Find the number b such that the line y = b divides the region bounded by the curves y = 4x2 and y = 16 into two regions with equal area. (Round your answer to two decimal places.)

Answer 1

Because of summetry, we only need to look at the first quadrant
Area of whole thing in quadrant I
= ∫(16 - 4x^2) dx from 0 to 2
= [16x - (4/3)x^3] from 0 to 2
= 32 - (4/3)(8) - 0
= 64/3

So the area created by y = b must be 32/3
the intersection of y = b and y = 4x^2:
4x^2 = b
x = √b/2

so we would have:
area = ∫( b - 4x^2) dx from 0 to √b/2
= [bx - (4/3)x^3] from 0 to √b
= b√b/2 - (4/3)b√b/8 - 0
= (1/3)b√b

(1/3)b√b = 32/3
b√b = 32
square both sides
b^3 = 1024
b = 10.079

b = appr 10.08

Proof
http://www.wolframalpha.com/input/?i=area+between+y+%3D+16+and+y+%3D+4x%5E2

http://www.wolframalpha.com/input/?i=area+between+y+%3D+10.079+and+y+%3D+4x%5E2

notice in my graph I did the whole thing, but
21.3322 = appr (1/2)(128/3)