y dA; R is the region in the first quadrant enclosed between the circle x2 + y2 = 25and the line x + y = 5.

To find the area of region R, we need to determine the points of intersection between the circle and the line.

First, let's solve the system of equations formed by the circle and the line:

x^2 + y^2 = 25 x + y = 5

We can solve the second equation for x:

x = 5 - y

Substituting this value of x into the first equation:

(5 - y)^2 + y^2 = 25

Expanding and simplifying:

25 - 10y + y^2 + y^2 = 25 2y^2 - 10y = 0

Factoring out 2y:

2y(y - 5) = 0

Setting each factor equal to zero:

2y = 0 or y - 5 = 0

Solving for y:

y = 0 or y = 5

Now, let's find the corresponding x-values for each y-value:

For y = 0: x = 5 - 0 = 5

For y = 5: x = 5 - 5 = 0

So, the points of intersection are (5, 0) and (0, 5).

To find the area of region R, we need to find the area between the circle and the line. This can be done by finding the area of the sector formed by the circle and subtracting the area of the triangle formed by the line and the x-axis.

The equation of the circle is x^2 + y^2 = 25. Since the circle is centered at the origin (0, 0), the radius is √25 = 5.

The equation of the line is x + y = 5. To find the x-intercept, we set y = 0:

x + 0 = 5 x = 5

So, the x-intercept is 5.

Now, we can calculate the area of the sector. The formula for the area of a sector is A = (θ/360) * π * r^2, where θ is the central angle and r is the radius.

The central angle can be found by finding the angle between the x-axis and the line connecting the origin to the point of intersection (5, 0). This angle can be found using trigonometry:

tan(θ) = opposite/adjacent = 5/5 = 1 θ = arctan(1) = π/4 radians

Now, we can calculate the area of the sector:

A_sector = (π/4) * π * 5^2 = (π/4) * 25π = 25π^2/4

Next, we need to calculate the area of the triangle. The base of the triangle is the x-intercept, which is 5. The height of the triangle is the y-intercept, which is also 5. So, the area of the triangle is:

A_triangle = (1/2) * base * height = (1/2) * 5 * 5 = 25/2

Finally, we can find the area of region R by subtracting the area of the triangle from the area of the sector:

A_R = A_sector - A_triangle = 25π^2/4 - 25/2

Therefore, the area of region R is 25π^2/4 - 25/2.