5.3 Example

From [1, p. 507, 21c]

Asked: Find the centroid of the first octant region inside $x^2+y^2=9$ and below $x+z=4$.

5.3.1 Approach

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

$$\hbox{\epsffile{mulintx21.eps}}$$

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

$$x=r\cos\theta \qquad y = r \sin \theta$$

Integrate $z$ first:

$$\hbox{\epsffile{mulintx22.eps}}$$

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

$$z_1=0 \qquad z_2 = 4-x = 4 - r\cos\theta$$

Next integrate $\theta$ and $r$:

$$\hbox{\epsffile{mulintx23.eps}}$$

$$\theta_1=0 \qquad \theta_2={\textstyle\frac{1}{2}}\pi$$

$$r_1=0 \qquad r_2=3$$

5.3.2 Results

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

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

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

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

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

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

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

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

hence $\bar x= (64 - 9\pi)/16(\pi-1)$

Etcetera.