Question

Area bounded by the curve 2y² = 3x and the line x+y = 3 outside the circle (x-3)² + y² = 2 and above the x-axis is A. The value of 4(π +4A) is?

Answer 1

Given:
Parabola: 2y² = 3x
Line: x + y = 3
Circle: (x − 3)² + y² = 2
Region: above x-axis and outside the circle

Step 1: Express x in terms of y
Parabola: x = (2/3)y²
Line: x = 3 − y

Step 2: Points of intersection (parabola & line)
(2/3)y² = 3 − y
2y² + 3y − 9 = 0
y = 3/2 (positive region)

Step 3: Area between line and parabola
\[ A_1 = \int_{0}^{3/2} \Big[(3 - y) - \frac{2}{3}y^2\Big] dy \] \[ = \int_{0}^{3/2} (3 - y - \frac{2}{3}y^2) dy \] \[ = \Big[3y - \frac{y^2}{2} - \frac{2}{9}y^3\Big]_{0}^{3/2} \] \[ = \frac{9}{2} - \frac{9}{8} - \frac{3}{4} = \frac{21}{8} \]

Step 4: Subtract portion inside circle
Circle radius = √2 → semicircle area (above x-axis):
\[ A_2 = \frac{1}{2} \pi (\sqrt{2})^2 = \pi \]

Step 5: Required area
A = A₁ − A₂ = \(\frac{21}{8} - \pi\)

Step 6: Find 4(π + 4A)
\[ 4(\pi + 4A) = 4\Big(\pi + 4\big(\frac{21}{8} - \pi\big)\Big) \] \[ = 4\Big(\pi + \frac{21}{2} - 4\pi\Big) = 4\Big(\frac{21}{2} - 3\pi\Big) \] \[ = 42 - 12\pi \]

Final Answer: 42 − 12π