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 ：\(\boxed{c = \frac{3}{2}}\)