Question

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

${\int }_0^{2\mathrm{\pi }}{\int }_0^4{\int }_0^{4-r}rdzdrd\theta$

Short answer

${\int }_0^{2\mathrm{\pi }}{\int }_0^4{\int }_0^{4-r}rdzdrd\theta =\frac{64}{3}\mathrm{\pi }$

Step-by-step solution

Given

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

${\int }_0^{2\mathrm{\pi }}{\int }_0^4{\int }_0^{4-r}rdzdrd\theta$

Evaluate the integral

$$\begin{gathered} V={\int }_0^{2\mathrm{\pi }}{\int }_0^4\left[{\int }_0^{4-r}rdz\right]drd\theta \\ V={\int }_0^{2\mathrm{\pi }}\left[{\int }_0^{4}r(4-r)dr\right]d\theta \\ V={\int }_0^{2\mathrm{\pi }}{\left[4\frac{r^2}{2}-\frac{r^3}{3}\right]}_0^{4}d\theta \\ V={\int }_0^{2\mathrm{\pi }}\left[32-\frac{64}{3}\right]d\theta \\ V=\frac{32}{3}{\int }_0^{2\mathrm{\pi }}d\theta \\ V=\frac{32}{3}(2\mathrm{\pi }-0) \\ V=\frac{64}{3}\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$