1. **State the problem:** Find the volume of the tetrahedron bounded by the coordinate planes $x=0$, $y=0$, $z=0$ and the plane $3x + 6y + 4z - 12 = 0$ using double integration. 2. **Rewrite the plane equation:** Solve for $z$ in terms of $x$ and $y$: $$3x + 6y + 4z = 12 \implies 4z = 12 - 3x - 6y \implies z = \frac{12 - 3x - 6y}{4} = 3 - \frac{3}{4}x - \frac{3}{2}y$$ 3. **Determine the projection region in the $xy$-plane:** Since $z \geq 0$, we have $$3 - \frac{3}{4}x - \frac{3}{2}y \geq 0 \implies \frac{3}{4}x + \frac{3}{2}y \leq 3$$ Divide both sides by 3: $$\frac{x}{4} + \frac{y}{2} \leq 1$$ 4. **Set up the double integral for volume:** Volume $V = \iint_R z \, dA = \iint_R \left(3 - \frac{3}{4}x - \frac{3}{2}y\right) \, dx \, dy$ 5. **Describe the region $R$:** $$0 \leq x \leq 4(1 - \frac{y}{2}) = 4 - 2y$$ $$0 \leq y \leq 2$$ 6. **Write the integral explicitly:** $$V = \int_0^2 \int_0^{4 - 2y} \left(3 - \frac{3}{4}x - \frac{3}{2}y\right) dx \, dy$$ 7. **Integrate with respect to $x$:** $$\int_0^{4 - 2y} \left(3 - \frac{3}{4}x - \frac{3}{2}y\right) dx = \left[3x - \frac{3}{8}x^2 - \frac{3}{2}yx\right]_0^{4 - 2y}$$ Substitute $x = 4 - 2y$: $$3(4 - 2y) - \frac{3}{8}(4 - 2y)^2 - \frac{3}{2}y(4 - 2y)$$ 8. **Simplify the expression:** Calculate each term: - $3(4 - 2y) = 12 - 6y$ - $(4 - 2y)^2 = 16 - 16y + 4y^2$ - $\frac{3}{8}(16 - 16y + 4y^2) = 6 - 6y + \frac{3}{2}y^2$ - $\frac{3}{2}y(4 - 2y) = 6y - 3y^2$ Putting it all together: $$12 - 6y - (6 - 6y + \frac{3}{2}y^2) - (6y - 3y^2) = 12 - 6y - 6 + 6y - \frac{3}{2}y^2 - 6y + 3y^2$$ Simplify: $$= (12 - 6) + (-6y + 6y - 6y) + (-\frac{3}{2}y^2 + 3y^2) = 6 - 6y + \frac{3}{2}y^2$$ 9. **Integrate with respect to $y$:** $$V = \int_0^2 \left(6 - 6y + \frac{3}{2}y^2\right) dy = \left[6y - 3y^2 + \frac{1}{2}y^3\right]_0^2$$ 10. **Evaluate the definite integral:** $$6(2) - 3(2)^2 + \frac{1}{2}(2)^3 = 12 - 12 + 4 = 4$$ **Final answer:** The volume of the tetrahedron is $4$ (rounded to a whole number).