Question

Find the area bounded by the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).

Hint:

Area = 4×(first-quadrant area) = (4b/a)∫√(a²−x²)dx from 0 to a.

Answer 1

Step 1: Understanding the Concept:
By symmetry, the ellipse is symmetric about both axes, so the total area is 4 times the area in the first quadrant.

Step 2: Expressing \(y\) in terms of \(x\):
From \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), solving for the first-quadrant branch:
\[ y = \frac{b}{a}\sqrt{a^2-x^2} \]

Step 3: Setting up the area integral:
\[ \text{Area} = 4\int_0^a y\,dx = 4\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx = \frac{4b}{a}\int_0^a\sqrt{a^2-x^2}\,dx \]

Step 4: Using the standard result \(\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx = \dfrac{\pi a^2}{4}\):
\[ \text{Area} = \frac{4b}{a}\cdot\frac{\pi a^2}{4} = \pi ab \]

Final Answer:
The area enclosed by the ellipse is \(\pi ab\) square units. \[ \boxed{\pi ab} \]