Q.Determine the area under the curve $y = \sqrt{a^2 - x^2}$ included between the lines $x = 0$ and $x = a$.

Concept: Area Under Curve — the definite integral of $y$ with respect to $x$ between the given limits.

The curve $y = \sqrt{a^2 - x^2}$ is the upper half of a circle of radius $a$ centered at the origin. The area from $x = 0$ to $x = a$ is one quarter of the full circle’s area.

Steps:

1. Required area:

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

The first thing to notice is the form of the equation. $y = \sqrt{a^2 - x^2}$ is not just any curve — it’s the upper semicircle of radius $a$ centered at the origin. Why? Because if you square both sides, you get $y^2 = a^2 - x^2$, or $x^2 + y^2 = a^2$, which is the equation of a full circle. The square root restricts $y$ to be non-negative, so we only get the top half.

The problem asks for the area under this curve between $x = 0$ and $x = a$. “Under the curve” means the region bounded above by the curve, below by the $x$-axis, and on the sides by the vertical lines $x=0$ and $x=a$. That’s exactly the region in the first quadrant under the semicircle — a quarter of the full circle.

So the area is simply one-fourth of the area of a circle of radius $a$:

But let’s also do it the calculus way, because that’s how you’d be expected to show it in an exam.

1. Set up the definite integral. The area under a curve $y = f(x)$ from $x = 0$ to $x = a$ is given by:

2. Recognise the integral as a standard form.

   The integral $\int \sqrt{a^2 - x^2} \, dx$ is a classic one. It’s most easily handled by a trigonometric substitution: let $x = a \sin \theta$. Then $dx = a \cos \theta \, d\theta$, and when $x = 0$, $\theta = 0$; when $x = a$, $\theta = \frac{\pi}{2}$.

3. Substitute and simplify.

So the integral becomes:

4. Use the double-angle identity. $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$, so: …

Method: Area of a quarter-circle by integration

This method finds the area under $y = \sqrt{a^2 - x^2}$ over $[0, a]$ — a quarter of a circle — and, more generally, any integral of the form $\int \sqrt{a^2 - x^2}\, dx$.

Steps

Step 1: Recognise the curve.

Squaring $y = \sqrt{a^2 - x^2}$ gives $x^2 + y^2 = a^2$, a circle of radius $a$; the radical keeps only the upper half. Restricting further to $0 \le x \le a$ isolates the first-quadrant quarter-circle.

Step 2: Write the definite integral.

Step 3: Use the standard antiderivative (or $x = a\sin\theta$).

$\int \sqrt{a^2 - x^2}\, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C.$ …

Common Mistakes

Mistake 1: Confusing the quarter-circle with the semicircle

Why it's wrong: the limits are $x = 0$ to $x = a$, which covers only the first-quadrant quarter of the disc; integrating from $-a$ to $a$ instead gives the half-disc area $\tfrac{\pi a^2}{2}$, double the required value. Correct approach: over $[0, a]$ the area is one quarter of $\pi a^2$, namely $\dfrac{\pi a^2}{4}$.

Mistake 2: Forgetting that $y = \sqrt{a^2 - x^2}$ is only the upper half …