Find the number b such that the line y = b divides the region bounded by the curves y=3x^2 and y = 2 into two regions with equal area.
Enter your answer either as a expression or as a decimal rounded to four places.
Hi,
for convenience I only consider the area in the first quadrant, that means x ≥ 0.
from y = 3x² you get x = √(y/3)
The area in question is: int from(0) to(2) √(y/3) dy = √(1/3)*((2/3)*y^(3/2)) from(0) to(2) = (4/3)*(√(2/3))
So the complete area is (8/3)*(√(2/3)) ≈ 2.1773... square-units
I consider now the half of the area in the first quadrant:
(2/3)*(√(2/3)) = int from(0) to(b) (√(y/3))dy = √(1/3)*((2/3)*y^(3/2)) from(0) to(b) = (2/3)*√(1/3)*√(b³).
Divide both sides by (2/3) and square both sides. Solve for b. I've got:
b = ³√(2) ≈ 1.2599..
EB