Sketch the region {(x,y):9x^{2} + 4y^{2} = 36} and the find the area of the region enclosed by it, using integration

Given equation:

9x^{2} + 4y^{2} = 36 ...... (1)

equation (1) represents an eclipse that is symmetrical about the x - axis and also about the y - axis, with center at origin and passes through (±2, 0) and (0, ±3).

A rough sketch is given as below: -

We have to find the area of the shaded region.

Required area

= shaded region ABCDA

= 4 (shaded region OBCO) ( as it is symmetrical about the x - axis as well as y - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,2) and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area of the region enclosed by it is equal to square units.

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