Question

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

${\int }_0^{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}{\int }_0^5{p}^2\sin \varphi dpd\theta d\varphi$

Short answer

${\int }_0^{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}{\int }_0^5{p}^2\sin \varphi dpd\theta d\varphi =\frac{500}{3}\mathrm{\pi }$

Step-by-step solution

Given

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

${\int }_0^{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}{\int }_0^5{p}^2\sin \varphi dpd\theta d\varphi$

Evaluate integral

$$\begin{gathered} V={\int }_0^{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}{\int }_0^5{p}^2\sin \varphi dpd\theta d\varphi \\ V={\int }_0^{\mathrm{\pi }}\sin \varphi d\varphi {\int }_0^{2\mathrm{\pi }}d\theta \left[{\int }_0^5{p}^{2}dp\right] \\ V={\left[\frac{p^3}{3}\right]}_0^5{\int }_0^{\mathrm{\pi }}\sin \varphi d\varphi {\int }_0^{2\mathrm{\pi }}d\theta \\ V=\frac{125}{3}{\int }_0^{\mathrm{\pi }}\sin \varphi d\varphi {\int }_0^{2\mathrm{\pi }}d\theta \\ V=\frac{125}{3}\left(2\mathrm{\pi }\right){\int }_0^{\mathrm{\pi }}\sin \varphi d\varphi \\ V=\frac{250}{3}\mathrm{\pi }{\left[-\cos \mathrm{\varphi }\right]}_0^{\mathrm{\pi }} \\ \mathrm{V}=\frac{250}{3}\mathrm{\pi }\left[1+1\right] \\ \mathrm{V}=\frac{500}{3}\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$