The area (in sq. units) of the region bounded by the curve \(y = 4-x^{2}\) and the lines y = 0 and y = 3 is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{28}{3}\)
Solving for x gives \(x = \pm\sqrt{4-y}\). The region lies between y = 0 and y = 3, and at each height its width is \(2\sqrt{4-y}\). So, integrating with respect to y,
\( \displaystyle A = \int_{0}^{3}2\sqrt{4-y}\,dy \)
\( \displaystyle = 2\left[-\frac{2}{3}(4-y)^{3/2}\right]_{0}^{3} \)
\( \displaystyle = -\frac{4}{3}(1-8) = \frac{28}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{14}{3}\) — Incorrect. This is \(\int_{0}^{3}\sqrt{4-y}\,dy\), only the half to the right of the y-axis.
Option (2) \(\displaystyle\frac{32}{3}\) — Incorrect. This is the whole area between the parabola and the x-axis, up to the vertex y = 4; the cap above y = 3 is not cut off.
Option (3) \(\displaystyle\frac{4}{3}\) — Incorrect. This is the cap between y = 3 and y = 4, the part that is outside the region.