Question

The area of the region \[ A = \{(x,y) : 4x^2 + y^2 \le 8 \;\text{and}\; y^2 \le 4x\} \] is

• A. \( \dfrac{\pi}{2} + 2 \)
• B. \( \pi + 4 \)
• C. \( \pi + \dfrac{2}{3} \)
• D. \( \dfrac{\pi}{2} + \dfrac{1}{3} \)

Hint:

When a region is bounded by two curves, always identify intersection points first, then integrate with respect to the variable that simplifies the limits.

Answer 1

The Correct Option is C

The given curves are: \[ 4x^2 + y^2 = 8 \Rightarrow x^2 + \frac{y^2}{4} = 2, \] an ellipse, and \[ y^2 = 4x, \] a parabola.
Step 1: Find the points of intersection.
Substitute \( y^2 = 4x \) into the ellipse: \[ 4x^2 + 4x = 8 \Rightarrow x^2 + x - 2 = 0. \] \[ x = 1,\; -2 \quad (\text{only } x=1 \text{ is valid}). \] Thus, \[ y^2 = 4 \Rightarrow y = \pm 2. \]
Step 2: Set up the area integral.
For \( x \in [0,1] \), \[ \text{Upper curve: } y = \sqrt{4x}, \quad \text{Lower curve: } y = -\sqrt{4x}. \] Area: \[ A = \int_0^1 \left(\sqrt{4x} - (-\sqrt{4x})\right) dx + \text{area under ellipse beyond parabola}. \] Evaluating the integrals gives: \[ A = \pi + \frac{2}{3}. \]
Final Answer: \[ \boxed{\pi + \frac{2}{3}} \]