Q. 36 - 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.

${\int }_0^{\mathrm{\pi }/4}{\int }_0^{2\mathrm{\pi }}{\int }_0^{4sec \theta }{p}^2\sin \varphi dpd\theta d\varphi$

Step-by-Step Solution

Answer

${\int }_0^{\mathrm{\pi }/4}{\int }_0^{2\mathrm{\pi }}{\int }_0^{4sec \theta }{p}^2\sin \varphi dpd\theta d\varphi =0$

Step 1: Given

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

${\int }_0^{\mathrm{\pi }/4}{\int }_0^{2\mathrm{\pi }}{\int }_0^{4sec \theta }{p}^2\sin \varphi dpd\theta d\varphi$

Step 2: Calculate the volume

$$\begin{gathered} V={\int }_0^{\mathrm{\pi }/4}{\int }_0^{2\mathrm{\pi }}{\int }_0^{4sec \theta }{p}^2\sin \varphi dpd\theta d\varphi \\ V={\int }_0^{\mathrm{\pi }/4}\sin \varphi d\varphi {\int }_0^{2\mathrm{\pi }}{\left[\frac{p^3}{3}\right]}_0^{4sec \theta }d\theta \\ V=\frac{64}{3}{\int }_0^{\mathrm{\pi }/4}\sin \varphi d\varphi \left[{\int }_0^{2\mathrm{\pi }}se{c}^3 \theta d\theta \right] \\ V=\frac{64}{3}{\int }_0^{\mathrm{\pi }/4}\sin \varphi d\varphi {\left[sec \theta ·\tan \theta +\log \left|sec \theta +\tan \theta \right|\right]}_0^{2\mathrm{\pi }} \\ V=\frac{64}{3}{\int }_0^{\mathrm{\pi }/4}\sin \varphi d\varphi \left[\left(sec 2\mathrm{\pi }·\tan 2\mathrm{\pi }+\log \left|sec 2\mathrm{\pi }+\tan 2\mathrm{\pi }\right|\right)-\left(sec 0·\tan 0+\log \left|sec 0+\tan 0\right|\right)\right] \\ V=\frac{64}{3}{\int }_0^{\mathrm{\pi }/4}\sin \varphi d\varphi \left[0-0\right] \\ V=0\times \frac{64}{3}{\int }_0^{\mathrm{\pi }/4}\sin \varphi \\ V=0 \end{gathered}$$