Question

The area of the region in the first quadrant which is above the parabola \(y = x^2\) and enclosed by the circle \(x^2 +y^2 = 2 \) and the y-axis is

• A. \(\dfrac{1}{6}+\dfrac{π}{4}\)
• B. \(\dfrac{1}{12}+\dfrac{π}{4}\)
• C. \(\dfrac{-1}{6}+\dfrac{\pi}{4}\)
• D. \(\dfrac{-1}{6}-\dfrac{π}{4}\)
• E. \(\dfrac{-π^2}{2} +4\)

Answer 1

The Correct Option is A

Step 1: Identify the region of interest: - Above the parabola \( y = x^2 \) - Inside the circle \( x^2 + y^2 = 2 \) - Bounded by the y-axis (\( x = 0 \)) - In the first quadrant

Step 2: Find intersection points: 1. Between parabola and circle: \[ x^2 + (x^2)^2 = 2 \] \[ x^2 + x^4 = 2 \] \[ x^4 + x^2 - 2 = 0 \] Let \( z = x^2 \): \[ z^2 + z - 2 = 0 \] \[ z = 1 \text{ or } z = -2 \] (only \( z = 1 \) valid) \[ x = 1 \] (in first quadrant)

2. Between circle and y-axis: \[ 0 + y^2 = 2 \] \[ y = \sqrt{2} \]

Step 3: Set up the integral for the area: The area can be calculated as: \[ \text{Area} = \left( \text{Area under circle from } x=0 \text{ to } x=1 \right) - \left( \text{Area under parabola from } x=0 \text{ to } x=1 \right) \]

Step 4: Calculate each part: 1. Area under circle: \[ \int_0^1 \sqrt{2 - x^2} \, dx \] This is a quarter-circle area minus a sector: \[ \frac{\pi (\sqrt{2})^2}{4} - \left( \frac{1}{2} (\sqrt{2})^2 \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) - \frac{1}{2} \cdot 1 \cdot \sqrt{2 - 1} \right) \] \[ = \frac{\pi}{2} - \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{4} + \frac{1}{2} \]

2. Area under parabola: \[ \int_0^1 x^2 \, dx = \left. \frac{x^3}{3} \right|_0^1 = \frac{1}{3} \]

Step 5: Compute net area: \[ \left( \frac{\pi}{4} + \frac{1}{2} \right) - \frac{1}{3} = \frac{\pi}{4} + \frac{1}{6} \]

Answer 2

Let the parabola be \( y = x^2 \) and the circle be \( x^2 + y^2 = 2 \). 

The region is in the first quadrant and is above the parabola \( y = x^2 \).

The parabola and circle intersect when \( x^2 + (x^2)^2 = 2 \), which gives \( x^4 + x^2 - 2 = 0 \). This factors as \( (x^2 - 1)(x^2 + 2) = 0 \), so \( x^2 = 1 \) (since \( x^2 \geq 0 \)). Thus, \( x = 1 \) 

Since we are in the first quadrant. The corresponding \( y \)-coordinate is \( y = x^2 = 1 \). The intersection point is \( (1, 1) \).

The area of the region is given by the integral:

\[ A = \int_0^1 (\sqrt{2 - x^2} - x^2) dx \]

We split the integral into two parts:

\[ A = \int_0^1 \sqrt{2 - x^2} dx - \int_0^1 x^2 dx \]

The second integral is easy:

\[ \int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]

For the first integral, we use the substitution \( x = \sqrt{2} \sin \theta \), so \( dx = \sqrt{2} \cos \theta \, d\theta \). When \( x = 0 \), \( \theta = 0 \). When \( x = 1 \), \( \sin \theta = \frac{1}{\sqrt{2}} \), so \( \theta = \frac{\pi}{4} \). Then

\[ \begin{aligned} \int_0^1 \sqrt{2 - x^2} dx &= \int_0^{\pi/4} \sqrt{2 - 2 \sin^2 \theta} (\sqrt{2} \cos \theta) \, d\theta \\ &= \int_0^{\pi/4} \sqrt{2} \cos \theta \sqrt{2} \cos \theta \, d\theta \\ &= 2 \int_0^{\pi/4} \cos^2 \theta \, d\theta \\ &= 2 \int_0^{\pi/4} \frac{1 + \cos(2\theta)}{2} \, d\theta \\ &= \int_0^{\pi/4} (1 + \cos(2\theta)) \, d\theta \\ &= \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_0^{\pi/4} \\ &= \frac{\pi}{4} + \frac{1}{2} \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{4} + \frac{1}{2} \end{aligned} \]

Therefore, the area is

\[ A = \frac{\pi}{4} + \frac{1}{2} - \frac{1}{3} = \frac{\pi}{4} + \frac{1}{6} \]

Final Answer: The final answer is \( {\frac{\pi}{4} + \frac{1}{6}} \).