Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is

A — The question asks for the area in the first quadrant bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2. This region is a quarter of the circle with radius 2. The area can be calculated using integration: ∫₀² √(4 - x²) dx. Using the standard integral formula from the context, this equals [x/2 √(4 - x²) + 2 sin⁻¹(x/2)]₀². Evaluating at x = 2 gives 0 + 2(π/2) = π, and at x = 0 gives 0. Therefore, the area is π square units.

Area lying in the first quadrant and bounded by the circle x² + y² =…

Question 3 of 9intermediate🔧 ApplyMCQ1 mark

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

Exercise: EXERCISE 8.1 | Q: 3 | (Page Page 5)

For More Understanding

Explanation

The question asks for the area in the first quadrant bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2. This region is a quarter of the circle with radius 2. The area can be calculated using integration: ∫₀² $\sqrt{4 - x²}$ dx. Using the standard integral formula from the context, this equals [x/2 $\sqrt{4 - x²}$ + 2 sin⁻¹(x/2)]₀². Evaluating at x = 2 gives 0 + 2(π/2) = π, and at x = 0 gives 0. Therefore, the area is π square units.

Solution Steps

Step 1: Identify the region - first quadrant bounded by circle x² + y² = 4, x = 0 (y-axis), and x = 2.
Step 2: Express y from the circle equation: y = $\sqrt{4 - x²}$ (positive in first quadrant).
Step 3: Set up the integral: Area = ∫₀² $\sqrt{4 - x²}$ dx.
Step 4: Apply the standard integral: ∫ $\sqrt{a² - x²}$ dx = (x/2) $\sqrt{a² - x²}$ + (a²/2)sin⁻¹(x/a).
Step 5: Evaluate: [(x/2) $\sqrt{4 - x²}$ + 2 sin⁻¹(x/2)]₀² = [0 + 2(π/2)] - [0 + 0] = π.

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