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Q2:
Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
Solution :
Step 1: Analyzing the Equation of the Ellipse
The given equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$. This is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where:
- $a^2 = 4 \implies a = 2$ (Semi-major axis along the $x$-axis is $2$)
- $b^2 = 9 \implies b = 3$ (Semi-major axis along the $y$-axis is $3$)
[Per standard conic section definitions, since $b > a$, the ellipse is elongated along the $y$-axis.]
Step 2: Expressing $y$ in terms of $x$
To find the area using integration, we isolate $y$ from the equation:
$\frac{y^2}{9} = 1 - \frac{x^2}{4}$
$y^2 = 9 \left(1 - \frac{x^2}{4}\right) = \frac{9}{4}(4 - x^2)$
$y = \pm \frac{3}{2}\sqrt{4 - x^2}$
Since the ellipse is symmetric about both axes, the total area $A$ is 4 times the area of the region in the first quadrant where $x$ ranges from $0$ to $2$.
Step 3: Setting up the Definite Integral
$A = 4 \int_{0}^{2} y \, dx = 4 \int_{0}^{2} \frac{3}{2} \sqrt{2^2 - x^2} \, dx$
$A = 6 \int_{0}^{2} \sqrt{2^2 - x^2} \, dx$
Step 4: Applying the Integration Formula
Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:
$A = 6 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$
$A = 6 \left[ \left( \frac{2}{2}\sqrt{4 - 4} + 2\sin^{-1}(1) \right) - \left( 0 + 2\sin^{-1}(0) \right) \right]$
$A = 6 \left[ 0 + 2(\frac{\pi}{2}) - 0 \right]$
$A = 6 \pi$
Visual Representation of the Ellipse
Summary of Calculation
| Component | Value |
|---|---|
| Integration Bounds | $[0, 2]$ |
| Integrand | $\frac{3}{2}\sqrt{4-x^2}$ |
| Multiplier (Symmetry) | $4$ |
Final Solution: The area of the region bounded by the ellipse is $6\pi$ square units.
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