Given the boundaries of the area to be found are,

• the circle, x² + y² = 32 ---(1)

• the line, y = x---- (2)

• Area should be in first quadrant.

From the equation, of the first circle, x² + y² = 32

• the vertex at (0,0) i.e. the origin

• the radius is 4√2 unit.

Now to find the point of intersection of (1) and (2), substitute y = x in (1)

x² + x² = 32

2x² = 32

x² = 16

x = ± 4

Substituting x in (2), we get y = ± 4

So the two points, A and B where (1) and (2) meet are A = (4,4) and B = (-4,-4)

As x and y have even powers for both the circles, they will be symmetrical about the x-axis and y-axis.

Consider the circle, x² + y² = 32, can be re-written as

----- (3)

Let us drop a perpendicular from A on to x-axis. The base of the perpendicular is C = (4, 0)

Now, the area to be found will be the area is

Area of the required region = Area of OADO.

Area of OADO= Area of OAC•+ Area of CADC

Area of OADOis

[Using the formula, and ]

[sin^-1(1) = 90° and ]

= 4π

Area of the required region is 4π sq. units.