Q.Find the area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Concept: Area of Ellipse — The ellipse is a scaled circle, so its area is $\pi$ times the product of its semi-axes.

Reasoning:

1. The ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is symmetric about both axes. The area in the first quadrant is $\int_0^a y \, dx$, where $y = b\sqrt{1 - x^2/a^2}$.

2. Total area $A = 4 \int_0^a b\sqrt{1 - \frac{x^2}{a^2}} \, dx$. Substitute $x = a\sin\theta$, so $dx = a\cos\theta\, d\theta$, and when $x=0$, $\theta=0$; when $x=a$, $\theta=\pi/2$. …

The formula for the area of a circle is $\pi r^2$. An ellipse is like a stretched circle — stretched by a factor of $a$ along the $x$-axis and $b$ along the $y$-axis. So intuitively, the area should be $\pi a b$, the product of the semi-axes times $\pi$. But let's derive it properly.

1. Set up the integral for area. The ellipse is symmetric about both axes. So the total area is twice the area of the upper half (where $y \ge 0$). From the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, solve for $y$ in the upper half:

This is valid for $x$ from $-a$ to $a$.

2. Write the area as an integral. The area of the upper half is $\int_{-a}^{a} y \, dx$. So the total area $A$ is:

3. Simplify using symmetry. The integrand is even (symmetric about $x=0$), so we can integrate from $0$ to $a$ and double:

4. Substitute to get a standard form. Let $x = a \sin \theta$. Then $dx = a \cos \theta \, d\theta$. When $x = 0$, $\theta = 0$; when $x = a$, $\theta = \frac{\pi}{2}$. The square root becomes:

(since $\cos \theta \ge 0$ in $[0, \pi/2]$).

The integral transforms to:

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

Method: Area of an ellipse from first principles (quarter + symmetry)

Use this to derive $\pi a b$ for a general ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ by integration, rather than quoting the formula.

Steps

Step 1: Exploit symmetry.

The ellipse is symmetric about both axes, so total area $= 4\times$ the first-quadrant area.

Step 2: Solve for $y$ and set up the integral.

Upper-half: $y = b\sqrt{1 - \dfrac{x^2}{a^2}}$, so

Step 3: Substitute $x = a\sin\theta$. …

Common Mistakes

Mistake 1: Forgetting the factor of 4 (or 2) for symmetry.

Why it's wrong: $\int_0^a b\sqrt{1-\tfrac{x^2}{a^2}}\,dx$ is only the first-quadrant area; the full ellipse needs $\times 4$. Correct approach: $\text{Area} = 4\int_0^a b\sqrt{1-\tfrac{x^2}{a^2}}\,dx = \pi a b$.

Mistake 2: Losing the $\tfrac{b}{a}$ or the $\cos^2$ average. …