MHB 17.7.07 other orders usually work well and are occasionally easier to evaluate

[karush]

The integrals we have seen so far suggest that there are preferred orders of integration for cylindricsl coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integral

\begin{align*}\displaystyle
dV&=\int_{0}^{2\pi}\int_{0}^{3}\int_{0}^{z/3}r^3 \, dr \, dz \, d\theta\\
\\
&=\color{red}{\frac{3\pi}{10}}
\end{align*}

ok I tried some rearrange but it just got worse
I would presume this is converting $r^3$ to rectangular coordinates
red is book answer

 

[HOI]

The integrals we have seen so far suggest that there are preferred orders of integration for cylindricsl coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integral

\begin{align*}\displaystyle
dV&=\int_{0}^{2\pi}\int_{0}^{3}\int_{0}^{z/3}r^3 \, dr \, dz \, d\theta\\
\\
&=\color{red}{\frac{3\pi}{10}}
\end{align*}

ok I tried some rearrange but it just got worse
I would presume this is converting $r^3$ to rectangular coordinates
red is book answer

First, because neither the integrand nor the limits of the other two integrals involve $\theta$ we can do that separately:
$\int_0^{2\pi} d\theta= 2\pi$
so that is simply
$dV= 2\pi \int_0^3 \int_0^{z/3} r^3 dr dz$
The integral of $r^3$ is $\frac{r^4}{4}$ and, evaluated between 0 and z/3, gives $\frac{z^4}{324}$. Integrating $\int_0^3 \frac{z^4}{324} dz= \left.\frac{z^5}{1620}\right|_0^3= \frac{243}{1620}= \frac{3}{20}$. Multiplying by $2\pi$ from the first integral that is $\frac{3\pi}{10}$.

"Changing the order of integration" does NOT mean "changing form cylindrical to Cartesian coordinates". Here, since, again, we could do the $\theta$ integral separately, we are looking at $\int_0^3\int_0^{z/3} r^3 drdz$ That means that we are letting z go from 0 to 3 and, for each z, letting r go from 0 to $z^3$. We can visualize that as the region under the cone $r= z^3$ but inside the cylinder r= 27. We could also cover that region by taking r from 0 to 27 and, for each r, z going from $\sqrt[3]{r}= r^{1/3}$ to 3:
$\int_0^{27} \int_{r^{1/3}}^3 r^3dzdr$.

 

[karush]

before I put in a homework pdf I have

$\tiny{244.15.7.7}$
$\textsf{Changing the Order of Integration In Cylindrical Coordinates}$
\begin{align*}\displaystyle
dV&=\int_{0}^{2\pi}\int_{0}^{3}\int_{0}^{z/3}r^3 \, dr \, dz \, d\theta\\
\end{align*}
$\textit{First, because neither the integrand}$
$\textit{nor the limits of the other two integrals involve $\theta$}$
$\textit{we can do that separately:}$
$$\displaystyle\int_0^{2\pi} d\theta= 2\pi$$
$\textit{so that is simply}$
$$\displaystyle dV= 2\pi \int_0^3 \int_0^{z/3} r^3 dr \, dz$$
$\textit{Then we can proceed with:}$
\begin{align*}\displaystyle
dV&= 2\pi\int_0^3
\biggr[\frac{r^4}{4} \biggr]_0^{z/3} dr \, dz \\
&=\int_0^3 \frac{z^4}{324} dz= \biggr[\frac{z^5}{1620}\biggr]_0^3= \frac{243}{1620}= \frac{3}{20}\\
&=2\pi \biggr[\frac{3}{20} \biggr]
=\color{red}{\frac{3\pi}{10}}
\end{align*}