Question

The area of the region in the first quadrant inside the circle \(x^2 + y^2 = 8\) and outside the parabola \(y^2 = 2x\) is equal to:

• A. \(\frac{\pi}{2} - \frac{1}{3}\)
• B. \(\pi - \frac{2}{3}\)
• C. \(\frac{\pi}{2} - \frac{2}{3}\)
• D. \(\pi - \frac{1}{3}\)

Answer 1

The Correct Option is B

We are required to find the area in the first quadrant inside the circle:

\( x^2 + y^2 = 8 \)

and outside the parabola:

\( y^2 = 2x \)

The points of intersection between the circle and the parabola can be found by substituting \( x = \frac{y^2}{2} \) into the circle's equation:

\( \left( \frac{y^2}{2} \right)^2 + y^2 = 8 \)

\( \frac{y^4}{4} + y^2 = 8 \)

Multiplying the entire equation by 4:

\( y^4 + 4y^2 = 32 \)

Rearranging terms:

\( y^4 + 4y^2 - 32 = 0 \)

Let \( z = y^2 \). Then:

\( z^2 + 4z - 32 = 0 \)

Solving this quadratic equation using the quadratic formula:

\( z = \frac{-4 \pm \sqrt{16 + 128}}{2} \)

\( z = \frac{-4 \pm \sqrt{144}}{2} \)

\( z = \frac{-4 \pm 12}{2} \)

This gives:

\( z = 4 \quad \text{or} \quad z = -8 \quad \text{(discarded as \( z \geq 0 \))} \)

Thus:

\( y^2 = 4 \implies y = 2 \quad \text{or} \quad y = -2 \)

In the first quadrant, we consider \( y = 2 \) and \( x = \frac{y^2}{2} = 2 \). The required area is given by the difference between the area under the circle from \( x = 0 \) to \( x = 2 \) and the area under the parabola from \( x = 0 \) to \( x = 2 \):

\( \text{Required Area} = \int_0^2 \sqrt{8 - x^2} \, dx - \int_0^2 \sqrt{2x} \, dx \)

Area under the circle:

\( \int_0^2 \sqrt{8 - x^2} \, dx \)

We use the substitution \( x = \sqrt{8} \sin \theta \), \( dx = \sqrt{8} \cos \theta d\theta \), with limits changing from \( x = 0 \) to \( x = 2 \), giving \( \theta = 0 \) to \( \theta = \frac{\pi}{4} \). The integral becomes:

\( \int_0^{\frac{\pi}{4}} \sqrt{8} \cos \theta \cdot \sqrt{8} \cos \theta d\theta = 8 \int_0^{\frac{\pi}{4}} \cos^2 \theta d\theta = 4 \int_0^{\frac{\pi}{4}} (1 + \cos 2\theta) d\theta \)

Evaluating:

\( 4 \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^{\frac{\pi}{4}} = 4 \left[ \frac{\pi}{8} + 0 \right] = \frac{\pi}{2} \)

Area under the parabola:

\( \int_0^2 \sqrt{2x} \, dx \)

Let \( u = 2x \), \( du = 2dx \):

\( \int_0^2 \sqrt{2x} \, dx = \int_0^4 \sqrt{u} \frac{du}{2} = \frac{1}{2} \int_0^4 u^{1/2} du \)

Evaluating:

\( \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_0^4 = \frac{1}{3} \left[ 4^{3/2} \right] = \frac{8}{3} \)

Required Area:

\( \text{Area} = \frac{\pi}{2} - \frac{8}{3} = \pi - \frac{2}{3} \)

Therefore:

\( \boxed{\pi - \frac{2}{3}} \)