The area (in sq. units) of the region bounded by the parabola \(x^{2} = 4y\) and the line y = 9 is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 72
The line y = 9 meets the parabola where
\(x^{2} = 36\), so \(x = \pm 6\)
Between x = −6 and x = 6, the line y = 9 is above the parabola \(y = \frac{x^{2}}{4}\). So the area is the integral of (upper − lower):
\( \displaystyle A = \int_{-6}^{6}\left(9-\frac{x^{2}}{4}\right)dx \)
The integrand is even, so
\( \displaystyle A = 2\int_{0}^{6}\left(9-\frac{x^{2}}{4}\right)dx \)
\( \displaystyle = 2\left[9x-\frac{x^{3}}{12}\right]_{0}^{6} \)
\(= 2(54-18) = 72\)
Shortcut: A parabola cut by a chord perpendicular to its axis encloses \(\frac{2}{3}\) of the surrounding rectangle. Here the rectangle is 12 wide and 9 high:
\( \displaystyle A = \frac{2}{3}\times 12\times 9 = 72 \)