Let's break down the problem step by step: ### **Step 1: Understanding the Integral** The given iterated integral is: \[ \int_0^1 \int_0^{1-x} \int_0^{8-8z} dy \, dz \, dx \] - The **outer integral** is over \( x \) from 0 to 1. - The **middle integral** is over \( z \) from 0 to \( 1-x \). - The **inner integral** is over \( y \) from 0 to \( 8-8z \). ### **Step 2: Region Description** #### **Bounds:** - \( 0 \leq x \leq 1 \) - \( 0 \leq z \leq 1-x \) - \( 0 \leq y \leq 8 - 8z \) #### **Geometric Interpretation:** - For a fixed \( x \), \( z \) ranges from 0 to \( 1-x \), so the region in the \( xz \)-plane is a right triangle with corners at \((0,0), (1,0), (0,1)\). - For a fixed \( z \), \( y \) ranges from 0 to \( 8-8z \), so as \( z \) increases, the upper bound for \( y \) decreases linearly. ### **Step 3: Faces and Base** - The **base in the xy-plane**: When \( z=0 \), \( x \) runs from 0 to 1, and for each \( x \), \( y \) runs from 0 to \( 8 \), forming a rectangle. But the actual base projection in the \( xy \)-plane is not a **trapezoid**. - The **face in the xz-plane**: Set \( y=0 \). The bounds are \( 0 \leq x \leq 1 \), \( 0 \leq z \leq 1-x \). That’s a triangle. - The **face in the yz-plane**: Set \( x=0 \). \( 0 \leq z \leq 1 \), \( 0 \leq y \leq 8-8z \). That’s also a triangle. - The **face in \( z=1-x \)**: Substitute in the bounds for \( z \), gives a triangle. - The **face in \( y = 8-8z \)**: Rearranged as \( z = 1 - \frac{y}{8} \); again, a triangle. ### **Step 4: Effect of Increasing Variables** - As \( x \) increases, the upper bound for \( z \) (**top of the region**) decreases, since \( z \leq 1-x \). - As \( y \) increases, for a fixed \( z \), the allowable \( z \) decreases, since \( y \leq 8-8z \Rightarrow z \leq 1-\frac{y}{8} \). --- ## **Final Answers (Based on the Options Given):** **Correct answers highlighted with ✓:** - The solid has **a trapezoidal base** in the xy-plane. (**✗**) - The solid has **a triangular face** in the xz-plane. (**✓**) - The solid has **a triangular face** in the yz-plane. (**✓**) - The solid has **a triangular face** in the plane \( z = 1 - x \). (**✓**) - The solid has **a triangular face** in the plane \( y = 8 - 8z \). (**✓**) - As \( x \) increases, the top of the region **decreases**. (**✓**) - As \( y \) increases, the top of the region **decreases**. (**✓**) --- ## **Summary Table (Copy of your selections with correct answers):** | Description | Answer | Correct? | |--------------------------------------------------------|------------------------ |:--------:| | The solid has a trapezoidal base in the xy-plane. | ✗ | | | The solid has a triangular face in the xz-plane. | ✓ | ✓ | | The solid has a triangular face in the yz-plane. | ✓ | ✓ | | The solid has a triangular face in \( z = 1-x \). | ✓ | ✓ | | The solid has a triangular face in \( y = 8-8z \). | ✓ | ✓ | | As \( x \) increases, the top of the region decreases. | ✓ | ✓ | | As \( y \) increases, the top of the region decreases. | ✓ | ✓ | --- ### **Explanation** The solid is a triangular prism-like region bounded by planes in the \( x, y, z \) axes, specifically defined by the given bounds in the iterated integral. The faces are triangular, and as \( x \) or \( y \) increases, the "top" of the region (in terms of \( z \)) decreases.

