Question

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

${\int }_0^4{\int }_0^{3-\frac{3}{4}x}{\int }_0^{2-\frac{1}{2}x-\frac{2}{3}y}dzdydx$

Short answer

${\int }_0^4{\int }_0^{3-\frac{3}{4}x}{\int }_0^{2-\frac{1}{2}x-\frac{2}{3}y}dzdydx=4$

Step-by-step solution

Given

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

${\int }_0^4{\int }_0^{3-\frac{3}{4}x}{\int }_0^{2-\frac{1}{2}x-\frac{2}{3}y}dzdydx$

Evaluate the integral

$$\begin{gathered} V={\int }_0^4{\int }_0^{3-\frac{3}{4}x}\left[{\int }_0^{2-\frac{1}{2}x-\frac{2}{3}y}dz\right]dydx \\ V={\int }_0^4\left[{\int }_0^{3-\frac{3}{4}x}\left(2-\frac{1}{2}x-\frac{2}{3}y\right)dy\right]dx \\ V={\int }_0^4{\left[2y-\frac{1}{2}xy-\frac{2}{3}\frac{y^2}{2}\right]}_0^{3-\frac{3}{4}x} \\ V={\int }_0^4\left[2\left(3-\frac{3}{4}x\right)-\frac{1}{2}x\left(3-\frac{3}{4}x\right)-\frac{1}{3}{\left(3-\frac{3}{4}x\right)}^2\right] \\ V={\int }_0^4\left(3-\frac{3}{2}x+\frac{3}{16}{x}^2\right)dx \\ V={\left[3x-\frac{3}{2}\frac{x^2}{2}+\frac{3}{16}\frac{x^3}{3}\right]}_0^4 \\ V=\left[12-12+4\right] \\ V=4\mathrm{cubic}\mathrm{units} \end{gathered}$$