Questions: Evaluate the triple integral [ iiintmathbfE x y z d V ] where E is the solid: (0 leq z leq 2), (0 leq y leq z), (0 leq x leq y).

Solution Steps

To evaluate the given triple integral, we need to set up the integral with the given bounds for $x$, $y$, and $z$. The bounds are:

• $0 \leq z \leq 2$
• $0 \leq y \leq z$
• $0 \leq x \leq y$

We will integrate in the order $x$, $y$, and then $z$.

We are given the triple integral: $$\iiint_{\mathbf{E}} x y z \, dV$$ where the region $E$ is defined by the bounds: $$0 \leq z \leq 2, \quad 0 \leq y \leq z, \quad 0 \leq x \leq y$$

First, we integrate the function $x y z$ with respect to $x$: $$\int_{0}^{y} x y z \, dx = y z \int_{0}^{y} x \, dx = y z \left[ \frac{x^2}{2} \right]_{0}^{y} = y z \left( \frac{y^2}{2} \right) = \frac{y^3 z}{2}$$

Next, we integrate the result with respect to $y$: $$\int_{0}^{z} \frac{y^3 z}{2} \, dy = \frac{z}{2} \int_{0}^{z} y^3 \, dy = \frac{z}{2} \left[ \frac{y^4}{4} \right]_{0}^{z} = \frac{z}{2} \left( \frac{z^4}{4} \right) = \frac{z^5}{8}$$

Finally, we integrate the result with respect to $z$: $$\int_{0}^{2} \frac{z^5}{8} \, dz = \frac{1}{8} \int_{0}^{2} z^5 \, dz = \frac{1}{8} \left[ \frac{z^6}{6} \right]_{0}^{2} = \frac{1}{8} \left( \frac{2^6}{6} \right) = \frac{1}{8} \left( \frac{64}{6} \right) = \frac{64}{48} = \frac{4}{3}$$

Final Answer