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Q3:
Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is

Solution :

Step 1: Analyzing the Geometric Constraints

The given equation is $x^2 + y^2 = 4$, which represents a circle centered at the origin $(0, 0)$ with radius $r = \sqrt{4} = 2$ units [By the standard form of circle equation $(x-h)^2 + (y-k)^2 = r^2$].

The boundary conditions are:

  • The circle: $y^2 = 4 - x^2 \implies y = \sqrt{4 - x^2}$ (taking the positive root as we are constrained to the first quadrant).
  • The vertical lines: $x = 0$ (the $y$-axis) and $x = 2$ (the rightmost edge of the circle).

Step 2: Defining the Integral Setup

The area $A$ in the first quadrant bounded by the curve $f(x) = \sqrt{4 - x^2}$ from $x = 0$ to $x = 2$ is given by the definite integral:

$A = \int_{0}^{2} \sqrt{4 - x^2} \, dx$

Step 3: Evaluating the Integral

Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$:

Substitute $a = 2$:

$A = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

$A = \left[ \frac{x}{2}\sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

Applying the Fundamental Theorem of Calculus:

$A = \left( \frac{2}{2}\sqrt{4 - 2^2} + 2 \sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{0}{2}\sqrt{4 - 0^2} + 2 \sin^{-1}\left(\frac{0}{2}\right) \right)$

$A = (1 \cdot 0 + 2 \cdot \sin^{-1}(1)) - (0 + 2 \cdot 0)$

$A = 2 \cdot \left(\frac{\pi}{2}\right) = \pi$

Step 4: Geometric Visualization

2 2 0

Final Solution: The area is $\pi$ square units.



CBSE Solutions for Class 12 Mathematics Application of Integrals


Chapters in CBSE - Class 12 Mathematics


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