Question

The area of the region lying between the curves \( y = \sqrt{4 - x^2} \), \( y^2 = 3x \) and the Y-axis is:

• A. \(\dfrac{\pi}{3} - \dfrac{1}{2\sqrt{3}}\)
• B. \(\dfrac{\pi}{6} + \dfrac{1}{2\sqrt{3}}\)
• C. \(\dfrac{\pi}{3} + \dfrac{1}{2\sqrt{3}}\)
• D. \(\dfrac{\pi}{6} - \dfrac{1}{2\sqrt{3}}\)

Hint:

When calculating area between two curves, always check the variable of integration that simplifies limits and expressions. Here, integrating with respect to \( y \) simplifies the bounds and expressions.

Answer 1

The Correct Option is A

Step 1: Interpret the curves 
We are given: 
- \( y = \sqrt{4 - x^2} \) which is the upper semicircle of radius 2 centered at origin. 
- \( y^2 = 3x \Rightarrow x = \dfrac{y^2}{3} \) is a rightward-opening parabola. 
- The region lies between the parabola and the circle, bounded by the Y-axis. 
Step 2: Find points of intersection 
Substitute \( x = \dfrac{y^2}{3} \) into \( y = \sqrt{4 - x^2} \Rightarrow y = \sqrt{4 - \left( \dfrac{y^2}{3} \right)^2} \) 
Solving this gives points of intersection at \( y = 0 \) and \( y = \sqrt{3} \) 
Step 3: Set up the area 
We'll integrate horizontally (along \( y \)) from 0 to \( \sqrt{3} \), and subtract the parabola from the circle: 
Area \( A = \int_0^{\sqrt{3}} \left( \sqrt{4 - y^2} - \dfrac{y^2}{3} \right) dy \) 
Step 4: Split the integral 
\[ A = \int_0^{\sqrt{3}} \sqrt{4 - y^2} \, dy - \int_0^{\sqrt{3}} \dfrac{y^2}{3} \, dy \] First integral: 
\[ \int_0^{\sqrt{3}} \sqrt{4 - y^2} \, dy \] is a standard integral, result is: 
\[ \left[ \dfrac{y}{2} \sqrt{4 - y^2} + 2 \sin^{-1} \left( \dfrac{y}{2} \right) \right]_0^{\sqrt{3}} = \dfrac{\pi}{3} \] Second integral: 
\[ \int_0^{\sqrt{3}} \dfrac{y^2}{3} dy = \dfrac{1}{3} \cdot \left[ \dfrac{y^3}{3} \right]_0^{\sqrt{3}} = \dfrac{1}{3} \cdot \dfrac{(\sqrt{3})^3}{3} = \dfrac{1}{2\sqrt{3}} \] Step 5: Final Answer 
\[ A = \dfrac{\pi}{3} - \dfrac{1}{2\sqrt{3}} \]