The curve to the right is the graph of the equation $y=-x \sqrt{9-x^{2}}$. Find the total area of the shaded regions in the graph.

The total area of the shaded regions is $\square$.
(Simplify your answer.)

Step 1 ：The function is defined for \(-3 \leq x \leq 3\).

Step 2 ：The integral of the absolute value of the function over this interval is \(\int_{-3}^{3} |-x \sqrt{9-x^{2}}| dx\).

Step 3 ：Since the function is symmetric about the y-axis, we can find the area of one of the shaded regions and then double it to get the total area. So, we have \(2 \int_{0}^{3} |x \sqrt{9-x^{2}}| dx\).

Step 4 ：Since \(x\) is non-negative in the interval \([0, 3]\), the absolute value can be removed: \(2 \int_{0}^{3} x \sqrt{9-x^{2}} dx\).

Step 5 ：To solve this integral, we can use the substitution method. Let \(u = 9 - x^{2}\), then \(du = -2x dx\).

Step 6 ：So, the integral becomes \(- \int \sqrt{u} du\).

Step 7 ：The integral of \(\sqrt{u}\) is \(\frac{2}{3} u^{3/2}\), so we have \(- \frac{2}{3} u^{3/2}\).

Step 8 ：Substituting \(u\) back in, we get \(- \frac{2}{3} (9 - x^{2})^{3/2}\).

Step 9 ：Evaluating this from \(0\) to \(3\), we get \(- \frac{2}{3} (9 - 3^{2})^{3/2} - (- \frac{2}{3} (9 - 0^{2})^{3/2})\).

Step 10 ：This simplifies to \(- \frac{2}{3} * 0 + \frac{2}{3} * 9^{3/2}\).

Step 11 ：Further simplifying, we get \(\frac{2}{3} * 27\).

Step 12 ：So, the total area of the shaded regions is \(\boxed{18}\) square units.