Evaluate the integral. \[ \int_{0}^{a} 9 x^{2} \sqrt{a^{2}-x^{2}} d x \] \( \square \)

To evaluate the integral \[ I = \int_{0}^{a} 9 x^{2} \sqrt{a^{2}-x^{2}} \, dx, \] we can use the method of integration by parts or a trigonometric substitution. Here, we will use the trigonometric substitution \( x = a \sin(\theta) \), which gives us \( dx = a \cos(\theta) \, d\theta \). The limits change as follows: when \( x = 0 \), \( \theta = 0 \), and when \( x = a \), \( \theta = \frac{\pi}{2} \). Now substituting into the integral, we have: \[ I = \int_{0}^{\frac{\pi}{2}} 9 (a \sin(\theta))^2 \sqrt{a^2 - (a \sin(\theta))^2} \cdot a \cos(\theta) \, d\theta. \] Simplifying the square root, we get: \[ \sqrt{a^2 - a^2 \sin^2(\theta)} = \sqrt{a^2 (1 - \sin^2(\theta))} = a \cos(\theta). \] Thus, the integral becomes: \[ I = \int_{0}^{\frac{\pi}{2}} 9 a^2 \sin^2(\theta) \cdot a \cos(\theta) \cdot a \cos(\theta) \, d\theta = 9 a^4 \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \cos^2(\theta) \, d\theta. \] Now we can use the identity \( \sin^2(\theta) \cos^2(\theta) = \frac{1}{4} \sin^2(2\theta) \): \[ I = 9 a^4 \cdot \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta. \] To find \( \int \sin^2(2\theta) \, d\theta \), we can use integration by parts or the formula: \[ \int \sin^2(x) \, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C. \] So, \[ \int_{0}^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta = \left[ \frac{2\theta}{2} - \frac{\sin(2 \cdot 2\theta)}{4} \right]_{0}^{\frac{\pi}{2}} = \left[ \theta - \frac{\sin(4\theta)}{4} \right]_{0}^{\frac{\pi}{2}}. \] Evaluating this gives: \[ \left[ \frac{\pi}{2} - 0 \right] - \left[ 0 - 0 \right] = \frac{\pi}{2}. \] Substituting back, we have: \[ I = 9 a^4 \cdot \frac{1}{4} \cdot \frac{\pi}{2} = \frac{9 \pi a^4}{8}. \] Therefore, the value of the integral is: \[ \boxed{\frac{9 \pi a^4}{8}}. \]