(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of \(y = x^2 \sqrt{1-x^2}\). Differentiate \(y\) with respect to \(x\) using the product and chain rule:

\(y = x^2 (1-x^2)^{1/2}\)

\(\frac{dy}{dx} = 2x \sqrt{1-x^2} + x^2 \cdot \frac{-x}{\sqrt{1-x^2}}\)

Set \(\frac{dy}{dx} = 0\) and solve for \(x \neq 0\):

\(2x \sqrt{1-x^2} = \frac{x^3}{\sqrt{1-x^2}}\)

\(2x(1-x^2) = x^3\)

\(2x - 2x^3 = x^3\)

\(2x = 3x^3\)

\(x^2 = \frac{2}{3}\)

\(x = \sqrt{\frac{2}{3}}\)

(ii) Use the substitution \(x = \sin \theta\), then \(dx = \cos \theta \ d\theta\). The integral becomes:

\(A = \int_0^1 x^2 \sqrt{1-x^2} \ dx = \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \theta \ d\theta\)

\(= \int_0^{\frac{\pi}{2}} \frac{1}{4} \sin^2 2\theta \ d\theta\)

\(= \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin^2 2\theta \ d\theta\)

(iii) To find the exact value of \(A\), integrate:

\(\int \sin^2 2\theta \ d\theta = \int \frac{1 - \cos 4\theta}{2} \ d\theta\)

\(= \frac{1}{2} \left( \theta - \frac{1}{4} \sin 4\theta \right)\)

Evaluate from \(0\) to \(\frac{\pi}{2}\):

\(A = \frac{1}{4} \left[ \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) \right] = \frac{1}{16} \pi\)