Question

${\mathbf{\int }}_{\mathbf{z}\mathbf{=}\mathbf{0}}^{\mathbf{2}}{\mathbf{\int }}_{\mathbf{x}\mathbf{=}\mathbf{z}}^{\mathbf{2}}{\mathbf{\int }}_{\mathbf{y}\mathbf{=}\mathbf{8}\mathbf{x}}^{\mathbf{z}}\mathbf{dydxdz}$

Short answer

The solution for the given integral is $-20$.

Step-by-step solution

Definition of integration

Integration is a way of finding the whole by adding or summing the parts.

Given Information

The integral${\int }_{\mathrm{z}=0}^2{\int }_{\mathrm{x}=\mathrm{z}}^2{\int }_{\mathrm{y}=8\mathrm{x}}^{\mathrm{z}}\mathrm{dydxdz}$.

To find the value for the given integral.

Finding the integral value

Now, Integrate the given value,

$$\begin{gathered} \mathrm{l}={\int }_0^1\mathrm{dz}{\int }_{\mathrm{z}}^2\mathrm{dx}{\int }_{8\mathrm{x}}^{\mathrm{z}}\mathrm{dy} \\ ={\int }_0^2\mathrm{dz}{\int }_{\mathrm{z}}^2\mathrm{dx}\left(\mathrm{z}-8\mathrm{x}\right) \\ ={\overline{){\int }_0^2\mathrm{dx}\left(\mathrm{zx}-4{\mathrm{x}}^2\right)}}_{\mathrm{z}}^2 \end{gathered}$$

Further Integrating

Then,

$$\begin{gathered} \mathrm{l}={\int }_0^2\mathrm{dz}\left(2\mathrm{z}-{\mathrm{z}}^2-16+4{\mathrm{z}}^2\right) \\ ={\int }_0^2\left(3{\mathrm{z}}^2+2\mathrm{z}-16\right)\mathrm{dz} \\ ={\overline{)\left({\mathrm{z}}^3+{\mathrm{z}}^2-16\mathrm{z}\right)}}_0^2 \\ =-20 \end{gathered}$$

Therefore, the required solution for the given integral is $-20$.