integrand by combining like terms:

V = ∫∫(400/100 + xy/100 - (20x + 20y)/100) dA

V = ∫∫(4 + xy/100 - 0.2x - 0.2y) dA

Now we can integrate with respect to y first:

V = ∫0^50 ∫0^√(50^2 - x^2) (4 + xy/100 - 0.2x - 0.2y) dy dx

V = ∫0^50 [(4y + xy^2/200 - 0.2xy - 0.1y^2)|y=0^√(50^2 - x^2)] dx

V = ∫0^50 [(4√(50^2 - x^2) + x(50^2 - x^2)/200 - 0.2x√(50^2 - x^2) - 0.1(50^2 - x^2)^2/200)|x=0^50]

Evaluating this integral, we get:

V ≈ 2,233.5 cubic feet

Therefore, the volume of the room is approximately 2,233.5 cubic feet, which is the amount of space that will need to be heated or cooled.