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Using Pappus' theorem to find the moment of a region 
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#1
Aug911, 06:09 PM

P: 205

1. The problem statement, all variables and given/known data
R is the region limited by the semicircumference [itex]\sqrt{r^2  x^2}[/itex] and the xaxis. 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 ycoordinate of the centroid of R, the volume V of the solid of revolution obtained by rotating R around the xaxis 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 ycoordinate 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 xcoordinates 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 ycoordinate of the centroid of a semicircular area limited by [itex]y = \sqrt{r^2  x^2}[/itex] and the xaxis 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^2x^2} \left (\frac{\sqrt{r^2x^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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