Q. 30 - Calculus

Question

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

${\int }_0^{2\mathrm{\pi }}{\int }_0^2{\int }_0^5 r dz dr d\theta$

Step-by-Step Solution

Answer

${\int }_0^{2\mathrm{\pi }}{\int }_0^2{\int }_0^5 r dz dr d\theta =20\mathrm{\pi }$

Step 1: Given

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

${\int }_0^{2\mathrm{\pi }}{\int }_0^2{\int }_0^5 r dz dr d\theta$

Step 2: Evaluate the integral

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