The statement **"The solid has a trapezoidal base in the xy-plane."** is incorrect based on the integral bounds. **Corrected sentence:** *The solid has a triangular base in the xy-plane.*

Let's analyze the given problem carefully and provide a detailed, step-by-step solution, including the interpretation of the integral, the region it describes, and the corresponding sketch description. --- ### Given Integral: \[ \iiint_{V} dy\, dz\, dx = \int_{x=0}^{1} \int_{z=0}^{1 - x} \int_{y=0}^{8 - 8z} dy\, dz\, dx \] --- ### Step 1: Understand the order of integration - **Innermost integral:** \( y \) from 0 to \( 8 - 8z \) - **Middle integral:** \( z \) from 0 to \( 1 - x \) - **Outer integral:** \( x \) from 0 to 1 --- ### Step 2: Interpret the bounds to find the region \(V\) **1. Bounds on \( y \):** \[ 0 \leq y \leq 8 - 8z \] This describes a region in the \( y - z \) plane, for each fixed \( z \). The upper bound for \( y \) depends linearly on \( z \). When: - \( z=0 \), \( y \leq 8 \), - \( z=1 \), \( y \leq 0 \). **2. Bounds on \( z \):** \[ 0 \leq z \leq 1 - x \] For each fixed \( x \), \( z \) ranges from 0 up to \( 1 - x \). **3. Bounds on \( x \):** \[ 0 \leq x \leq 1 \] --- ### Step 3: Visualize the region - For fixed \( x \), the \( z \)-coordinate runs from 0 to \( 1 - x \), forming a line decreasing from 1 to 0 as \( x \) increases. - For fixed \( z \), \( y \) runs from 0 to \( 8 - 8z \). When \( z=0 \), the maximum \( y \) is 8; when \( z=1 \), \( y=0 \). --- ### Step 4: Describe the cross-sections - **In the \( yz \)-plane** (for fixed \( x \)): \[ 0 \leq y \leq 8 - 8z, \quad 0 \leq z \leq 1 - x \] which describes a right triangle with vertices at: \[ ( y, z ) = (0, 0), \quad (8, 0), \quad (0, 1 - x) \] - **In the \( xz \)-plane** (for fixed \( y \)): The bounds are \( 0 \leq z \leq 1 - x \), with \( x \) from 0 to 1, forming a triangular region in the \( xz \)-plane bounded by \( x=0 \), \( z=0 \), and \( z=1-x \). --- ### Step 5: Interpret the shape of the solid - The base in the \( xy \)-plane: At \( z=0 \), \( y \) ranges from 0 to 8. For each \( x \), the \( z \)-range depends on \( x \), so the projection in the \( xy \)-plane is a rectangle extending from \( (0,0) \) to \( (1,8) \). - The top surface in \( z \): Given \( z \leq 1 - x \), as \( x \) increases, the maximum value of \( z \) decreases, meaning the "top" surface slopes downward along the \( x \)-axis. - The \( y \)-bound depends on \( z \), decreasing linearly as \( z \) increases. --- ### **Summary of the Region:** - In the \( xy \)-plane, the projection is a rectangle \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 8 \). - The upper surface in the \( z \)-direction is sloped from \( z=1 \) at \( x=0 \) to \( z=0 \) at \( x=1 \). - For each fixed \( z \), the \( y \)-coordinate runs from 0 up to \( 8 - 8z \), which is a line decreasing from 8 to 0 as \( z \) increases. --- ### **Sketch description:** - The base in the \( xy \)-plane is a rectangle with vertices at \((0,0), (1,0), (1,8), (0,8)\). - The top surface is a triangular plane sloping downward from \((x,z) = (0,1)\) to \((1,0)\). - The surface in \( y \) extends from 0 up to the line \( y=8 -8z \), which decreases as \( z \) increases. --- ### **Final conclusion about the shape:** The solid is bounded: - **In \( x \):** from 0 to 1. - **In \( z \):** from 0 to \( 1 - x \), decreasing linearly as \( x \) increases. - **In \( y \):** from 0 to \( 8 - 8z \), decreasing linearly as \( z \) increases. --- ### **Answer to the description options:** - The base in the \( xy \)-plane is a **rectangle** (not trapezoid). - The face in the \( xz \)-plane is a **triangle** with vertices at \((0,0)\), \((1,0)\), \((0,1)\). - The face in the \( yz \)-plane (at \( x=0 \)) is also a **triangle** with vertices at \((0,0)\), \((0,8)\), \((0,0)\). - The face in the plane \( z = 1 - x \) is a **triangle** connecting the points \((0,1)\), \((1,0)\), and \((0,0)\). - The face in \( y = 8 - 8z \) is a **triangle** in the \( y-z \) plane. - As \( x \) increases, the maximum \( z \) decreases, so the top "slopes downward" in the \( x \)-direction. - As \( y \) increases, the maximum \( z \) decreases, indicating the top surface slopes downward in \( y \)-direction as well. --- ### **Summary:** - The **solid** is bounded above by the plane \( z = 1 - x \), which slopes downward from \( z=1 \) at \( x=0 \) to \( z=0 \) at \( x=1 \). - Its **base** in the \( xy \)-plane is a rectangle from \((0,0)\) to \((1,8)\). - The **side faces** are triangles in planes \( x=0 \), \( y=0 \), and the sloped planes \( z=1-x \) and \( y=8 - 8z \). --- **Note:** The initial statement about a trapezoidal base is **incorrect**; the base is rectangular. --- **Final answer:** **The solid has a rectangular base in the xy-plane, with triangular faces in the xz- and yz-planes, bounded above by the sloped plane \( z=1 - x \), and the top sloping downward as \( x \) increases.**