Find the Area of the Region in the First Quadrant Enclosed by the X-Axis, the Line y = x, and the Circle x² + y² = 32

To find the area of the region in the first quadrant enclosed by the x-axis, the line y = x, and the circle x² + y² = 32, we will follow these steps:

1. **Identify the Intersections:** We first need to determine where the line intersects the circle. We substitute y = x into the circle’s equation:

x² + (x)² = 32

This simplifies to:

2x² = 32

From this, we find:

x² = 16

Thus, x = 4 (only considering the positive root since we are in the first quadrant). Therefore, we have the intersection point at (4, 4).

2. **Determine the Area:** The area we’re interested in is bounded by the x-axis, the line, and the circle. We can find this area by integrating between the limits where the line and the circle intersect.

The area under the line from x = 0 to x = 4 is given by:

Area_line = ∫(from 0 to 4) x dx = [1/2 * x²] | from 0 to 4 = [1/2 * 16 - 0] = 8

3. **Area under the Circle:** The area under the curve of the circle from x = 0 to x = 4 can be found by solving for y in the circle’s equation:

y = √(32 - x²)

The area under the circle is then:

Area_circle = ∫(from 0 to 4) √(32 - x²) dx

This integral can be computed using trigonometric substitution or numerical methods, but for simplicity, let’s compute it directly as:

Area_circle = (1/4)πr² where r² = 32 ⇒ = (1/4)π(32) = 8π

4. **Final Area Calculation:** The area of the region we are interested in is given by:

Area = Area_circle - Area_line = (8π - 8)

Thus, the final area of the region in the first quadrant enclosed by the x-axis, the line y = x, and the circle x² + y² = 32 is:

Area = 8(π - 1)