Question

Area enclosed by \( 4x^2 + y^2 \le 8 \) and \( y^2 \le 4x \) (in square unit) is:

• A. \( \left(\pi + \frac{4}{3}\right) \) sq. unit
• B. \( \left(\pi - \frac{4}{3}\right) \) sq. unit
• C. \( \left(\pi + \frac{2}{3}\right) \) sq. unit
• D. \( \left(\pi - \frac{2}{3}\right) \) sq. unit

Hint:

For area between curves:
Use symmetry whenever possible
Always subtract left curve from right curve

Answer 1

The Correct Option is C

Concept: The required region lies \emph{inside} the ellipse \[ 4x^2 + y^2 = 8 \] and to the \emph{right} of the parabola \[ y^2 = 4x \] Area is obtained by integrating with respect to \(y\), using: \[ \text{Area} = \int (x_{\text{right}} - x_{\text{left}})\, dy \]
Step 1: Express both curves as functions of \(x\). Ellipse: \[ 4x^2 = 8 - y^2 \Rightarrow x = \frac{1}{2}\sqrt{8 - y^2} \] Parabola: \[ x = \frac{y^2}{4} \]
Step 2: Find limits of integration. Points of intersection satisfy: \[ \frac{y^2}{4} = \frac{1}{2}\sqrt{8 - y^2} \] Squaring: \[ \frac{y^4}{16} = \frac{8 - y^2}{4} \] \[ y^4 + 4y^2 - 32 = 0 \] Let \( t = y^2 \): \[ t^2 + 4t - 32 = 0 \Rightarrow t = 4 \] \[ y = \pm 2 \]
Step 3: Set up the area integral. By symmetry about the \(x\)-axis: \[ \text{Area} = 2 \int_{0}^{2} \left( \frac{1}{2}\sqrt{8 - y^2} - \frac{y^2}{4} \right) dy \]
Step 4: Evaluate the integrals. \[ \int_{0}^{2} \sqrt{8 - y^2}\, dy = 2 + \pi \] \[ \int_{0}^{2} \frac{y^2}{4}\, dy = \frac{2}{3} \] Substituting: \[ \text{Area} = 2\left[\frac{1}{2}(2+\pi) - \frac{2}{3}\right] \] \[ = 2\left(1 + \frac{\pi}{2} - \frac{2}{3}\right) = \pi + \frac{2}{3} \] \[ \boxed{\text{Area} = \left(\pi + \frac{2}{3}\right)\ \text{sq. unit}} \]