Question

Using integration find the area of the region {(x, y) : x² + y² ≤ 2ax, y² ≥ ax, x, y ≥ 0}.

Answer 1

Given region R is

Obviously, x² + y² = 2ax ⇒ (x - a)² + (y - 0)² = a² is a circle having centre at (a, 0) and radius r = a.

Therefore the region R₁ ≡ {(x, y): x² + y² ≤ 2ax} is the region inside the circle with centre (a, 0) and radius a.

Also y² = ax is right handed parabola with vertex at origin.

So, region R₂ ≡ {(x, y): y² ≥ ax} is the region out side parabola.

Also, R₃ ≡ {(x, y): x ≥ 0, y ≥ 0} is region in first quadrant.

Hence, R = R₁ ∩ R₂ ∩ R₃ is the shaded region shown above in figure.

Now for co-ordinate of A, we solve y² = ax and

x² + y² = 2ax as follows

[Putting y² = ax]

Hence co- ordinate of A is (a, a)

\(\therefore\) Required area