${\int }_0^{\mathrm{\π}}{\int }_0^2{\int }_0^{r \sin \left(\mathrm{\θ}\right)}r \mathrm{dz} \mathrm{dr} d\mathrm{\θ}$ = $\frac{16}{3}$ 

(The integration in the $z$-direction is from $z\=0$ to $z\=y\=r \sin \left(\mathrm{\θ}\right)$.)

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cylindrical

Table 8.1.5(a)   Task template integrating in cylindrical coordinates

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Multiple Integration≻Cylindrical

Table 8.1.5(b)   Task template implementing the MultiInt command iterating in the order $\mathrm{dz} \mathrm{dr} d\mathrm{\θ}$

${\int }_0^{\mathrm{\π}}{\int }_0^2{\int }_0^{r \sin \left(\mathrm{\θ}\right)}r \mathit{\ⅆ}z \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}$ = $\frac{16}{3}$$$

Table 8.1.5(c)   Integration via first principles

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\int }_0^{\mathrm{\pi }}{\int }_0^2{\int }_0^{r\sin \left(\mathrm{\theta }\right)}r\ⅆz\ⅆr\ⅆ\mathrm{\theta }$$\=$$\frac{16}{3}$$$

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

$\mathrm{MultiInt}\left(1\,z\=0..r \sin \left(\mathrm{\theta }\right)\,r\=0..2\,\mathrm{\θ}\=0.. \mathrm{\π}\,\mathrm{coordinates}\=\mathrm{cylindrical}\left[r\,\mathrm{\θ}\,z\right]\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_0^{\mathrm{\pi }}{\int }_0^2{\int }_0^{r\sin \left(\mathrm{\theta }\right)}r\ⅆz\ⅆr\ⅆ\mathrm{\theta }$

$\mathrm{MultiInt}\left(1\,z\=0..r \sin \left(\mathrm{\θ}\right)\,r\=0..2\,\mathrm{\θ}\=0..\mathrm{\π}\,\mathrm{coordinates}\=\mathrm{cylindrical}\left[r\,\mathrm{\θ}\,z\right]\right)$ = $\frac{16}{3}$$$

Table 8.1.5(d)   MultiInt command iterating in cylindrical coordinates in the order $\mathrm{dz} \mathrm{dr} d\mathrm{\θ}$

$$\begin{gathered} \mathrm{Int}\left(r\,\left[z\=0..r \sin \left(\mathrm{\theta }\right)\,r\=0..2\,\mathrm{\θ}\=0.. \mathrm{\π}\right]\right)\= \\ \mathrm{int}\left(r\,\left[z\=0..r \sin \left(\mathrm{\theta }\right)\,r\=0..2\,\mathrm{\θ}\=0.. \mathrm{\π}\right]\right) \end{gathered}$$

${\int }_0^{\mathrm{\pi }}{\int }_0^2{\int }_0^{r\sin \left(\mathrm{\theta }\right)}r\ⅆz\ⅆr\ⅆ\mathrm{\theta }=\frac{16}{3}$

Table 8.1.5(e)   Top-level Int and int commands