Question

Evaluating triple integrals: Each of the triple integrals that follows represents the volume of a solid. Sketch the solid and evaluate the integral.

${\int }_0^3{\int }_0^{2-\frac{2}{3}y}{\int }_0^{4-\frac{4}{3}y-2z}dxdzdy$

Short answer

${\int }_0^3{\int }_0^{2-\frac{2}{3}y}{\int }_0^{4-\frac{4}{3}y-2z}dxdzdy=4$

Step-by-step solution

Given 

We have to find the volume by using the triple integral.

${\int }_0^3{\int }_0^{2-\frac{2}{3}y}{\int }_0^{4-\frac{4}{3}y-2z}dxdzdy$

Evaluate the integral

$$\begin{gathered} V={\int }_0^3{\int }_0^{2-\frac{2}{3}y}{\int }_0^{4-\frac{4}{3}y-2z}dxdzdy \\ V={\int }_0^3\left[{\int }_0^{2-\frac{2}{3}y}\left(4-\frac{4}{3}y-2z\right)dz\right]dy \\ V={\int }_0^3{\left[4z-\frac{4}{3}yz-z^2\right]}_0^{2-\frac{2}{3}y}dy \\ V={\int }_0^3\left[4\left(2-\frac{2}{3}y\right)-\frac{4}{3}y\left(2-\frac{2}{3}y\right)-{\left(2-\frac{2}{3}y\right)}^2\right]dy \\ V={\int }_0^3\left(4-\frac{8}{3}y+\frac{4}{9}{y}^2\right)dy \\ V={\left[4y-\frac{8}{3}\frac{y^2}{2}+\frac{4}{9}\frac{y^3}{3}\right]}_0^3 \\ V=\left[12-12+4\right] \\ V=4\mathrm{cubic}\mathrm{units} \end{gathered}$$