Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}


The area bounded by the curves, {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}, is shown by shaded region as OABCO.


The points of intersection of both the curves are


We can observed that area OABCO is symmetrical about x-axis.


Thus, Area of OABCO = 2 × Area OBC


Now,


Area OBCO = Area OMC + Area MBC




Put 2x = t


dx =


So, when x = , t = 3 and x = , t = 1, we get,









Therefore, the required area is



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