1. **State the problem:** We want to evaluate the triple integral $$\iiint_V \phi \, dV$$ where $$\phi = 4xz$$ and $$V$$ is the volume bounded by the planes $$2z + 4y + x = 2$$ and the coordinate planes $$x=0, y=0, z=0$$. 2. **Set up the region and limits:** The region $$V$$ is a tetrahedron in the first octant bounded by the coordinate planes and the plane $$2z + 4y + x = 2$$. 3. **Express the limits:** - From the plane equation, solve for $$x$$: $$x = 2 - 4y - 2z$$. - Since $$x, y, z \geq 0$$, the limits are: - $$x$$ from $$0$$ to $$2 - 4y - 2z$$ - $$y$$ from $$0$$ to the value where $$x=0$$, so $$2 - 4y - 2z \geq 0 \Rightarrow 4y + 2z \leq 2$$ - $$z$$ from $$0$$ to the value where $$y=0$$, so $$2 - 2z \geq 0 \Rightarrow z \leq 1$$ 4. **Find limits for $$y$$ and $$z$$:** - For fixed $$z$$, $$y$$ ranges from $$0$$ to $$\frac{2 - 2z}{4} = \frac{1 - z}{2}$$. - $$z$$ ranges from $$0$$ to $$1$$. 5. **Set up the integral:** $$ \iiint_V 4xz \, dV = \int_0^1 \int_0^{\frac{1 - z}{2}} \int_0^{2 - 4y - 2z} 4xz \, dx \, dy \, dz $$ 6. **Integrate with respect to $$x$$:** $$ \int_0^{2 - 4y - 2z} 4xz \, dx = 4z \int_0^{2 - 4y - 2z} x \, dx = 4z \left[ \frac{x^2}{2} \right]_0^{2 - 4y - 2z} = 2z (2 - 4y - 2z)^2 $$ 7. **Substitute and integrate with respect to $$y$$:** $$ \int_0^{\frac{1 - z}{2}} 2z (2 - 4y - 2z)^2 \, dy $$ Let $$u = 2 - 4y - 2z$$, then $$du = -4 dy$$, so $$dy = -\frac{du}{4}$$. When $$y=0$$, $$u = 2 - 0 - 2z = 2 - 2z$$. When $$y = \frac{1 - z}{2}$$, $$u = 2 - 4 \cdot \frac{1 - z}{2} - 2z = 2 - 2(1 - z) - 2z = 2 - 2 + 2z - 2z = 0$$. So the integral becomes: $$ 2z \int_{u=2 - 2z}^0 u^2 \left(-\frac{1}{4}\right) du = -\frac{z}{2} \int_{2 - 2z}^0 u^2 du = \frac{z}{2} \int_0^{2 - 2z} u^2 du $$ Calculate the integral: $$ \frac{z}{2} \left[ \frac{u^3}{3} \right]_0^{2 - 2z} = \frac{z}{2} \cdot \frac{(2 - 2z)^3}{3} = \frac{z (2 - 2z)^3}{6} $$ 8. **Integrate with respect to $$z$$:** $$ \int_0^1 \frac{z (2 - 2z)^3}{6} dz = \frac{1}{6} \int_0^1 z (2 - 2z)^3 dz $$ Let $$w = 2 - 2z$$, then $$dw = -2 dz$$, so $$dz = -\frac{dw}{2}$$. When $$z=0$$, $$w=2$$. When $$z=1$$, $$w=0$$. Also, $$z = \frac{2 - w}{2}$$. Rewrite the integral: $$ \frac{1}{6} \int_{w=2}^0 \frac{2 - w}{2} w^3 \left(-\frac{1}{2}\right) dw = \frac{1}{6} \cdot \frac{1}{4} \int_0^2 (2 - w) w^3 dw = \frac{1}{24} \int_0^2 (2w^3 - w^4) dw $$ 9. **Calculate the integral:** $$ \int_0^2 (2w^3 - w^4) dw = \left[ \frac{2 w^4}{4} - \frac{w^5}{5} \right]_0^2 = \left[ \frac{w^4}{2} - \frac{w^5}{5} \right]_0^2 $$ Evaluate at $$w=2$$: $$ \frac{2^4}{2} - \frac{2^5}{5} = \frac{16}{2} - \frac{32}{5} = 8 - 6.4 = 1.6 $$ 10. **Final value:** $$ \frac{1}{24} \times 1.6 = \frac{1.6}{24} = 0.066666... \approx 0.07 $$ **Answer:** $$\boxed{0.07}$$ (rounded to 2 decimal places)