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 

Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.