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...

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.
