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

Answer: (B)

$\pi - \frac{2}{3}$

Answer 2

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:

$\boxed{\pi - \frac{2}{3}}$