Q.Find the value of the following: 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 (A) $\pi$ (B) $\frac{\pi}{2}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{4}$

The region is the quarter of the circle $x^2 + y^2 = 4$ (radius $2$) in the first quadrant, between $x=0$ and $x=2$.

Step 1 – The required area is given by the definite integral

where from the circle equation, $y = \sqrt{4 - x^2}$ (positive in first quadrant).

Step 2 – So

This is a standard form: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$, with $a=2$.

Step 3 – Evaluating from $0$ to $2$: …

The problem asks for the area in the first quadrant bounded by the circle $x^2 + y^2 = 4$, the y-axis ($x = 0$), and the vertical line $x = 2$.

The circle $x^2 + y^2 = 4$ has centre at the origin and radius $2$. The first quadrant is the region where $x \ge 0$ and $y \ge 0$. The line $x = 2$ is the rightmost point of the circle (where the circle touches the x-axis at $(2,0)$). So the region is exactly the quarter of the circle lying in the first quadrant.

Area under a curve is found by integrating $y$ with respect to $x$ between the given limits. Here, from the circle equation, the upper half is $y = \sqrt{4 - x^2}$. The area in the first quadrant is the integral of this from $x = 0$ to $x = 2$.

1. Set up the integral The area $A$ is given by

2. Recognise the geometric meaning

   The integral $\int_{0}^{2} \sqrt{4 - x^2} \, dx$ is the area of a quarter-circle of radius $2$. The full circle area is $\pi r^2 = \pi (2)^2 = 4\pi$. One quarter of that is $\pi$.

3. Evaluate the integral (standard trigonometric substitution)

   Use $x = 2 \sin \theta$, so $dx = 2 \cos \theta \, d\theta$. When $x = 0$, $\theta = 0$; when $x = 2$, $\theta = \frac{\pi}{2}$. Then

The integral becomes

4. Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ …

Method: Area of a circular sector by integration (quarter-circle in the first quadrant)

Use this when the region is a slice of a circle $x^2 + y^2 = r^2$ cut off by axes or vertical lines — here the first-quadrant quarter.

Steps

Step 1: Solve the circle for the relevant half.

In the first quadrant, $y = \sqrt{r^2 - x^2}$ (the positive root). The region under this arc between two $x$-values is an area-under-curve problem.

Step 2: Set up the definite integral with the correct limits.

Choosing $a=0,\, b=r$ captures exactly the first-quadrant quarter. …

Common Mistakes

Mistake 1: Computing the whole semicircle instead of the first-quadrant quarter.

Why it's wrong: integrating $\sqrt{4-x^2}$ from $-2$ to $2$ gives the half-disc area $2\pi$, not the quarter asked for. Correct approach: use the stated limits $x=0$ to $x=2$, which enclose exactly one quadrant, giving $\pi$.

Mistake 2: Squaring errors on the radius. …