8. Find the volume of the solid bounded by the planes \( z=x, y=x, x+y=2 \), and \( z=0 \).

**Step 1. Describe the region:** The solid is bounded by the planes \[ z=x,\quad y=x,\quad x+y=2,\quad \text{and} \quad z=0. \] The top of the solid is given by \( z=x \) and the bottom by \( z=0 \). The vertical walls are given by \( y=x \) and \( x+y=2 \). **Step 2. Determine the projection onto the \(xy\)–plane:** Since the planes \( y=x \) and \( x+y=2 \) are vertical (independent of \(z\)), they form the boundary of the region \( R \) in the \(xy\)–plane. For a point \((x,y)\) in \(R\) the value of \(z\) is allowed to vary between \(z=0\) and \(z=x\). The boundaries in the \(xy\)–plane are: - The line \( y=x \). - The line \( x+y=2 \), which can be written as \( y=2-x \). Note that the two lines intersect when \[ x=x,\quad \text{and} \quad x+ x=2 \quad \Longrightarrow \quad 2x=2,\quad x=1,\quad y=1. \] At \( x=0 \): - On \( y=x \) we have \( y=0 \). - On \( x+y=2 \) we have \( y=2 \). Thus, the projected region \( R \) in the \(xy\)–plane is the triangle with vertices \((0,0)\), \((0,2)\), and \((1,1)\). For a fixed \(x\), the variable \(y\) runs from the line \( y=x \) (the lower boundary) up to the line \( y=2-x \) (the upper boundary). Also, \(x\) runs from \(0\) to \(1\) (since at \(x=1\) the two lines meet). **Step 3. Set up the integral for the volume:** The volume \(V\) is given by integrating the height \(z\) from \(z=0\) to \(z=x\) over the region \(R\): \[ V = \int_{x=0}^{1} \int_{y=x}^{2-x} \int_{z=0}^{x} dz\,dy\,dx. \] **Step 4. Integrate with respect to \(z\):** The innermost integral is: \[ \int_{z=0}^{x} dz = x. \] So the volume becomes: \[ V = \int_{x=0}^{1} \int_{y=x}^{2-x} x\,dy\,dx. \] **Step 5. Integrate with respect to \(y\):** For fixed \(x\), \[ \int_{y=x}^{2-x} x\,dy = x\Big[(2-x)-x\Big] = x\,(2-2x). \] Thus, \[ V = \int_{x=0}^{1} x(2-2x)\,dx. \] **Step 6. Integrate with respect to \(x\):** Simplify the integrand: \[ x(2-2x)=2x-2x^2. \] Then, \[ V = \int_{0}^{1} (2x-2x^2)\,dx = 2\int_{0}^{1} (x-x^2)\,dx. \] Now compute the integral: \[ \int_{0}^{1} (x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}. \] Thus, \[ V = 2\cdot\frac{1}{6} = \frac{1}{3}. \] **Final Answer:** The volume of the solid is \[ \boxed{\frac{1}{3}}. \]