Answer
$c=2\sqrt[3] 2$
Work Step by Step
Step 1. See figure. We need to find the $c$ value such that $area1=area2$.
Step 2. Integrate over $dy$ and consider the symmetry to use the function $x=\sqrt y$; we have
$A_1=2\int_{c}^4 (\sqrt y)dy =\frac{4}{3}x^{3/2}|_{c}^4=\frac{4}{3}(4^{3/2}-c^{3/2})=\frac{4}{3}(8-c^{3/2})$
Step 3. Similarly, $A_2=2\int_{0}^c (\sqrt y)dy =\frac{4}{3}x^{3/2}|_{0}^c=\frac{4}{3}c^{3/2}$
Step 4. Letting $A_1=A_2$, we have $8-c^{3/2}=c^{3/2}$, which gives $c=2\sqrt[3] 2$
Step 5. The areas are $A_1=A_2=\frac{16}{3}$ and the total area is $A=2A_1=\frac{32}{3}$
