Given the boundaries of the area to be found are,

• The ellipse,

• y = 0 (x-axis)

From the equation, of the ellipse

• the vertex at (0,0) i.e. the origin,

• the minor axis is the x-axis and the ellipse intersects the x- axis at A(-2,0) and B(2,0).

• the major axis is the y-axis and the ellipse intersects the y- axis at C(3,0) and D(-3,0).

As x and y have even powers, the area of the ellipse will be symmetrical about the x-axis and y-axis.

Here the ellipse, , can be re-written as

•

• ----- (1)

As given, the boundaries of the re to be found will be

• The ellipse, with vertex at (0,0).

• The x-axis.

Now, the area to be found will be the area under the ellipse which is above the x-axis.

Area of the required region = Area of ABC.

Area of ABC = Area of AOC + Area of BOC

[area of AOC = area of BOC as the ellipse is symmetrical about the y-axis]

Area of ABC = 2 Area of BOC

[Using the formula, ]

[sin^-1(1) = 90° and sin^-1(0) = 0° ]

The Area of the required region 3π sq. units