# Using Pappus' theorem to find the moment of a region

by pc2-brazil
Tags: moment, pappus, region, theorem
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 P: 205 1. The problem statement, all variables and given/known data R is the region limited by the semi-circumference $\sqrt{r^2 - x^2}$ 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 $\bar{y}$ 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: $$V = 2\pi\bar{y}A$$ The moment of a plane region with respect to the x axis is given by: $$M_x = \bar{y}A$$ where $\bar{y}$ is the y-coordinate of the centroid of the region, or: $$M_x = \int_a^b f(x)\times \frac{f(x)}{2} dx$$ 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 $\frac{f(x)}{2}$ 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 $y = \sqrt{r^2 - x^2}$ and the x-axis is $\bar{y} = \frac{4r}{3\pi}$. So, the vertical coordinate of the centroid with respect to the line y = -4 would be $\bar{y} = \frac{4r}{3\pi}+4$. Since the area A is equal to $\frac{\pi r^2}{2}$, the moment should be obtained by: $$M = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$$ I also get this same value if I calculate it by the following integral, which comes from the definition of moment of a region: $$M = \int_{-r}^r \sqrt{r^2-x^2} \left (\frac{\sqrt{r^2-x^2}}{2} + 4 \right ) dx$$ But this doesn't lead to the correct answer, which is: $$\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )$$ Thank you in advance.

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