Using Pappus' theorem to find the moment of a region


by pc2-brazil
Tags: moment, pappus, region, theorem
pc2-brazil
pc2-brazil is offline
#1
Aug9-11, 06:09 PM
P: 205
1. The problem statement, all variables and given/known data
R is the region limited by the semi-circumference [itex]\sqrt{r^2 - x^2}[/itex] and the x-axis. Use Pappus' theorem to find the moment of R with respect to the line y = -4.

2. Relevant equations
Pappus' theorem:
If R is the region limited by the functions f(x) and g(x), then, if A is the area of R and [itex]\bar{y}[/itex] is the y-coordinate of the centroid of R, the volume V of the solid of revolution obtained by rotating R around the x-axis is given by:
[tex]V = 2\pi\bar{y}A[/tex]
The moment of a plane region with respect to the x axis is given by:
[tex]M_x = \bar{y}A[/tex]
where [itex]\bar{y}[/itex] is the y-coordinate of the centroid of the region, or:
[tex]M_x = \int_a^b f(x)\times \frac{f(x)}{2} dx[/tex]
where a and b are the x-coordinates that limit the region R, f(x) is the height of the region at x (therefore, f(x)dx is the area of each element of area) and [itex]\frac{f(x)}{2}[/itex] is the centroid of the infinitesimal rectangle.
Moment of region is analogous to moment of mass; the center of mass is calculated by dividing moment of mass by the total mass; the centroid is calculated by dividing the moment of region by the total area of the region.

3. The attempt at a solution
The y-coordinate of the centroid of a semicircular area limited by [itex]y = \sqrt{r^2 - x^2}[/itex] and the x-axis is [itex]\bar{y} = \frac{4r}{3\pi}[/itex]. So, the vertical coordinate of the centroid with respect to the line y = -4 would be [itex]\bar{y} = \frac{4r}{3\pi}+4[/itex]. Since the area A is equal to [itex]\frac{\pi r^2}{2}[/itex], the moment should be obtained by:
[tex]M = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}[/tex]
I also get this same value if I calculate it by the following integral, which comes from the definition of moment of a region:
[tex]M = \int_{-r}^r \sqrt{r^2-x^2} \left (\frac{\sqrt{r^2-x^2}}{2} + 4 \right ) dx[/tex]
But this doesn't lead to the correct answer, which is:
[tex]\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )[/tex]

Thank you in advance.
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