Question

The area bounded by the parabola \( x^2 = 4y \), the lines \( y = 2 \), \( y = 4 \) and the Y-axis is

• A. \( \dfrac{4}{3}(8 - 2\sqrt{2}) \) sq. units
• B. \( \dfrac{8}{3}(8 - 2\sqrt{2}) \) sq. units
• C. \( \dfrac{8}{3}(8 + 2\sqrt{2}) \) sq. units
• D. \( (8 - 2\sqrt{2}) \) sq. units

Hint:

When a region is bounded by a curve and the Y-axis, integrate with respect to \( y \) using \( x \) as a function of \( y \).

Answer 1

The Correct Option is B

Step 1: Express \( x \) in terms of \( y \).
From \( x^2 = 4y \), \[ x = 2\sqrt{y} \] The Y-axis corresponds to \( x = 0 \).

Step 2: Set up the area integral.
Required area \[ A = \int_{y=2}^{4} (2\sqrt{y} - 0)\,dy = 2\int_{2}^{4} \sqrt{y}\,dy \]

Step 3: Integrate.
\[ A = 2\left[ \frac{2}{3}y^{3/2} \right]_{2}^{4} = \frac{4}{3}(8 - 2\sqrt{2}) \]

Step 4: Multiply by 2 (both sides of Y-axis symmetry).
\[ A = \frac{8}{3}(8 - 2\sqrt{2}) \]

Step 5: Conclusion.
\[ \boxed{\dfrac{8}{3}(8 - 2\sqrt{2}) \text{ sq. units}} \]