Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in the plane z = f(x, y) = 3 - x - y.

Steps to Solve

1. Identify the region of integration
   The base of the prism is defined by the triangle bounded by the x-axis, the line $y = x$, and the line $x = 1$. The vertices of the triangle are (0,0), (1,0), and (1,1).

2. Set up the double integral for volume
   The volume $V$ of the prism can be found using the double integral over the region $R$ and the function $z = f(x,y) = 3 - x - y$.

   The integral is given by: $$ V = \iint_R (3 - x - y) , dA $$

3. Define the limits of integration
   The limits of integration for $y$ will be from the x-axis ($y = 0$) to the line $y = x$. The limits for $x$ will be from 0 to 1. Thus, the integral becomes: $$ V = \int_0^1 \int_0^x (3 - x - y) , dy , dx $$

4. Evaluate the inner integral
   Evaluate the inner integral with respect to $y$: $$ \int_0^x (3 - x - y) , dy $$ The result of this integral is: $$ (3 - x)y - \frac{y^2}{2} \bigg|_0^x = (3 - x)x - \frac{x^2}{2} $$

5. Simplify the expression
   This results in: $$ V = \int_0^1 \left((3 - x)x - \frac{x^2}{2}\right) , dx $$

6. Evaluate the outer integral
   Now evaluate the outer integral: $$ V = \int_0^1 \left(3x - x^2 - \frac{x^2}{2}\right) , dx $$
   Simplifying this yields: $$ V = \int_0^1 (3x - \frac{3x^2}{2}) , dx $$

7. Calculate the integral
   Calculate: $$ \int_0^1 \left(3x - \frac{3x^2}{2}\right) , dx = \frac{3x^2}{2} - \frac{3x^3}{6} \bigg|_0^1 = \frac{3}{2} - \frac{1}{2} = 1 $$