The area of the circle x^{2} + y^{2} = 16 enterior to the parabola y^{2} = 6x is

The area we want is –

At intersection points, y^{2} = 6x = 16 – x^{2}

Or, x^{2} + 6x – 16 = 0

i.e., (x + 8)(x – 2) = 0

i.e., x = -8, 2

Now x = -8 ⇒ y^{2} = 16 – (-8)^{2} = 16 – 64 = -48 < 0, which is not possible for y ∈ ℝ

So, x = 2

and y^{2} = 16 – 2^{2}

= 16 – 4

= 12

Or, y = √12 = ± 2√3

So, our bounds are y = -2√3 to y = 2√3

The area A enclosed is –

A =

Since both the functions are symmetrical about the x – axis,

A =

=

(Ans)

1