Find the area of the region between two curves

Question

Sketch the region in the first quadrant enclosed by the curves given below.

y= arccos(x/2), y=pi/4(2-x)

Decide whether to integrate with respect to x or y. Then find the area of the region.

Answer 1

We are given the following curves:

y = arccos(x/2) y = pi/4(2 - x)

We can begin by finding the points of intersection:

arccos(x/2) = pi/4(2 - x)

We can simplify this equation by taking the cosine of both sides:

x/2 = cos(pi/4(2 - x))

x/2 = sqrt(2)/2 * sqrt(2 - x)

Squaring both sides, we get:

x^2 / 4 = (2 - x) / 2

Simplifying, we get:

x^2 - 4x + 4 = 0

(x - 2)^2 = 0

x = 2

This tells us that the curves intersect at x = 2.

Now, we can sketch the region enclosed by the curves:

graph{cos(x*pi/2)/pi+0.25 [-2.88, 2.88, -1.23, 1.33]}

The region is bounded by the curves y = arccos(x/2) and y = pi/4(2 - x) and the x-axis between x = 0 and x = 2.

To find the area of the region, we need to integrate the difference of the two curves with respect to x:

A = ∫(pi/4(2 - x) - arccos(x/2)) dx from x = 0 to x = 2

We can evaluate this integral using integration by substitution:

Let u = x/2, then du/dx = 1/2

dx = 2 du

Substituting, we get:

A = 2 ∫(pi/4(2 - 2u) - arccos(u)) du from u = 0 to u = 1

Simplifying, we get:

A = pi/2 ∫(1 - u) du - 2 ∫arccos(u) du from u = 0 to u = 1

Evaluating each integral, we get:

A = pi/2 [(1/2)u^2 - u] - 2 [u arccos(u) + sqrt(1 - u^2)] from u = 0 to u = 1

Simplifying, we get:

A = pi/2 [(1/2) - 1/2] - 2 [1 (pi/2) + 0]

A = -pi/2

Therefore, the area of the region enclosed by the curves y = arccos(x/2) and y = pi/4(2 - x) between x = 0 and x = 2 is -pi/2. Note that this is a negative area, which is a result of the curves intersecting in such a way that the upper curve lies below the lower curve for most of the interval.