3-8 evaluate the iterated integral.\n3. $int_{0}^{2}int_{0}^{z^{2}}int_{0}^{y - z}(2x - y)dx dy dz$

Explanation:

Step1: Integrate with respect to $x$

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

Step2: Integrate the result with respect to $y$

Now, integrate $\int_{0}^{z^{2}}(z^{2}-yz)dy$. [ \begin{align*} \int_{0}^{z^{2}}(z^{2}-yz)dy&=\left[z^{2}y-\frac{1}{2}y^{2}z\right]_{0}^{z^{2}}\ &=z^{2}\cdot z^{2}-\frac{1}{2}(z^{2})^{2}z\ &=z^{4}-\frac{1}{2}z^{5} \end{align*} ]

Step3: Integrate the result with respect to $z$

Finally, integrate $\int_{0}^{2}(z^{4}-\frac{1}{2}z^{5})dz$. [ \begin{align*} \int_{0}^{2}(z^{4}-\frac{1}{2}z^{5})dz&=\left[\frac{1}{5}z^{5}-\frac{1}{12}z^{6}\right]_{0}^{2}\ &=\frac{1}{5}\times2^{5}-\frac{1}{12}\times2^{6}\ &=\frac{32}{5}-\frac{64}{12}\ &=\frac{32}{5}-\frac{16}{3}\ &=\frac{96 - 80}{15}\ &=\frac{16}{15} \end{align*} ]

Answer:

$\frac{16}{15}$