Q. 34 - 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.  

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

Step-by-Step Solution

Answer

${\int }_0^{\mathrm{\pi }}{\int }_0^2{\int }_0^{r\sin \theta } rdzdrd\theta =\frac{16}{3}$

Step 1: Given

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

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

Step 2: Evaluate the integral

$$\begin{gathered} V={\int }_0^{\mathrm{\pi }}{\int }_0^2\left[{\int }_0^{r\sin \theta } rdz\right]drd\theta \\ V={\int }_0^{\mathrm{\pi }}{\int }_0^2\left[ r\left(r\sin \theta -0\right)\right]drd\theta \\ V={\int }_0^{\mathrm{\pi }}{\int }_0^2{r}^2 \sin \theta drd\theta \\ V={\int }_0^{\mathrm{\pi }}\sin \theta d\theta {\int }_0^2{r}^{2}dr \\ V={\int }_0^{\mathrm{\pi }}\sin \theta d\theta {\left[\frac{r^3}{3}\right]}_0^2 \\ V=\frac{8}{3}{\int }_0^{\mathrm{\pi }}\sin \theta d\theta \\ V=\frac{8}{3}{\left[-\cos \theta \right]}_0^{\mathrm{\pi }} \\ V=\frac{8}{3}\left[-\cos \mathrm{\pi }+\cos 0\right] \\ V=\frac{8}{3}\left[1+1\right] \\ V=\frac{16}{3} \mathrm{cubic} \mathrm{units} \end{gathered}$$