2 for the following regions determine the orders of integration (in c (1) f(x,y,z)=xz d is the set of…

Explanation:

Step1: Set up the triple - integral

The triple - integral is (\iiint_{D}f(x,y,z)dV=\int_{0}^{2}\int_{0}^{z}\int_{0}^{y}xz\ dx\ dy\ dz)

Step2: Integrate with respect to (x)

First, integrate (\int_{0}^{y}xz\ dx). Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n = 1) here), we have: [ \begin{align*} \int_{0}^{y}xz\ dx&=z\int_{0}^{y}x\ dx\ &=z\left[\frac{x^{2}}{2}\right]_{0}^{y}\ &=\frac{z y^{2}}{2} \end{align*} ]

Step3: Integrate the result with respect to (y)

Now, integrate (\int_{0}^{z}\frac{z y^{2}}{2}dy). Using the power rule (\int y^{n}dy=\frac{y^{n + 1}}{n + 1}+C) ((n=2) here): [ \begin{align*} \int_{0}^{z}\frac{z y^{2}}{2}dy&=\frac{z}{2}\int_{0}^{z}y^{2}dy\ &=\frac{z}{2}\left[\frac{y^{3}}{3}\right]_{0}^{z}\ &=\frac{z^{4}}{6} \end{align*} ]

Step4: Integrate the new result with respect to (z)

Finally, integrate (\int_{0}^{2}\frac{z^{4}}{6}dz). Using the power rule (\int z^{n}dz=\frac{z^{n+1}}{n + 1}+C) ((n = 4) here): [ \begin{align*} \int_{0}^{2}\frac{z^{4}}{6}dz&=\frac{1}{6}\int_{0}^{2}z^{4}dz\ &=\frac{1}{6}\left[\frac{z^{5}}{5}\right]_{0}^{2}\ &=\frac{1}{6}\times\frac{32}{5}\ &=\frac{16}{15} \end{align*} ]

Answer:

(\frac{16}{15})