Find by double integration the area of the region enclosed by curves2 2 2 ,x y a x y a    in the first quadrant.

To find the area of the region enclosed by the curves, we can use double integration.

First, let's find the points of intersection between the two curves. Setting the equations equal to each other, we have:

x^2 + y^2 = ax + ay

Rearranging the equation, we get:

x^2 - ax + y^2 - ay = 0

Now, we can complete the square for both x and y terms:

(x^2 - ax + a^2/4) + (y^2 - ay + a^2/4) = a^2/2

Simplifying further, we have:

(x - a/2)^2 + (y - a/2)^2 = a^2/2

This equation represents a circle with center (a/2, a/2) and radius sqrt(a^2/2).

To find the points of intersection, we need to solve the equation for x and y. Since we are only interested in the first quadrant, we can set x and y to be positive.

Setting x = a/2 and y = a/2, we get one point of intersection at (a/2, a/2).

Next, we can solve for the other point of intersection by substituting y = a - x into the equation:

x^2 + (a - x)^2 = ax + a(a - x)

Expanding and simplifying, we have:

2x^2 - 2ax + a^2 - ax = 0

Rearranging the equation, we get:

2x^2 - 3ax + a^2 = 0

Using the quadratic formula, we can solve for x:

x = (3a ± sqrt(9a^2 - 8a^2))/4

Simplifying further, we have:

x = (3a ± sqrt(a^2))/4

Since we are only interested in the positive value of x, we have:

x = (3a + a)/4 = a

Therefore, the other point of intersection is (a, 0).

Now, we can set up the double integral to find the area enclosed by the curves. Since the curves are symmetric about the y-axis, we can integrate from y = 0 to y = a/2, and then multiply the result by 2.

The integral is given by:

2 * ∫[0 to a/2] ∫[0 to a] dx dy

Simplifying, we have:

2 * ∫[0 to a/2] a dy

Integrating, we get:

2 * a * [y] [0 to a/2]

Evaluating the integral, we have:

2 * a * (a/2 - 0) = a^2

Therefore, the area of the region enclosed by the curves is a^2.