The parabola y^2=4x divides the area of the circle x^2+... | Area of Bounded Region (Mathematics)

The parabola

y^2=4x

divides the area of the circle

x^2+y^2=5

in two parts. The area of the smaller part is equal to:

\frac{1}{3} + 5sin^{-1} (\frac{2}{\sqrt5})

\frac{2}{3} + \sqrt5sin^{-1} (\frac{2}{\sqrt5})

\frac{2}{3} + 5sin^{-1} (\frac{2}{\sqrt5})

\frac{1}{3} + \sqrt5sin^{-1} (\frac{2}{\sqrt5})

Correct Answer:

\frac{2}{3} + 5sin^{-1} (\frac{2}{\sqrt5})

(c):

The points of intersection of

y^2 = 4x

and

x^2 + y^2 = 5

are

(1, 2)

and

(1, -2)

\therefore

Required area

= 2 \{ \text{area of } OACO + \text{area of } CABC \}

= 2 \left[ \int_{0}^{1} 2\sqrt{x} dx + \int_{1}^{\sqrt{5}} \sqrt{5 - x^2} dx \right]

= 2 \left[ \left| \frac{4}{3} x^{3/2} \right|_{0}^{1} + \left( \frac{1}{2} x \sqrt{5 - x^2} + \frac{5}{2} \sin^{-1} \frac{x}{\sqrt{5}} \right)_{1}^{\sqrt{5}} \right]

= \frac{2}{3} + 5 \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{2}{3} + 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} \right)

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