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.

Thank you in advance.

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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.