Calculus

posted by Lindsay

Find the area of the region enclosed by
y−2x^2 ≥ 0 and |x|+|y| ≤ 1

1. Reiny

the graph of |x|+|y| ≤ 1 is a square with vertices
(1,0), (0,1), (-1,0) and (0,-1)
y ≥ 2x^2 is the region above y = 2x^2
so we need their intersection:
The line of the square in the first quadrant is
y = -x + 1
then 2x^2 = -x + 1
2x^2 + x - 1 = 0
(2x-1)(x+1) = 0
x = 1/2 or x = -1

from the diagram we want x=1/2, then y = 1/2)

From symmetry we can take the region form x = 0 to x = 1/2, then double that answer.

Area = ∫(-x + 1 - 2x^2) dx from 0 to 1/2
= -x^2/2 + x - 2x^3/3 | from 0 to 1/2
= -1/8 + 1/2 - 1/12 - 0
= 7/24

2. Reiny

oops, forgot to double our answer ...

since this was only the area in the first quadrant, the area in II would be the same, so the
total area = 14/24 or

7/12 square units

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