Area Calculation for Ellipse and Parabola Region

The question asks for the area of the region $A$ defined by the inequalities:

• $4x^2 + y^2 \leq 8$
• $y^2 \leq 4x$

The first inequality describes the region inside an ellipse centered at the origin. The second inequality describes the region inside a parabola opening to the right, also centered at the origin.

Finding Intersection Points

To determine the bounds of integration, we find the points where the boundaries of the ellipse and parabola intersect. We solve the system of equations:

$ 4x^2 + y^2 = 8 \quad (\text{Ellipse}) $

$ y^2 = 4x \quad (\text{Parabola}) $

Substitute $y^2 = 4x$ into the ellipse equation:

$ 4x^2 + (4x) = 8 $

Simplify the equation:

$ 4x^2 + 4x - 8 = 0 $

Divide by 4:

$ x^2 + x - 2 = 0 $

Factor the quadratic:

$ (x + 2)(x - 1) = 0 $

From $y^2 = 4x$, we know $x \ge 0$. Therefore, the valid solution is $x=1$. Substitute $x=1$ back into the parabola equation:

$ y^2 = 4(1) \implies y = \pm 2 $

The intersection points are $(1, 2)$ and $(1, -2)$.

Setting Up the Area Integral

We can compute the area by integrating with respect to $y$. The region is bounded on the right by the ellipse and on the left by the parabola.

From the ellipse equation $4x^2 + y^2 = 8$, we solve for $x$ (right side): $4x^2 = 8 - y^2 \implies x = \frac{\sqrt{8-y^2}}{2}$. Let this be $x_E$.

From the parabola equation $y^2 = 4x$, we solve for $x$: $x = \frac{y^2}{4}$. Let this be $x_P$.

The area $A$ is the integral of the difference between the right boundary ($x_E$) and the left boundary ($x_P$) from $y=-2$ to $y=2$:

$ A = \int_{-2}^{2} (x_E - x_P) \, dy $

$ A = \int_{-2}^{2} \left( \frac{\sqrt{8-y^2}}{2} - \frac{y^2}{4} \right) \, dy $

Split the integral into two parts:

$ A = \int_{-2}^{2} \frac{\sqrt{8-y^2}}{2} \, dy - \int_{-2}^{2} \frac{y^2}{4} \, dy $

Calculating the First Integral ($I_1$)

Let $I_1 = \int_{-2}^{2} \frac{\sqrt{8-y^2}}{2} \, dy$. This integral represents the area bounded by the ellipse curve $x = \frac{\sqrt{8-y^2}}{2}$, the y-axis ($x=0$), and the lines $y=-2$ and $y=2$. Use the substitution $y = 2\sqrt{2}\sin\theta$, so $dy = 2\sqrt{2}\cos\theta \, d\theta$. The limits change from $y=\pm 2$ to $\theta=\pm\frac{\pi}{4}$. $ I_1 = \int_{-\pi/4}^{\pi/4} \frac{\sqrt{8 - (2\sqrt{2}\sin\theta)^2}}{2} (2\sqrt{2}\cos\theta \, d\theta) $

$ I_1 = \int_{-\pi/4}^{\pi/4} \frac{2\sqrt{2}\cos\theta}{2} (2\sqrt{2}\cos\theta \, d\theta) = \int_{-\pi/4}^{\pi/4} 4\cos^2\theta \, d\theta $

Using the identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$:

$ I_1 = 4 \int_{-\pi/4}^{\pi/4} \frac{1+\cos(2\theta)}{2} \, d\theta = 2 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{-\pi/4}^{\pi/4} $

$ I_1 = 2 \left[ \left(\frac{\pi}{4} + \frac{1}{2}\right) - \left(-\frac{\pi}{4} - \frac{1}{2}\right) \right] = 2 \left[ \frac{\pi}{2} + 1 \right] = \pi + 2 $

Calculating the Second Integral ($I_2$)

Let $I_2 = \int_{-2}^{2} \frac{y^2}{4} \, dy$. This integral represents the area bounded by the parabola $x = \frac{y^2}{4}$, the y-axis ($x=0$), and the lines $y=-2$ and $y=2$. $ I_2 = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{-2}^{2} = \frac{1}{4} \left( \frac{2^3}{3} - \frac{(-2)^3}{3} \right) $

$ I_2 = \frac{1}{4} \left( \frac{8}{3} - \left(-\frac{8}{3}\right) \right) = \frac{1}{4} \left( \frac{16}{3} \right) = \frac{4}{3} $

Calculating the Total Area

Subtract the second integral from the first to find the total area $A$:

$ A = I_1 - I_2 = (\pi + 2) - \frac{4}{3} $

$ A = \pi + \frac{6}{3} - \frac{4}{3} = \pi + \frac{2}{3} $

The calculated area is $\pi + \frac{2}{3}$.