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Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cylindrical

Table 8.5.3(a)   Integration in cylindrical coordinates via a visualization task template

Initialize

Loading Student:-MultivariateCalculus

Assign the transformation equations the name $T$ 

$\left[x\=\frac{u \cos \left(w\right)}{v}\,y\=\frac{u \sin \left(w\right)}{v}\,z\=u^2\right]$$\stackrel{\text{assign to a name}}{\to }$$T$

Obtain the Jacobian $\frac{\partial \left(x\,y\,z\right)}{\partial \left(u\,v\,w\right)}$ 

$\genfrac{}{}{0ex}{}{\left[x\,y\,z\right]}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{T}$

$\left[\frac{u\cos \left(w\right)}{v}\,\frac{u\sin \left(w\right)}{v}\,u^2\right]$

$\stackrel{\text{Jacobian}}{\to }$

$\stackrel{\text{determinant}}{\to }$

$-\frac{2u^3\left({\cos \left(w\right)}^2\+{\sin \left(w\right)}^2\right)}{v^3}$

$\stackrel{\text{simplify}}{\=}$

$-\frac{2u^3}{v^3}$

Apply the change of coordinates to the bounding surfaces

$\genfrac{}{}{0ex}{}{z\=9\left(x^2\+y^2\right)}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{T}$

$u^2\=\frac{9u^2{\cos \left(w\right)}^2}{v^2}\+\frac{9u^2{\sin \left(w\right)}^2}{v^2}$

$\stackrel{\text{simplify}}{\=}$

$u^2\=\frac{9u^2}{v^2}$

$\stackrel{\text{solve for v}}{\to }$

$\left[\left[v\=3\right]\,\left[v\=-3\right]\right]$

$\genfrac{}{}{0ex}{}{z\=4\left(x^2\+y^2\right)}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{T}$

$u^2\=\frac{4u^2{\cos \left(w\right)}^2}{v^2}\+\frac{4u^2{\sin \left(w\right)}^2}{v^2}$

$\stackrel{\text{simplify}}{\=}$

$u^2\=\frac{4u^2}{v^2}$

$\stackrel{\text{solve for v}}{\to }$

$\left[\left[v\=2\right]\,\left[v\=-2\right]\right]$

$\genfrac{}{}{0ex}{}{z\=2}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{T}$

$u^2\=2$

$\stackrel{\text{solve}}{\to }$

$\left\{u\=\sqrt{2}\right\}\,\left\{u\=-\sqrt{2}\right\}$

$\genfrac{}{}{0ex}{}{z\=3}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{T}$

$u^2\=3$

$\stackrel{\text{solve}}{\to }$

$\left\{u\=\sqrt{3}\right\}\,\left\{u\=-\sqrt{3}\right\}$

Write and evaluate an appropriate iterated triple-integral

${\int }_0^{2 \mathrm{\π}}{\int }_2^3{\int }_{{\sqrt{2}}_{}}^{{\sqrt{3}}_{}}2{\left(u\/v\right)}^3 \mathit{\ⅆ}u \mathit{\ⅆ}v \mathit{\ⅆ}w$ = $\frac{25}{72}\mathrm{\π}$$$$$

Solution in cylindrical coordinates

$$\begin{gathered} \mathrm{Int}\left(r\,\left[r\=\sqrt{z}\/3..\sqrt{z}\/2\,z\=2..3\,\mathrm{\θ}\=0..2 \mathrm{\π}\right]\right)\= \\ \mathrm{int}\left(r\,\left[r\=\sqrt{z}\/3..\sqrt{z}\/2\,z\=2..3\,\mathrm{\theta }\=0..2 \mathrm{\pi }\right]\right) \end{gathered}$$

${\∫}_0^{2\mathrm{\π}}{\∫}_2^3{\∫}_{\frac{1}{3}\sqrt{z}}^{\frac{1}{2}\sqrt{z}}r\ⅆr\ⅆz\ⅆ\mathrm{\θ}\=\frac{25}{72}\mathrm{\π}$

Initialize

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

$T≔\left[x\=\frac{u \cos \left(w\right)}{v}\,y\=\frac{u \sin \left(w\right)}{v}\,z\=u^2\right]\:$

Obtain the Jacobian $\frac{\partial \left(x\,y\,z\right)}{\partial \left(u\,v\,w\right)}$ 

$\mathrm{simplify}\left(\mathrm{Jacobian}\left(\mathrm{eval}\left(\left[x\,y\,z\right]\,T\right)\,\left[u\,v\,w\right]\,\mathrm{output}\=\mathrm{determinant}\right)\right)$ = $-\frac{2u^3}{v^3}$$$

Apply the change of coordinates to the bounding surfaces

$s_1≔\mathrm{simplify}\left(\mathrm{eval}\left(z\=4\left(x^2\+y^2\right)\,T\right)\right)$

$u^2\=\frac{4u^2}{v^2}$

$\mathrm{solve}\left(s_1\,v\right)$ = $2\,-2$$$

$s_3≔\mathrm{eval}\left(z\=2\,T\right)$ = $u^2\=2$$\mathrm{solve}\left(s_3\,u\right)$$\sqrt{2}\,-\sqrt{2}$$$

$s_2≔\mathrm{simplify}\left(\mathrm{eval}\left(z\=9\left(x^2\+y^2\right)\,T\right)\right)$

$u^2\=\frac{9u^2}{v^2}$

$\mathrm{solve}\left(s_2\,v\right)$ = $3\,-3$$$

$s_4≔\mathrm{eval}\left(z\=3\,T\right)$ = $u^2\=3$$\mathrm{solve}\left(s_4\,u\right)$$\sqrt{3}\,-\sqrt{3}$$$

Write and evaluate the appropriate iterated triple-integral

$$\begin{gathered} \mathrm{Int}\left(2{\left(u\/v\right)}^3\,\left[u\=\sqrt{2}..\sqrt{3}\,v\=2..3\,w\=0..2 \mathrm{\π}\right]\right)\= \\ \mathrm{int}\left(2{\left(u\/v\right)}^3\,\left[u\=\sqrt{2}..\sqrt{3}\,v\=2..3\,w\=0..2 \mathrm{\pi }\right]\right) \end{gathered}$$

${\∫}_0^{2\mathrm{\π}}{\∫}_2^3{\∫}_{\sqrt{2}}^{\sqrt{3}}\frac{2u^3}{v^3}\ⅆu\ⅆv\ⅆw\=\frac{25}{72}\mathrm{\π}$