# Use Pappus' theorem to find the moment of a region limited by a semi-circunference.

This is part of self-study; I found this question in the book "The Calculus with Analytic Geometry" (Leithold).

$R$ 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 $y = -4$.

Pappus' theorem (also referred to as Guldinus theorem or Pappus-Guldinus theorem) is as follows:

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: $V = 2\pi\bar{y}A$.

Also, what I'm calling moment (I'm not sure if this is a common term) is the quantity that is divided by the area of the region in order to find a coordinate of the centroid of the region: for example, to find the x-coordinate of the centroid ($\bar{x}$), first one finds the moment around the y-axis ($M_y$), then divides it by the area of the region ($A$).

Since the region $R$ is symmetric with respect to the y-axis, the x-coordinate of the centroid ($\bar{x}$) is zero (therefore, the moment around the y-axis ($M_y$) is zero, because $M_y = \bar{x}A$, where $A$ is the area of the region). So, I only need to find the vertical coordinate of the centroid (with respect to the line $y = -4$) and the moment around the line $y = -4$.

The book's answer for the moment around the line $y = -4$ is: $\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )$. I included two attempts below; both arrive at a same result, which is different from the result of the book.

Attempt 1

First I tried to use Pappus' theorem to find the vertical coordinate of the centroid of the semi-circular region limited by $\sqrt{r^2 - x^2}$, with respect to the x-axis (not yet with respect to the line y = -4). I will call it $\bar{y}_x$.

Since the solid of revolution obtained by rotating this semi-circular region around the x-axis is a sphere, its volume is $V = \frac{4}{3}\pi r^3$. The area of the semi-circular region is $A=\frac{\pi r^2}{2}$. Substituting $V$ and $A$ into Pappus' theorem:

$\frac{4}{3}\pi r^3 = 2\pi\bar{y}_x\frac{\pi r^2}{2}$

$\bar{y}_x = \frac{4r}{3\pi}$.

This is the vertical coordinate of the centroid with respect to the x-axis. The vertical coordinate of the centroid with respect to the line $y = -4$ is:

$\bar{y} = \frac{4r}{3\pi} + 4$.

To find the moment around the line $y = -4$, I use the fact that $\bar{y} = \frac{M_x}{A}$:

$M_x = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$.

Attempt 2

The moment of a plane region with respect to the line $y = -4$ can be found by dividing this region into infinitesimal elements of area, then multiplying the area of each element of area by the vertical coordinate of the centroid.

So, if $f(x) = \sqrt{r^2 - x^2}$, then, if I divide the semi-circular region into several rectangles of length $dx$, the area of each rectangle is $f(x) dx$ and the vertical coordinate of the centroid of each rectangle with respect to the line $y = -4$ is $\frac{f(x)}{2} + 4$. So, the moment with respect to the line $y = -4$ is:

$M_x = \int_{-r}^r \left [ f(x)\times \left ( \frac{f(x)}{2} + 4 \right ) dx \right ]$.

Solving this integral gives the following result:

$M_x = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$,

which is the same result I found in the first attempt.

Both your attempts lead to the correct result, and the solution given in the book is wrong. To see that the book's solution cannot be correct do the following "Gedankenexperiment": Assume that we are told to compute the moment with respect to the line $y=-\eta$ for some given $\eta$ instead of $4$. For an $\eta \gg r$ this moment would be approximatively proportional to $r^2$ and to $\eta$, as is the case in your solution but not in the solution given by the book: The $O(r^3)$ dependency is unwarranted.