Q.Sketch the region $\{(x, 0) : y = \sqrt{4 - x^2}\}$ and x-axis. Find the area of the region using integration.

Concept: Area Under Curve – the area bounded by $y = \sqrt{4 - x^2}$ and the x-axis is the region between the curve and the axis from $x = -2$ to $x = 2$.

Steps:

1. The curve $y = \sqrt{4 - x^2}$ is the upper half of a circle $x^2 + y^2 = 4$ (radius 2). The region is the semicircle above the x-axis.

2. Area is given by

3. Use the standard formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$ with $a = 2$:

4. Evaluate: at $x = 2$, $\sqrt{4 - 4} = 0$, $\sin^{-1}(1) = \frac{\pi}{2}$; at $x = -2$, $\sqrt{4 - 4} = 0$, $\sin^{-1}(-1) = -\frac{\pi}{2}$.

The problem asks us to sketch the region bounded by $y = \sqrt{4 - x^2}$ and the x-axis, then find its area using integration. Let’s first understand what this curve is.

The equation $y = \sqrt{4 - x^2}$ is not just any curve — it’s the upper half of a circle. Why? Because if you square both sides, you get $y^2 = 4 - x^2$, which rearranges to $x^2 + y^2 = 4$. That’s a circle of radius 2 centred at the origin. But since $y$ is defined as the positive square root (the $\sqrt{\ }$ symbol always gives the non-negative value), we only get the top half: $y \ge 0$. The x-axis ($y = 0$) is the lower boundary. So the region is exactly the semicircle above the x-axis, from $x = -2$ to $x = 2$.

Now, the area under a curve $y = f(x)$ from $x = a$ to $x = b$ is given by the definite integral $\int_a^b f(x) \, dx$. Here, $f(x) = \sqrt{4 - x^2}$, and the region runs from the leftmost point of the semicircle ($x = -2$) to the rightmost ($x = 2$). So the area is:

This integral is a classic one. It represents the area of a semicircle of radius 2, so we already know the answer should be $\frac{1}{2} \pi (2)^2 = 2\pi$. But let’s evaluate it properly using integration, as the problem demands.

1. Set up the integral. The area is $A = \int_{-2}^{2} \sqrt{4 - x^2} \, dx$. The integrand is an even function (since $\sqrt{4 - (-x)^2} = \sqrt{4 - x^2}$), so we can simplify by integrating from 0 to 2 and doubling:

This saves a bit of work.

2. Use a trigonometric substitution.

   The expression $\sqrt{4 - x^2}$ suggests the substitution $x = 2 \sin \theta$, because then $4 - x^2 = 4 - 4\sin^2 \theta = 4\cos^2 \theta$, and $\sqrt{4 - x^2} = 2|\cos \theta|$. For $x$ from 0 to 2, $\theta$ goes from $0$ to $\pi/2$, where $\cos \theta \ge 0$, so we can drop the absolute value: $\sqrt{4 - x^2} = 2\cos \theta$.

   Also, $dx = 2\cos \theta \, d\theta$. When $x = 0$, $\theta = 0$; when $x = 2$, $\theta = \pi/2$.

3. Transform the integral.

   Substitute everything in:

4. Evaluate the trigonometric integral. Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:

Integrate term by term:

Since $\sin \pi = 0$ and $\sin 0 = 0$, this simplifies to:

Method: Area under a curve that is really half a circle

This method handles any "find the area under $y = \sqrt{a^2 - x^2}$" (or similar semicircular) problem, where the curve turns out to be part of a circle.

Steps

Step 1: Recognise the shape by squaring.

Whenever you see $y = \sqrt{a^2 - x^2}$, square both sides to reveal the hidden conic. Here $y^2 = a^2 - x^2$, i.e.

That is a circle of radius $a$ centred at the origin. Because the square root sign only returns non-negative values, the curve is the upper half — a semicircle above the $x$-axis.

Step 2: Read off the limits from the geometry.

The semicircle meets the $x$-axis where $y = 0$, i.e. at $x = -a$ and $x = a$. Those become the limits of integration. Always let the picture, not guesswork, fix the limits.

Step 3: Write the area as a definite integral.

Step 4: Evaluate with the standard result.

Use the memorised antiderivative

or equivalently the substitution $x = a\sin\theta$. Substituting the limits, the square-root terms vanish at both ends and only the $\sin^{-1}$ terms survive.

Step 5: Sanity-check against known area.

A full circle has area $\pi a^2$, so a semicircle must give $\tfrac{1}{2}\pi a^2$. If your integral disagrees, you have most likely mishandled a limit or forgotten that the radical only gives the top half.

Common Mistakes

Mistake 1: Treating $y = \sqrt{4 - x^2}$ as the whole circle

Why it's wrong: the square-root symbol returns only the non-negative value, so this curve is just the upper semicircle of $x^2 + y^2 = 4$. Integrating as if both halves were included (or writing $y = \pm\sqrt{4-x^2}$) doubles the region and gives $4\pi$ instead of $2\pi$. Correct approach: the region is the half-disc above the $x$-axis, area $\tfrac12\pi r^2 = 2\pi$.

Mistake 2: Wrong limits after the substitution $x = 2\sin\theta$

Why it's wrong: at $x = 2$ the correct angle is $\theta = \tfrac{\pi}{2}$, not $\theta = \pi$; pushing $\theta$ to $\pi$ makes $\cos\theta$ negative and corrupts the sign of the integrand. Correct approach: map $x = 0 \to \theta = 0$ and $x = 2 \to \theta = \tfrac{\pi}{2}$, where $\cos\theta \ge 0$ so $\sqrt{4-x^2} = 2\cos\theta$.