let the region r be the area enclosed the function f(x)=√x + 2, the horizontal line y = 4, and the y - axis…

let the region r be the area enclosed the function f(x)=√x + 2, the horizontal line y = 4, and the y - axis. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth.

let the region r be the area enclosed the function f(x)=√x + 2, the horizontal line y = 4, and the y - axis. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth.

Answer

Explanation:

Step1: Find intersection point with (y = 4)

Set (\sqrt{x}+2 = 4), then (\sqrt{x}=2), so (x = 4).

Step2: Integral in terms of (x)

The upper - curve is (y = 4) and the lower - curve is (y=\sqrt{x}+2) for (x\in[0,4]). The area (A_x=\int_{0}^{4}(4 - (\sqrt{x}+2))dx=\int_{0}^{4}(2-\sqrt{x})dx).

Step3: Solve for (x) in terms of (y)

From (y=\sqrt{x}+2), we get (x=(y - 2)^2) for (y\in[2,4]). The left - curve is (x = 0) and the right - curve is (x=(y - 2)^2). The area (A_y=\int_{2}^{4}(y - 2)^2dy).

Answer:

Integral in terms of (x): (\int_{0}^{4}(2-\sqrt{x})dx) Integral in terms of (y): (\int_{2}^{4}(y - 2)^2dy)