Final answer:
To find the area of the curtain wall below the graph of z = 4y, y ≥ 0, and above the parabola y^2 = x, 0 ≤ x ≤ 2, we can find the intersection points of the two curves and calculate the area between them. The intersection points are y = 0 and y = 4. By integrating the function (4y - y^2) with respect to y from y = 0 to y = 4, we find that the area of the curtain wall is 32 - 64/3 square units.
Step-by-step explanation:
To find the area of the curtain wall below the graph of z = 4y, y ≥ 0, and above the parabola y^2 = x, 0 ≤ x ≤ 2, we need to find the intersection points of the two curves and calculate the area between them.
To find the intersection points, we set z = 4y equal to y^2 = x.
Substituting z = 4y into y^2 = x, we get:
4y^2 = x
Now we can find the intersection points by solving the equation:
y^2 = x = 4y
y^2 - 4y = 0
y(y - 4) = 0
So the intersection points are y = 0 and y = 4.
Now we can calculate the area between the curves using the formula for finding the area between two curves:
Area = ∫(4y - y^2) dx from x = 0 to x = 2
Area = ∫(4y - y^2) dx from x = 0 to x = 2
Since the area is between y = 0 and y = 4, we integrate with respect to y:
Area = ∫(4y - y^2) dy from y = 0 to y = 4
Integrating, we get:
Area = [2y^2 - (y^3/3)] from y = 0 to y = 4
Area = 2(4^2) - (4^3/3) - 0 = 32 - 64/3
So the area of the curtain wall is 32 - 64/3 square units.