We'll put y in function of x:
\(
y^2=4-x\\\\
y=\pm\sqrt{4-x}\)
Look the graph of the function in the attached figure. The areas above and below the x-axis are equal. So, we can represent the area bounded by the graph as:
\(A=2\displaystyle\int \sqrt{4-x}\,dx}\)
Using that to calculate the area bounded is equal to calculate the double of area above the x-axis.
The line x=z divides the region into two regions of equal area with 0 ≤ x ≤ 2, then:
\(
A_1=A_2\\\\
2\displaystyle\int^z_0\sqrt{4-x}\,dx}=2\displaystyle\int^2_z\sqrt{4-x}\,dx}\\\\
\displaystyle\int^z_0\sqrt{4-x}\,dx}=\displaystyle\int^2_z\sqrt{4-x}\,dx}\\\\
\left[-\dfrac{2}{3}(4-x)^{\frac{3}{2}}\right]^z_0=\left[-\dfrac{2}{3}(4-x)^{\frac{3}{2}}\right]^2_z\\\\
\left[(4-x)^{\frac{3}{2}}\right]^z_0=\left[(4-x)^{\frac{3}{2}}\right]^2_z\\\\\)
\((4-z)^{\frac{3}{2}}-(4-0)^{\frac{3}{2}}=(4-2)^{\frac{3}{2}}-(4-z)^{\frac{3}{2}}\\\\
(4-z)^{\frac{3}{2}}-4^{\frac{3}{2}}=2^{\frac{3}{2}}-(4-z)^{\frac{3}{2}}\\\\
2(4-z)^{\frac{3}{2}}=4^{\frac{3}{2}}+2^{\frac{3}{2}}\\\\
2(4-z)^{\frac{3}{2}}=8+2\sqrt2\\\\
(4-z)^{\frac{3}{2}}=4+\sqrt2\\\\
((4-z)^{\frac{3}{2}})^2=(4+\sqrt2)^2\\\\
(4-z)^3=16+8\sqrt2+2\\\\
(4-z)^3=18+8\sqrt2\\\\
4-z=\sqrt[3]{18+8\sqrt2}\\\\
\boxed{z=4-\sqrt[3]{18+8\sqrt2}}\)
