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...
 A: We can find the area by integrating with respect to $x$ or with respect to $y$. 
With respect to $y$ is easier. I would prefer to use symmetry and say that the area is equal to
$$2\int_0^4 y^{1/2}\,dy.$$
Integrate. We get $(2)(2/3)4^{3/2}=32/3$, your answer.
So we want the area from $y=0$ to $b$ to be half of $\frac{32}{3}$, that is, $\frac{16}{3}$.
The area from $y=0$ to $y=b$ is equal to
$$2\int_0^b y^{1/2}\,dy.$$
This is equal to $(2)(2/3)b^{3/2}$.
Thus  we need to find the  $b$ such that
$$(2)(2/3)b^{3/2}=\frac{16}{3}.$$
Simplify. We get the equation $b^{3/2}=4$.
Solve for $b$. We get $b=4^{2/3}$. This can be written in various other ways, such as  $b=2\sqrt[3]{2}$. 
Remark: We could have bypassed the initial computation of the area, and written that we want to find $b$ such that 
$$2\int_0^b y^{1/2}\,dy=2\int_b^4 y^{1/2}\,dy.$$
But I think it is better to find first the area from $0$ to $4$, to get some concrete feeling about the situation.
For a "reality check" note that our answer must be $\gt 2$: we have to go more than half of the way to $4$ to capture half the area.
A: 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}
$.
A: The intersection of $y=4$ and $y=x^2$ is at $(-2, 4)$ and $(2, 4)$ so the whole area between the two curves is (by the symmetry of $y=x^2$)
$$
\int_{-2}^2 4-x^2dx=2\int_0^24-x^2dx=\frac{32}{3}
$$
as you noted. Now if the upper boundary is $y=b$ the intersection will be at $(-\sqrt{b}, b)$ and $(\sqrt{b}, b)$ so you'll have the area
$$\begin{align}
\int_{-\sqrt{b}}^{\sqrt{b}} b-x^2dx&=2\int_0^{\sqrt{b}}b-x^2dx\\
  &= 2(bx-\frac{1}{3}x^3)\left|_0^\sqrt{b}\right.\\
  &=2(b^{3/2}-\frac{1}{3}b^{3/2})=\frac{4}{3}b^{3/2}
\end{align}$$
and you want to find $b$ so that
$$
\frac{4}{3}b^{3/2}=\frac{16}{3}
$$
which I'll bet you can do.
