Let's break down the problem step by step: ### 1. **Find the intercepts** The tetrahedron is bounded by the coordinate planes (\(x=0\), \(y=0\), \(z=0\)) and the plane \(8x + y + z = 5\). - **x-intercept:** Set \(y=0, z=0\), solve \(8x=5 \implies x = \frac{5}{8}\) - **y-intercept:** Set \(x=0, z=0\), solve \(y=5\) - **z-intercept:** Set \(x=0, y=0\), solve \(z=5\) ### 2. **Set up the bounds for the triple integral** - \(x\) ranges from \(0\) to \(\frac{5}{8}\). - For a fixed \(x\), \(y\) ranges from \(0\) to \(5-8x\). - For fixed \(x\) and \(y\), \(z\) ranges from \(0\) to \(5-8x-y\). ### 3. **Fill in the boxes:** From the given integral: \[ V = \int_{0}^{A} \int_{B}^{C} \int_{0}^{D} dz\,dy\,dx \] #### - \(A = \frac{5}{8}\) #### - \(B = 0\) #### - \(C = 5-8x\) #### - \(D = 5-8x-y\) ### 4. **Write the triple integral:** \[ V = \int_{0}^{\frac{5}{8}} \int_{0}^{5-8x} \int_{0}^{5-8x-y} dz\,dy\,dx \] ### 5. **Evaluate the volume** First, integrate with respect to \(z\): \[ \int_{0}^{5-8x-y} dz = 5-8x-y \] So now, \[ V = \int_{0}^{\frac{5}{8}} \int_{0}^{5-8x} (5-8x-y) \, dy\, dx \] Now integrate with respect to \(y\): \[ \int_{0}^{5-8x} (5-8x-y) dy = \left[(5-8x)y - \frac{1}{2}y^2\right]_{0}^{5-8x} \] Plug in \(y = 5-8x\): \[ (5-8x)(5-8x) - \frac{1}{2}(5-8x)^2 = (5-8x)^2 - \frac{1}{2}(5-8x)^2 = \frac{1}{2}(5-8x)^2 \] So, \[ V = \int_{0}^{\frac{5}{8}} \frac{1}{2}(5-8x)^2 dx \] Expand and integrate: \[ = \frac{1}{2} \int_{0}^{\frac{5}{8}} (25 - 80x + 64x^2) dx \] \[ = \frac{1}{2} \left[25x - 40x^2 + \frac{64}{3}x^3\right]_{0}^{\frac{5}{8}} \] Plug in \(x = \frac{5}{8}\): - \(25x = 25 \times \frac{5}{8} = \frac{125}{8}\) - \(40x^2 = 40 \times \left(\frac{25}{64}\right) = \frac{1000}{64} = \frac{125}{8}\) - \(\frac{64}{3}x^3 = \frac{64}{3} \times \left(\frac{125}{512}\right) = \frac{64 \times 125}{3 \times 512} = \frac{8000}{1536} = \frac{125}{24}\) So, \[ V = \frac{1}{2} \left( \frac{125}{8} - \frac{125}{8} + \frac{125}{24} \right ) = \frac{1}{2} \left( \frac{125}{24} \right ) = \frac{125}{48} \] --- ## **Final Answers:** \[ \boxed{ \begin{aligned} A &= \frac{5}{8} \\ B &= 0 \\ C &= 5 - 8x \\ D &= 5 - 8x - y \\ \text{Volume} &= \frac{125}{48} \end{aligned} } \]