Question

Using integration, find the area bounded by the curve \( 9x^2 + 16y^2 = 144 \) and the lines \( x = 2 \), \( x = -2 \).

Hint:

Write the ellipse as \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) and find \( 4\int_0^2 \frac{3}{4}\sqrt{16 - x^2}\,dx \).

Answer 1

Step 1: Write the ellipse in standard form:
Divide \( 9x^2 + 16y^2 = 144 \) by 144.
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
This is an ellipse with \( a = 4 \) and \( b = 3 \).
Solving for \( y \) gives \( y = \pm\frac{3}{4}\sqrt{16 - x^2} \).

Step 2: Set up the area:
The region between \( x = -2 \) and \( x = 2 \) is symmetric about both axes.
So the area is 4 times the first-quadrant part from \( x = 0 \) to \( x = 2 \).
\[ A = 4\int_{0}^{2} \frac{3}{4}\sqrt{16 - x^2}\,dx = 3\int_{0}^{2}\sqrt{16 - x^2}\,dx \]

Step 3: Integrate with the standard formula:
Use \( \int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \) with \( a = 4 \).
\[ A = 3\left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4} \right]_{0}^{2} \]
At \( x = 2 \): \( \frac{2}{2}\sqrt{12} = 2\sqrt{3} \) and \( 8\sin^{-1}\frac{1}{2} = 8\cdot\frac{\pi}{6} = \frac{4\pi}{3} \).
At \( x = 0 \) the value is 0.

Step 4: Simplify:
\[ A = 3\left( 2\sqrt{3} + \frac{4\pi}{3} \right) = 6\sqrt{3} + 4\pi \]

Final Answer:
The required area is \( (4\pi + 6\sqrt{3}) \) square units. \[ \boxed{4\pi + 6\sqrt{3}\ \text{sq. units}} \]