Question

I have a question concerning the application of Pappus' theorem; details in the reply.

Answer 1

R is the region limited by the semi-circumference \[y=\sqrt{r^2 - x^2}\] and the x-axis. Use Pappus' theorem to find the moment of R with respect to the line y = -4.
Just for reference, Pappus' 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 y (with a bar above) 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 the y coordinate of the centroid multiplied by the area of the region.
So, the "moment of a region" (M) is that which is divided by the total area to find the centroid of the region.

Answer 2

I've tried to solve this as follows:
I calculated that the y-coordinate of the centroid of a semicircular area limited by the x-axis and the curve \[y = \sqrt{r^2 - x^2}\] is given by: \[\bar{y} = \frac{4r}{3\pi}\]
So, the vertical coordinate of the centroid with respect to the line y = -4 would be just
\[\bar{y} = \frac{4r}{3\pi}+4\]
So, the moment should be obtained by:
\[M = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}\]
and I also get the exact same result when doing it without using Pappus' theorem; that is, by directly evaluating the integral below:
\[M = \int_{-r}^r \sqrt{r^2-x^2} \left (\frac{\sqrt{r^2-x^2}}{2} + 4 \right ) dx\]
So, this should be the result. But the book I'm using says it should give:
\[\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )\]