Subsections


8.3 Example

From [1, p. 507, 21c]

Asked: Find the centroid of the first octant region inside $x^2+y^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9 and below $x+z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4.


8.3.1 Approach

The region inside $x^2+y^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9 is the inside of a cylinder of radius 3 around the z-axis. The equation $x+z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 describes a plane through the $y$-axis under 45 degrees with the $x$-axis:


\begin{displaymath}
\hbox{\epsffile{mulintx21.eps}}
\end{displaymath}

Use cylindrical coordinates $r$, $\theta$, and $z$:

\begin{displaymath}
x=r\cos\theta \qquad y = r \sin \theta
\end{displaymath}

Integrate $z$ first:


\begin{displaymath}
\hbox{\epsffile{mulintx22.eps}}
\end{displaymath}

(Why not $r$ first? Why not $\theta$?). Boundaries are

\begin{displaymath}
z_1=0 \qquad z_2 = 4-x = 4 - r\cos\theta
\end{displaymath}

Next integrate $\theta$ and $r$:


\begin{displaymath}
\hbox{\epsffile{mulintx23.eps}}
\end{displaymath}


\begin{displaymath}
\theta_1=0 \qquad \theta_2={\textstyle\frac{1}{2}}\pi
\end{displaymath}


\begin{displaymath}
r_1=0 \qquad r_2=3
\end{displaymath}


8.3.2 Results

For the volume $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\int\int\int {\rm d}V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\int\int\int
r\;{\rm d}z {\rm d}r {\rm d}\theta$:


\begin{displaymath}
V = \int_{\theta=0}^{\pi/2}
\int_{r=0}^{3}
\left[
\i...
...cos\theta} r \; {\rm d}z
\right]
{\rm d}r
{\rm d}\theta
\end{displaymath}


\begin{displaymath}
= \int_{\theta=0}^{\pi/2}
\left[
\int_{r=0}^{3}
(4-r\cos\theta) r
{\rm d}r
\right]
{\rm d}\theta
\end{displaymath}


\begin{displaymath}
= \int_{\theta=0}^{\pi/2}
18-9\cos\theta
{\rm d}\theta = 9 (\pi -1)
\end{displaymath}

For $V \bar x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\int\int\int x \;{\rm d}V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\int\int\int
x r {\rm d}z {\rm d}r {\rm d}\theta$:


\begin{displaymath}
V \bar x = \int_{\theta=0}^{\pi/2}
\int_{r=0}^{3}
\lef...
... \cos \theta \; {\rm d}z
\right]
{\rm d}r
{\rm d}\theta
\end{displaymath}


\begin{displaymath}
= \int_{\theta=0}^{\pi/2}
\left[
\int_{r=0}^{3}
4r^2\cos\theta -r^3\cos^2\theta
{\rm d}r
\right]
{\rm d}\theta
\end{displaymath}


\begin{displaymath}
= \int_{\theta=0}^{\pi/2}
36\cos\theta-\frac{81}4\cos^2\theta
{\rm d}\theta = \frac9{16}(64-9\pi)
\end{displaymath}

hence $\bar x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(64 - 9\pi)/16(\pi-1)$

Etcetera.