Find the volume of the given solid. Bounded by the planes \(z = x,y = x,z = 0,x + y = 2\)

Consider the planes \(z = x,y = x,z = 0,x + y = 2\)

Let\(\begin{aligned}{l}x + y = 2\,\,\,\,\,and\,\,\,\,y = x\\ \Rightarrow x + x = 2\\ \Rightarrow 2x = 2\\ \Rightarrow x = 1\end{aligned}\)

And\(\begin{aligned}{l}x + y = 2\\ \Rightarrow y = 2 - x\end{aligned}\)

The region of integration is shown below: \(D = \{ (x,y)|0 \le x \le 1,x \le y \le 2 - x\} \)

The volume can be given by: \(V = \int\limits_0^1 {\int\limits_x^{2 - x} {xdy} } dx\)

\(\begin{aligned}{l}V = \int\limits_0^1 {\int\limits_x^{2 - x} {xdydx} } \\V = \int\limits_0^1 {x\left[ y \right]_x^{2 - x}dx} \\V = \int\limits_0^1 {x\left[ {2 - x - x} \right]dx} \\V = \int\limits_0^1 {x\left[ {2 - 2x} \right]dx} \\V = \int\limits_0^1 {\left[ {2x - 2{x^2}} \right]dx} \\V = \left[ {2\frac{{{x^2}}}{2} - 2\frac{{{x^3}}}{3}} \right]_0^1\\V = \left[ {1 - \frac{2}{3} - 0} \right]\\V = \frac{1}{3}\end{aligned}\)

Hence, The volume of the given solid can be evaluated as:\(V = \frac{1}{3}\).