Sketch the region {(x,y):9x2 + 4y2 = 36} and the find the area of the region enclosed by it, using integration
Given equation:
9x2 + 4y2 = 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,

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