1. **State the problem:** Find the volume of the solid tetrahedron enclosed by the plane $2x + y + z = 4$ and the coordinate planes $x=0$, $y=0$, and $z=0$ using a triple integral. 2. **Set up the problem:** The volume is bounded by the coordinate planes and the plane $2x + y + z = 4$. Since $x,y,z \geq 0$, solve for $z$: $$z = 4 - 2x - y$$ 3. **Determine the limits of integration:** - For $z$: from $0$ to $4 - 2x - y$ - For $y$: from $0$ to the line where $z=0$, so $y = 4 - 2x$ - For $x$: from $0$ to where $y=0$, so $x = 2$ 4. **Write the triple integral for volume $V$:** $$V = \int_0^2 \int_0^{4-2x} \int_0^{4-2x-y} dz\, dy\, dx$$ 5. **Integrate with respect to $z$:** $$\int_0^{4-2x-y} dz = 4 - 2x - y$$ 6. **Substitute and integrate with respect to $y$:** $$\int_0^{4-2x} (4 - 2x - y) dy = \left[(4 - 2x)y - \frac{y^2}{2}\right]_0^{4-2x} = (4 - 2x)(4 - 2x) - \frac{(4 - 2x)^2}{2} = \frac{(4 - 2x)^2}{2}$$ 7. **Simplify the expression:** $$(4 - 2x)^2 = 16 - 16x + 4x^2$$ So the integral becomes: $$\frac{16 - 16x + 4x^2}{2} = 8 - 8x + 2x^2$$ 8. **Integrate with respect to $x$:** $$\int_0^2 (8 - 8x + 2x^2) dx = \left[8x - 4x^2 + \frac{2x^3}{3}\right]_0^2 = (16 - 16 + \frac{16}{3}) - 0 = \frac{16}{3}$$ 9. **Final answer:** The volume of the tetrahedron is $$\boxed{\frac{16}{3}}$$.