<s>[INST] Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area. Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$  and $y = 4$ into two regions with equal area.
I could find the area of this region but i have no clue how to split it parallel to the $x$ axis perfectly... this is too abstract for me...
I found the whole area and got $\dfrac{32}{3}$... I tried plugging that back into the definite integral but that definitely wasn't right...
 [/INST] If $y=b$,
then $y=x^2$ intersects that
at $x=\sqrt{b}$.
That whole area is a rectangle
with vertices
$
( \sqrt{b}, 0),
( \sqrt{b}, b)
( -\sqrt{b}, b),
( -\sqrt{b}, b)
$.
Its area is
$2b\sqrt{b}
=2b^{3/2}
$.
The area below the parabola is
$2\int_0^{\sqrt{b}} x^2 dx
=2 \frac{x^3}{3}\big|_0^{\sqrt{b}}
=\frac23 b^{3/2}
$.
Therefore,
the area between the parabola
and the line $y=b$
is
$2b^{3/2}-\frac23 b^{3/2}
=\frac43 b^{3/2}
$.
If we put $b=4$,
the area between the parabola
and the line $y=4$
is
$\frac43 4^{3/2}
=\frac43 8
=\frac{32}{3}
$.
We want the area between
the parabola and the line
$y=b$ to be half of this,
which is
$\frac{16}{3}
$.
Therefore,
$\frac43 b^{3/2}
=\frac{16}{3}
$
or
$b^{3/2}
=4
$
or
$b = 4^{2/3}
=\sqrt[3]{16}
$.
</s>