The area (in sq. units) of the region in the first quadrant bounded by the parabola \(y = x^{2}\), the y-axis and the line y = 9 is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 18
The region lies between the curve and the y-axis, so it is found by integrating in y. In the first quadrant the parabola is
\(x = \sqrt{y}\)
For each y from 0 to 9, x runs from the y-axis to the curve:
\( \displaystyle A = \int_{0}^{9}\sqrt{y}\,dy \)
\( \displaystyle = \frac{2}{3}\left[y^{3/2}\right]_{0}^{9} \)
\( \displaystyle = \frac{2}{3}\times27 = 18 \)
Shortcut: The line y = 9 meets the parabola at x = 3. The rectangle with sides 3 and 9 has area 27, and the part of it under the curve is \(\displaystyle\int_{0}^{3}x^{2}\,dx = 9\), so the region has area 27 − 9 = 18.
Why the other options are incorrect
Option (1) 9 — Incorrect. This is \(\displaystyle\int_{0}^{3}x^{2}\,dx\), the region between the curve and the x-axis, not the y-axis.
Option (2) 27 — Incorrect. This is the whole rectangle with sides 3 and 9; the part under the curve is not removed.
Option (4) 36 — Incorrect. This counts the region on both sides of the y-axis; only the first quadrant is asked.