Let $f(x)=x^2-c^2$ and $g(x)=c^2-x^2$.

Find $c>0$ such that the area of the region enclosed by the parabolas $f(x)$ and $g(x)$ is 9 .
$$c=$$

Step 1 ：Set $f(x)=g(x)$ to find the points of intersection: $x^2-c^2=c^2-x^2$

Step 2 ：Simplify to get $2x^2=2c^2$

Step 3 ：Solve for x to get $x=\pm c$

Step 4 ：Calculate the area between the parabolas from $x=-c$ to $x=c$ using the integral of the upper function minus the lower function: $A={\int }_{-c}^c(g(x)-f(x))dx$

Step 5 ：Substitute $g(x)$ and $f(x)$ into the integral: $A={\int }_{-c}^c((c^2-x^2)-(x^2-c^2))dx$

Step 6 ：Simplify the integral: $A={\int }_{-c}^c(2c^2-2x^2)dx$

Step 7 ：Separate the integral into two parts: $A=2c^2{\int }_{-c}^{c}dx-2{\int }_{-c}^c{x}^{2}dx$

Step 8 ：Evaluate the integrals: $A=2c^2[x{]}_{-c}^c-2[\frac{x^3}{3}{]}_{-c}^c$

Step 9 ：Simplify the expression: $A=4c^3-2(\frac{c^3}{3}-\frac{(-c{)}^3}{3})$

Step 10 ：Further simplify to get $A=4c^3-\frac{4c^3}{3}$

Step 11 ：Combine like terms: $A=\frac{12c^3}{3}-\frac{4c^3}{3}$

Step 12 ：Final simplification: $A=\frac{8c^3}{3}$

Step 13 ：Set the area equal to 9: $\frac{8c^3}{3}=9$

Step 14 ：Solve for $c^3$ to get $c^3=\frac{27}{8}$

Step 15 ：Take the cube root of both sides: $c=\sqrt[3]{\frac{27}{8}}$

Step 16 ：Simplify the cube root: $c=\sqrt[3]{\frac{3^3}{2^3}}$

Step 17 ：Final answer: $c=\frac{3}{2}$

Step 18 ：$\boxed{{\displaystyle c=\frac{3}{2}}}$