If the area (in sq. units) of the region \(\{(x, y): x^{2}+y^{2} \le r^{2},\ y \ge 0\}\), where r > 0, is \(18\pi\), then r is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6
The region is the part of the disc of radius r on or above the x-axis, that is, the upper half-disc. Its area is half of \(\pi r^{2}\):
\( \displaystyle \frac{1}{2}\pi r^{2} = 18\pi \)
\(r^{2} = 36\)
\(r = 6\) (since r > 0)
Why the other options are incorrect
Option (1) \(3\sqrt{2}\) — Incorrect. This comes from taking \(18\pi\) as the area of the whole disc, \(\pi r^{2} = 18\pi\); the region is only half of the disc.
Option (2) 36 — Incorrect. 36 is \(r^{2}\), not r.
Option (3) 3 — Incorrect. With r = 3 the half-disc has area \(\frac{9\pi}{2}\), not \(18\pi\).