Q.Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

The region is an ellipse centred at the origin with semi-major axis $a = 4$ (along $x$) and semi-minor axis $b = 3$ (along $y$).

The area of an ellipse is given by the formula $A = \pi a b$.

Substituting the values:

The problem asks for the area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$. This is a standard result, but let's build it from first principles so you see why the formula $\pi a b$ works — and so you can handle any ellipse question in your exam.

An ellipse is essentially a stretched circle. If you take a circle of radius $r$ and stretch it horizontally by a factor $a/r$ and vertically by a factor $b/r$, you get an ellipse with semi-axes $a$ and $b$. Since area scales by the product of the stretch factors, the area of the ellipse is $\pi r^2 \cdot (a/r)(b/r) = \pi a b$. That's the intuition.

Now let's do it rigorously using integration — the method your exam expects.

1. Identify the semi-axes.

   The given equation is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Comparing with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get $a^2 = 16$ so $a = 4$, and $b^2 = 9$ so $b = 3$. The ellipse is centered at the origin, symmetric about both axes.

2. Set up the area using symmetry.

   The ellipse is symmetric about the $x$-axis and the $y$-axis. So the total area is 4 times the area in the first quadrant.

   In the first quadrant, $x$ goes from $0$ to $a = 4$, and for each $x$, $y$ goes from $0$ to the upper half of the ellipse. Solve for $y$ from the equation:

So the area in the first quadrant is

3. Evaluate the integral. The integral $\int \sqrt{16 - x^2}\,dx$ is a standard form. Use the substitution $x = 4\sin\theta$, so $dx = 4\cos\theta\,d\theta$. When $x = 0$, $\theta = 0$; when $x = 4$, $\theta = \frac{\pi}{2}$. Then

The integral becomes

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

So the first-quadrant area is $3\pi$.

4. Multiply by 4 for the total area. Total area $= 4 \times 3\pi = 12\pi$.

Method: Area of an ellipse by symmetry and integration

This is the technique for finding the whole area enclosed by a conic centred at the origin (ellipse or circle) using a definite integral, rather than just quoting a formula.

Steps

Step 1: Read the semi-axes off the standard form.

Write the curve as

The number under $x^2$ is $a^2$ and the number under $y^2$ is $b^2$, so $a$ and $b$ are just the square roots. (For a circle $a = b = r$.)

Step 2: Use the two axes of symmetry.

The curve is symmetric about both the $x$- and $y$-axes, so the total area is four times the first-quadrant piece. Solve the upper half for $y$:

which runs over $x$ from $0$ to $a$.

Step 3: Integrate using the standard square-root form.

Set up

and evaluate with the standard result

Between $0$ and $a$ this quarter-circle integral is $\dfrac{\pi a^2}{4}$, so the area collapses to $A = \pi a b$ (which becomes $\pi r^2$ for a circle). Quote $\pi a b$ directly only if the question does not demand the integration steps.

Common Mistakes

Mistake 1: Using the denominators directly as $a$ and $b$

A student reads $\frac{x^2}{16}+\frac{y^2}{9}=1$ and writes area $=\pi\times 16\times 9$. Why it's wrong: $16$ and $9$ are $a^2$ and $b^2$, not $a$ and $b$. Correct approach: take square roots first — $a=\sqrt{16}=4$, $b=\sqrt{9}=3$, so area $=\pi ab=12\pi$.

Mistake 2: Forgetting the factor of $4$ when integrating

When deriving the area by integration, some compute only the first-quadrant piece $\int_0^4 \frac{3}{4}\sqrt{16-x^2}\,dx=3\pi$ and report $3\pi$ as the answer. Why it's wrong: that integral covers just one of the four symmetric quadrants. Correct approach: multiply by $4$ to get $12\pi$.

Mistake 3: Mishandling the $\sqrt{16-x^2}$ integral

Students forget the standard form $\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}$ (or the substitution $x=4\sin\theta$) and try to integrate the square root term by term. Correct approach: use the standard result or the trig substitution — never distribute the square root over the subtraction.