Q.Find the area enclosed by the circle $x^2 + y^2 = a^2$

The key idea is that the area of a circle is a special case of the area of an ellipse, or can be found directly by integration.

Step 1: The circle $x^2 + y^2 = a^2$ is symmetric about both axes. The area in the first quadrant is one-fourth of the total area.

Step 2: In the first quadrant, $y = \sqrt{a^2 - x^2}$ for $0 \le x \le a$. The area of one quadrant is:

Step 3: This integral evaluates to $\frac{\pi a^2}{4}$ (using the standard formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$, or by recognising it as a quarter-circle).

Step 4: Total area is $4 \times \frac{\pi a^2}{4} = \pi a^2$.

The problem asks for the area inside a circle of radius $a$ centered at the origin. This is a classic result, but deriving it from first principles using integration is a great way to build intuition for how area works in Cartesian coordinates.

The equation $x^2 + y^2 = a^2$ describes a circle. If you solve for $y$, you get $y = \pm \sqrt{a^2 - x^2}$. The positive square root gives the upper half of the circle; the negative gives the lower half. The circle is symmetric about the $x$-axis, so the total area is twice the area of the upper half.

The key idea: the area under a curve $y = f(x)$ from $x = a$ to $x = b$ is $\int_a^b f(x) \, dx$. Here, the upper semicircle runs from $x = -a$ to $x = a$. So the area of the upper half is $\int_{-a}^{a} \sqrt{a^2 - x^2} \, dx$. The total area is twice that.

1. Set up the integral for the total area. The total area $A$ is:

The integrand $\sqrt{a^2 - x^2}$ is an even function (symmetric about $x=0$), so we can simplify:

This avoids dealing with negative limits.

2. Use a trigonometric substitution.

   The expression $\sqrt{a^2 - x^2}$ suggests the substitution $x = a \sin \theta$. Why? Because $\sqrt{a^2 - a^2 \sin^2 \theta} = a \sqrt{1 - \sin^2 \theta} = a \cos \theta$, which is simpler.

   When $x = 0$, $\theta = 0$. When $x = a$, $\theta = \frac{\pi}{2}$. Also, $dx = a \cos \theta \, d\theta$.

   Substitute into the integral:

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

Integrate term by term:

So:

Method: Area of a full circle via the first-quadrant quarter

Use this to derive the area enclosed by $x^2 + y^2 = a^2$ from integration — compute one symmetric quadrant and scale up.

Steps

Step 1: Use symmetry to reduce the work.

The circle is symmetric about both axes, so the total area is $4\times$ the first-quadrant area (where $x, y \ge 0$).

Step 2: Express the arc and set up the quadrant integral.

In the first quadrant, $y = \sqrt{a^2 - x^2}$, so

Step 3: Evaluate with the substitution $x = a\sin\theta$.

Then $\sqrt{a^2 - x^2} = a\cos\theta$ and $dx = a\cos\theta\,d\theta$, giving $4a^2\int_0^{\pi/2}\cos^2\theta\,d\theta$. Using $\cos^2\theta = \tfrac{1+\cos 2\theta}{2}$ yields $\tfrac{\pi}{4}$ for the integral, so

Keep the factor of 4 (or 2, if you halve) explicit — dropping it is the classic error.

Common Mistakes

Mistake 1: Dropping the symmetry factor.

Why it's wrong: $\int_0^a \sqrt{a^2-x^2}\,dx$ is only one quarter of the circle ($\tfrac{\pi a^2}{4}$); reporting that as the whole area misses the factor of 4. Correct approach: total area $= 4\int_0^a\sqrt{a^2-x^2}\,dx = \pi a^2$.

Mistake 2: Mishandling the substitution limits.

Why it's wrong: with $x=a\sin\theta$, $x=0\to\theta=0$ and $x=a\to\theta=\tfrac{\pi}{2}$; carrying over the old $x$-limits corrupts the $\cos^2\theta$ integral. Correct approach: change the limits with the variable, then $4a^2\cdot\tfrac{\pi}{4} = \pi a^2$.