Q.Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Concept: Area of Ellipse — The area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi a b$.

Reasoning:

1. The given ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$, so $a^2 = 4$ and $b^2 = 9$. Thus $a = 2$, $b = 3$.
2. The ellipse is symmetric about both axes. The total area is $4$ times the area in the first quadrant.
3. Using the standard formula for the area of an ellipse: $\text{Area} = \pi a b = \pi \cdot 2 \cdot 3 = 6\pi$.

The problem asks for the area bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$. This is a standard ellipse centered at the origin, with its major axis along the $y$-axis (since $9 > 4$). The semi-major axis length is $b = 3$ and the semi-minor axis length is $a = 2$.

The area of an ellipse is one of those results that feels intuitive once you see the connection to a circle. A circle of radius $r$ has area $\pi r^2$. If you stretch that circle horizontally by a factor of $a$ and vertically by a factor of $b$, you get an ellipse, and its area becomes $\pi a b$. That’s the core idea — scaling changes area multiplicatively.

Let’s verify this using integration, which also reinforces why the formula works.

1. Set up the integral for area. The ellipse is symmetric about both axes. So the total area is 4 times the area in the first quadrant. From the equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$, solve for $y$ in the first quadrant:

The $x$-coordinate runs from $0$ to $2$ (where $y = 0$). So the area in the first quadrant is:

2. Evaluate the integral. The integral $\int \sqrt{a^2 - x^2} \, dx$ is a standard form. Here $a = 2$. Recall the formula:

Applying it:

At $x = 2$: $\frac{2}{2}\sqrt{0} + 2\sin^{-1}(1) = 0 + 2 \cdot \frac{\pi}{2} = \pi$

At $x = 0$: $0 + 2\sin^{-1}(0) = 0$

So the definite integral equals $\pi$.

Therefore:

3. Multiply by 4 for total area.

Method: Area enclosed by a standard ellipse

Use this whenever the region is the full interior of an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, regardless of which axis is longer.

Steps

Step 1: Identify the semi-axes.

The denominator under $x^2$ is $a^2$ and under $y^2$ is $b^2$. Take square roots to obtain $a$ and $b$ — never use the denominators themselves as the semi-axes.

Step 2: Apply the area formula.

The product $ab$ is symmetric, so it does not matter which of $a,b$ is the larger (semi-major) axis.

Step 3: Substitute and simplify.

Multiply the two semi-axes by $\pi$. This formula is derived by integrating a quarter of the ellipse and multiplying by 4, but for the whole ellipse the result $\pi a b$ can be quoted directly.

Common Mistakes

Mistake 1: Using the denominators $4$ and $9$ as the semi-axes.

Why it's wrong: those are $a^2$ and $b^2$; the semi-axes are $a=2,\ b=3$. Multiplying $\pi\cdot 4\cdot 9$ is wrong. Correct approach: take square roots first, then $\text{Area} = \pi a b = \pi\cdot 2\cdot 3 = 6\pi$.

Mistake 2: Worrying about which axis is "major".

Why it's wrong: since $9>4$ the major axis is along $y$, but the area formula $\pi a b$ is symmetric, so the labelling does not change the answer. Correct approach: just multiply the two semi-axes — the product $ab=6$ gives $6\pi$ either way.