91Ó°ÊÓ

First, evaluate the inner integral with respect to \( z \). The integral is \( \int_{0}^{\sqrt{y}} 4 x^{2} z^{3} \, dz \). Using the power rule for integration, we have:\[\int 4 x^{2} z^{3} \, dz = 4 x^{2} \cdot \frac{z^4}{4} = x^{2} z^4\]Evaluate it from \( z = 0 \) to \( z = \sqrt{y} \).\[x^2 (\sqrt{y})^4 - x^2 (0)^4 = x^2 y^2\]

Now integrate the result from Step 1 with respect to \( y \):\[\int_{0}^{1-x} x^{2} y^{2} \, dy\]This can again be solved using the power rule:\[\int x^{2} y^{2} \, dy = x^{2} \cdot \frac{y^3}{3}\]Evaluate this from \( y = 0 \) to \( y = 1-x \):\[x^2 \cdot \frac{(1-x)^3}{3} - x^2 \cdot \frac{0^3}{3} = \frac{x^2 (1-x)^3}{3}\]

Lastly, we integrate the result from Step 2 with respect to \( x \):\[\int_{0}^{1} \frac{x^{2} (1-x)^{3}}{3} \, dx\]Expanding \((1-x)^3\), we have:\[(1-x)^3 = 1 - 3x + 3x^2 - x^3\]Thus, the integral becomes:\[\frac{1}{3}\int_{0}^{1} x^2 (1 - 3x + 3x^2 - x^3) \, dx\]Which expands to:\[\frac{1}{3}\int_{0}^{1} (x^2 - 3x^3 + 3x^4 - x^5) \, dx\]Solving each term separately:\[\frac{1}{3}\left[\frac{x^3}{3} - \frac{3x^4}{4} + \frac{3x^5}{5} - \frac{x^6}{6}\right]\text{ from } 0 \text{ to } 1\]Which evaluates to:\[\frac{1}{3}\left(\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6}\right)\]Finding a common denominator, which is 60, the expression inside the bracket becomes\[\frac{20}{60} - \frac{45}{60} + \frac{36}{60} - \frac{10}{60} = \frac{1}{60}\]Thus, divide by 3 to get the final answer:\[\frac{1}{3} \times \frac{1}{60} = \frac{1}{180}\]

The final value of the iterated integral is \( \frac{1}{180} \).