The area of the region enclosed between the circles x2 + y2 = 4 and x2 + (y

Solution & Explanation

Related Formula

\int \sqrt(a² - x²) dx = (x)/(2)\sqrt(a² - x²) + (a²)/(2) ⁻¹((x)/(a))

Core Logic

Area Between Two Curves diagram for Q12 - JEE Main 2026 Evening

The intersection points of

x² + y² = 4

and

x² + (y-2)² = 4

: Subtracting the equations gives

y² - (y-2)² = 0 y² - (y² - 4y + 4) = 0 4y = 4 y = 1

. Substitute

y=1

back:

x² + 1 = 4 x = \pm \sqrt(3)

. The points of intersection are

(\sqrt(3), 1)

and

(-\sqrt(3), 1)

The area is symmetric about the y-axis, so we integrate from $x = 0$ $x = 0$ to $x = \sqrt(3)$ $x = \sqrt{3}$ and double the result. Upper curve is the lower arc of the top circle: $y = 2 - \sqrt(4-x²)$ $y = 2 - \sqrt{4-x^2}$ -- wait, the area is bounded by the top arc of the bottom circle and the bottom arc of the top circle. Upper curve for area: $y = \sqrt(4-x²)$ $y = \sqrt{4-x^2}$ (from $x²+y²=4$ $x^2+y^2=4$) Lower curve for area: $y = 2 - \sqrt(4-x²)$ $y = 2 - \sqrt{4-x^2}$ (from $x²+(y-2)²=4$ $x^2+(y-2)^2=4$)

A = 2\int₀\sqrt(3) [\sqrt(4-x²) - (2 - \sqrt(4-x²))] dx

Step 1: Integration

A = 2\int₀\sqrt(3) (2\sqrt(4-x²) - 2) dx = 4\int₀\sqrt(3) (\sqrt(4-x²) - 1) dx

A = 4[ (1)/(2)(x\sqrt(4-x²) + 4 ⁻¹(x)/(2)) - x ]₀\sqrt(3)

Evaluating the limits:

= 4[ (1)/(2)(\sqrt(3)(1) + 4 ⁻¹( \sqrt(3)2)) - \sqrt(3) - (0) ]

= 4[ (1)/(2)(\sqrt(3) + 4((π)/(3))) - \sqrt(3) ]

= 4[ \sqrt(3)2 + (2π)/(3) - \sqrt(3) ]

= 4[ (2π)/(3) - \sqrt(3)2 ]

= (8π)/(3) - 2\sqrt(3) = (2)/(3)(4π - 3\sqrt(3)) (Sq. units)

Pattern Recognition

For intersecting identical circles with centres on an axis, symmetry simplifies the integration drastically. Recognize that integrating the circle arc function handles the bulk of the calculation.

Chapter Mix

Class 12 Maths: Area Under Curves

More Area Under Curves Previous-Year Questions — Page 6

Q11 jee_main_2024_31_jan_evening Area Under Curves

The area of the region enclosed by the parabola

y = 4x - x²

and

3y = (x - 4)²

is equal to

A. $(32)/(9)$ $\frac{32}{9}$
B. $4$ $4$
C. $6$ $6$
D. $(14)/(3)$ $\frac{14}{3}$

Solution

Related Formula

Area = \intₐb (yupper - ylower) dx

Core Logic

Area Under Curves diagram for Q11 - JEE Main 2024 Evening

Find intersection points of $y = 4x - x²$ $y = 4x - x^2$ and $3y = (x - 4)²$ $3y = (x - 4)^2$:

3(4x - x²) = x² - 8x + 16

12x - 3x² = x² - 8x + 16

4x² - 20x + 16 = 0 x² - 5x + 4 = 0

Roots are $x = 1, 4$ $x = 1, 4$.

Area integral:

Area = \int₁⁴ [ (4x - x²) - ((x - 4)²)/(3) ] dx

= [ (4x²)/(2) - (x³)/(3) - ((x - 4)³)/(9) ]₁⁴

= [ 2(16) - (64)/(3) - 0 ] - [ 2(1) - (1)/(3) - ((-3)³)/(9) ]

= ( 32 - (64)/(3) ) - ( 2 - (1)/(3) + 3 )

= (32)/(3) - ( 5 - (1)/(3) ) = (32)/(3) - (14)/(3) = (18)/(3) = 6

Chapter Mix

Class 12 Maths: Applications of the Integrals

Q5 jee_main_2024_31_jan_morning Area bounded by Parabolas and Inequalities

The area of the region

(x,y): y² \leq 4x, x < 4, (xy(x - 1)(x - 2))/((x - 3)(x - 4)) > 0, x \neq 3

A. $(16)/(3)$ $$\frac{16}{3}$$
B. $(64)/(3)$ $$\frac{64}{3}$$
C. $(8)/(3)$ $$\frac{8}{3}$$
D. $(32)/(3)$ $$\frac{32}{3}$$

Solution

Core Logic

Given $y² \leq 4x$ $y^2 \le 4x$ and $x < 4$ $x < 4$. Analyze the inequality $(xy(x-1)(x-2))/((x-3)(x-4)) > 0$ $\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ considering $y > 0$ $y > 0$ and $y < 0$ $y < 0$ separately.

Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning

Step 1: Case I (y > 0)

If $y > 0$ $y > 0$, the inequality reduces to $(x(x-1)(x-2))/((x-3)(x-4)) > 0$ $\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$. Using wavy curve method and given $x in (0, 4)$ $x \in (0, 4)$: $x in (0, 1) (2, 3)$ $x \in (0, 1) \cup (2, 3)$.

Step 2: Case II (y < 0)

If $y < 0$ $y < 0$, the inequality reduces to $(x(x-1)(x-2))/((x-3)(x-4)) < 0$ $\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$. Using wavy curve method and given $x in (0, 4)$ $x \in (0, 4)$: $x in (1, 2) (3, 4)$ $x \in (1, 2) \cup (3, 4)$.

Step 3: Area Computation

Because the regions map perfectly without overlap in opposite quadrants relative to the x-axis, they form complete parabolic strips when combined: Area $= 2 \int₀⁴ \sqrt(x) dx = 2 \cdot (2)/(3)[x3/2]₀⁴ = (4)/(3) \cdot 8 = (32)/(3)$ .

Chapter Mix

Class 12 Maths: Area Under Curves Class 11 Maths: Linear Inequalities

The area of the region enclosed between the circles $x² + y² = 4$ $x^{2} + y^{2} = 4$ and $x² + (y - 2)² = 4$ $x^{2} + (y - 2)^{2} = 4$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Integration

Pattern Recognition

Chapter Mix

Reference Study Guides

More Area Under Curves Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

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The area of the region enclosed between the circles x² + y² = 4$x^{2} + y^{2} = 4$ and x² + (y - 2)² = 4$x^{2} + (y - 2)^{2} = 4$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Integration

Pattern Recognition

Chapter Mix

Reference Study Guides

More Area Under Curves Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

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The area of the region enclosed between the circles $x² + y² = 4$ $x^{2} + y^{2} = 4$ and $x² + (y - 2)² = 4$ $x^{2} + (y - 2)^{2} = 4$ is: