The region bounded by the parabola \(y = x^{2}\) and the line y = 8 is divided into two regions of equal area by the line y = c. Then c is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4\sqrt[3]{2}\)
For a height t > 0, the part of the region below the line y = t is found by integrating in y. At height y the region runs from \(x = -\sqrt{y}\) to \(x = \sqrt{y}\), so its width is \(2\sqrt{y}\):
\( \displaystyle A(t) = \int_{0}^{t}2\sqrt{y}\,dy = \frac{4}{3}\,t^{3/2} \)
The whole region is A(8), and the part below y = c must be half of it:
\( \displaystyle \frac{4}{3}\,c^{3/2} = \frac{1}{2}\times\frac{4}{3}\times8^{3/2} \)
\( \displaystyle c^{3/2} = \frac{8^{3/2}}{2} \)
Squaring both sides:
\( \displaystyle c^{3} = \frac{8^{3}}{4} = 128 \)
\(c = \sqrt[3]{128} = \sqrt[3]{64\times2} = 4\sqrt[3]{2}\)
The answer is not the midway height 4: the region is wider near the top, so the line must sit above y = 4 to leave half the area below it.