Question

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

Hint:

The area of an ellipse with semi-major axis \(a\) and semi-minor axis \(b\) is given by the formula \(A = \pi ab\). This is a standard result worth memorizing. Notice that if \(a=b=r\), the ellipse becomes a circle, and the area is \(\pi r^2\), as expected.

Answer 1

Step 1: Understanding the Concept:
The area of a region bounded by a curve can be found using a definite integral. The ellipse given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is symmetric with respect to both the x-axis and the y-axis. We can calculate the area of the region in the first quadrant and then multiply it by 4 to get the total area of the ellipse.
Step 2: Key Formula or Approach:
The area \(A\) of a region bounded by a curve \(y = f(x)\), the x-axis, and the lines \(x = c\) and \(x = d\) is given by:
\[ A = \int_{c}^{d} y \, dx \] First, we need to express \(y\) in terms of \(x\) from the equation of the ellipse.
From \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we get:
\[ \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2} \] \[ y^2 = \frac{b^2}{a^2}(a^2 - x^2) \] \[ y = \pm \frac{b}{a}\sqrt{a^2 - x^2} \] For the first quadrant, \(y\) is positive, so we take \(y = \frac{b}{a}\sqrt{a^2 - x^2}\). The limits of integration for \(x\) in the first quadrant are from 0 to \(a\).
Step 3: Detailed Explanation:
The total area \(A\) of the ellipse is 4 times the area of the part in the first quadrant.
\[ A = 4 \int_{0}^{a} y \, dx = 4 \int_{0}^{a} \frac{b}{a}\sqrt{a^2 - x^2} \, dx \] To evaluate this integral, we use the trigonometric substitution \(x = a\sin\theta\).
Then \(dx = a\cos\theta \, d\theta\).
We also need to change the limits of integration:
When \(x = 0\), \(a\sin\theta = 0 \implies \sin\theta = 0 \implies \theta = 0\).
When \(x = a\), \(a\sin\theta = a \implies \sin\theta = 1 \implies \theta = \frac{\pi}{2}\).
Substituting these into the integral:
\[ A = 4 \frac{b}{a} \int_{0}^{\pi/2} \sqrt{a^2 - a^2\sin^2\theta} \cdot (a\cos\theta) \, d\theta \] \[ A = 4 \frac{b}{a} \int_{0}^{\pi/2} \sqrt{a^2(1 - \sin^2\theta)} \cdot (a\cos\theta) \, d\theta \] Using the identity \(1 - \sin^2\theta = \cos^2\theta\):
\[ A = 4 \frac{b}{a} \int_{0}^{\pi/2} \sqrt{a^2\cos^2\theta} \cdot (a\cos\theta) \, d\theta \] \[ A = 4 \frac{b}{a} \int_{0}^{\pi/2} (a\cos\theta) \cdot (a\cos\theta) \, d\theta \] \[ A = 4ab \int_{0}^{\pi/2} \cos^2\theta \, d\theta \] Now, use the identity \(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\):
\[ A = 4ab \int_{0}^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta \] \[ A = 2ab \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{0}^{\pi/2} \] \[ A = 2ab \left[ \left(\frac{\pi}{2} + \frac{\sin(\pi)}{2}\right) - \left(0 + \frac{\sin(0)}{2}\right) \right] \] \[ A = 2ab \left[ \left(\frac{\pi}{2} + 0\right) - (0 + 0) \right] \] \[ A = 2ab \left(\frac{\pi}{2}\right) = \pi ab \] Step 4: Final Answer:
The area of the region bounded by the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(\pi ab\).