91Ó°ÊÓ

First, we will integrate the function \(f(x, y, z) = 4yz\) with respect to \(z\) over the interval \([0, 2-x]\). To do this, we'll calculate the antiderivative of \(4yz\) with respect to \(z\), and then evaluate it at the limits of integration: $$ \int_{0}^{2-x} 4yz \, dz = 2yz^2 \Big|_0^{2-x} $$ Now, let's plug in the bounds: $$ 2y(2-x)^2 - 2y(0)^2 = 2y(4 - 4x + x^2) $$

Now, we have a double integral with respect to \(y\) and \(x\). Let's find the antiderivative of our current function \(2y(4 - 4x + x^2)\) with respect to \(y\) as follows: $$ \int_{0}^{\sqrt{1-x^2}} 2y(4 - 4x + x^2) \, dy = (4 - 4x + x^2)y^2 \Big|_0^{\sqrt{1-x^2}} $$ Next, plug in the bounds: $$ (4 - 4x + x^2)(1-x^2) - (4 - 4x + x^2)(0)^2 = (4 - 4x + x^2)(1-x^2) $$

Finally, we will integrate our remaining function \((4 - 4x + x^2)(1-x^2)\) with respect to \(x\) over the interval \([0, 1]\): $$ \int_{0}^{1} (4 - 4x + x^2)(1-x^2) \, dx $$ To make it easier, first expand the expression: $$ \int_{0}^{1} (4-4x^3-4x+x^2+x^4) \, dx $$ Now integrate term by term: $$ \left[4x - x^4 + \frac{-4x^2}{2} + \frac{x^3}{3} + \frac{x^5 }{5}\right]_0^1 = 4 - 1 + (-2) +( \frac{1}{3}) +(\frac{1}{5}) $$ Finally, simplify the expression and find the value of the integral: $$ 4 - 1 - 2 + \frac{1}{3} + \frac{1}{5} = 1 + \frac{1}{3} + \frac{1}{5} = \frac{15}{15} + \frac{5}{15} + \frac{3}{15} = \frac{23}{15} $$ Thus, the value of the triple integral is: $$ \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{2-x} 4 y z \, dz \, dy \, dx = \frac{23}{15} $$