Question

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

${\int }_0^{2\mathrm{\pi }}{\int }_0^2{\int }_0^{5}rdzdrd\theta$

Short answer

${\int }_0^{2\mathrm{\pi }}{\int }_0^2{\int }_0^{5}rdzdrd\theta =20\mathrm{\pi }$

Step-by-step solution

Given

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

${\int }_0^{2\mathrm{\pi }}{\int }_0^2{\int }_0^{5}rdzdrd\theta$

Evaluate the integral

$$\begin{gathered} V={\int }_0^{2\mathrm{\pi }}{\int }_0^2\left[{\int }_0^{5}rdz\right]drd\theta \\ V={\int }_0^{2\mathrm{\pi }}\left[{\int }_0^{2}5rdr\right]d\theta \\ V=5{\int }_0^{2\mathrm{\pi }}{\left[\frac{r^2}{2}\right]}_0^{2}d\theta \\ V=5\times 2{\int }_0^{2\mathrm{\pi }}d\theta \\ V=10(2\mathrm{\pi }-0) \\ V=20\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$