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in Definite Integrals by (28.7k points)
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The area of the circle x2 + y2 = 16 enterior to the parabola y2 = 6x is

A. \(\frac{4}{3}[4\pi - \sqrt{3}]\)

B. \(\frac{4}{3}[4\pi + \sqrt{3}]\)

C. \(\frac{4}{3}[8\pi - \sqrt{3}]\)

D. \(\frac{4}{3}[8\pi + \sqrt{3}]\)

1 Answer

+1 vote
by (29.1k points)
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Best answer

Correct answer is B.

The area we want is –

At intersection points, y2 = 6x = 16 – x2

Or, x2 + 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 ∈ \(\mathbb R\)

So, x = 2

and y2 = 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 –

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

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