Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area. Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area. 
I know how to graph it out but I have no idea what to do after that.
 A: You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
$$\int_0^b\sqrt{4-y}\,dy=\int_b^4\sqrt{4-y}\,dy$$
and you can proceeded here.
A: The area between $y=4-x^2$ and $y=0$ is $\int_{-2}^{2}(4-x^2)\,dx = \frac{32}{3}$. 
We have to find some constant $b\in(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $\frac{16}{3}$. That leads to the equation
$$ \int_{-\sqrt{4-b}}^{\sqrt{4-b}}(4-b-x^2)\,dx = \frac{16}{3} $$
and ultimately to $b=\color{red}{2(2-\sqrt[3]{2})}$.
A: The graph  of $f(x)=4-x^2$ intersects $y=0$ at $x=\pm2$. As you can probably tell, $f(x)$is a quadratic function. because of this, it is symmetric about the vertical line containing it's vertex. As you can tell from looking at the graph, the vertical line containing the vertex of $f(x)$ is the line $x=0$. This can be verified by setting $f'(x)=0$, and solving for $x$. Thus $b=0$. You can re-verify this by computing $\int_{-2}^{0}f(x)dx=\frac{16}{3}$ and $\int_{0}^{2}f(x)dx=\frac{16}{3}$.
