3. Evaluate the triple integral $\iiint_{\mathcal{R}} x+2 y z d V$ where $\mathcal{R}$ is bounded by the planes $x+y+z=5, y=x$, and $x=1$ in the first octant.
Answer
Final Answer: The value of the triple integral is \(\boxed{\frac{541}{15}}\).
Step 1 ：First, we need to set up the limits of integration for the triple integral. The region \(\mathcal{R}\) is bounded by the planes \(x+y+z=5\), \(y=x\), and \(x=1\) in the first octant. This means that \(x\) varies from 0 to 1, \(y\) varies from \(x\) to \(5-x\), and \(z\) varies from 0 to \(5-x-y\).
Step 2 ：Then we can set up the triple integral as \(\iiint_{\mathcal{R}} x+2 y z d V\), where \(f = x + 2*y*z\).
Step 3 ：Evaluating the integral, we find that the value is \(\frac{541}{15}\).
Step 4 ：Final Answer: The value of the triple integral is \(\boxed{\frac{541}{15}}\).
