The area of region \( \{(x,y):9x^2+16y^2\le 144\ ,x\ge 0\ ,y\ge 0\} \) is-
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3\pi\ Sq.\ units \)
Dividing by 144, \( \frac{x^2}{16}+\frac{y^2}{9}\le 1 \), an ellipse with a = 4 and b = 3. The region is its first-quadrant part.
\( \displaystyle \text{Area}=\int_0^4\frac{3}{4}\sqrt{16-x^2}\,dx \)
\( \displaystyle =\frac{3}{4}\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\frac{x}{4}\right]_0^4 \)
\( \displaystyle =\frac{3}{4}\cdot 8\cdot\frac{\pi}{2} \)
\( \displaystyle =3\pi \)
Shortcut: The whole ellipse has area \( \pi ab=12\pi \), and the first quadrant is one quarter of it, i.e. \( 3\pi \).
Why the other options are incorrect
Option (2) \( 6\pi\ Sq.\ units \) — Incorrect. This is half the ellipse (two quadrants).
Option (3) \( 9\pi\ Sq.\ units \) — Incorrect. This is three quarters of the ellipse.
Option (4) \( 12\pi\ Sq.\ units \) — Incorrect. This is the area of the whole ellipse; only the first quadrant is required.