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