\section{Pappus's Centroid Theorem}

\begin{figure}[htp]
	\centering
	\includegraphics[width=0.6\linewidth]{img/pappus_1.png}
	\caption{}
	\label{pappus1}
\end{figure}

\remark{}
\textit{Centroids} are closely related to, and often synonymous with, centres of mass. A centroid is the geometric centre of an object, regardless of the mass distribution. Thus, the centroid and centre of mass coincide when the mass is uniformly distributed.

\remark{}
Figure \ref{pappus1} depicts three different surfaces constructed by revolving a line segment (in red) about a central axis. These are often called \textit{surfaces of revolution}.

\problem{}
\textit{Pappus's First Centroid Theorem} allows you to determine the area of a surface of revolution using information about the line segment and the axis of rotation.
Can you intuitively come up with Pappus's First Centroid Theorem for yourself? Figure \ref{pappus1} is very helpful. It may also help to draw from surface area formulae you already know. What limitations are there on the theorem?

\begin{solution}
	\url{https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem}
	\url{https://mathworld.wolfram.com/PappussCentroidTheorem.html}
\end{solution}

\vfill
\pagebreak

\problem{}
\textit{Pappus's Second Centroid Theorem} simply extends this concept to \textit{solids of revolution}, which are exactly what you think they are.


\problem{}
Now that you've done the first theorem, what do you think Pappus's Second Centroid Theorem states?

\vfill


\problem{}
The centroid of a semi-circular line segment is already given in Figure \ref{pappus1}, but what about the centroid of a filled semi-circle? (Hint: For a sphere of radius $r$, $V=\frac{4}{3}\pi r^3$)

\begin{figure}[htp]
	\centering
	\includegraphics[width=0.4\linewidth]{img/arc.png}
	\caption{}
	\label{arc}
\end{figure}

\problem{}
Given arc $AB$ with radius $r$ and subtended by $2\alpha$, determine $OG$, the distance from the centre of the circle to the centroid of the arc.

\begin{solution}
	\url{https://mathspanda.com/A2FM/Lessons/Centres_of_mass_of_standard_shapes_LESSON.pdf}
\end{solution}

\vfill

\pagebreak

\problem{}<sector>
Where is the centroid of the \textit{sector} of the circle in Figure \ref{arc}?
\hint{cut it up.}


\vfill

\problem{}
Seeing your success with his linear staff, the wizard challenges you with another magical staff to balance.
It looks identical to the first one, but you're told that the density decreases from $\lambda_0$ to $0$
according to the function $\lambda(x) = \lambda_0\sqrt{1-x^2}$.

\begin{solution}
	This is equivalent to finding the x-coordinate of the centroid of a quarter-circle. See \ref{sector}.
\end{solution}

\vfill
\problem{}
Infinitely many masses $m_i$ are placed at $x_i$ along the positive $x$-axis, starting with $m_0 = 1$ placed at $x_0 = 1$. Each successive mass is placed twice as far from the origin compared to the previous one. But also, each successive mass has a quarter the weight of the previous one. Find the CoM if it exists.

\begin{solution}
	We have $m_i = 1/4^i$ and $x_i = 2^i$ so
	\begin{align*}
		\sum_{i=0}^\infty m_i x_i &= \sum \frac{2^i}{4^i} \\
		&= \sum \frac{1}{2^i} \\
		&= \frac{1}{1-1/2} = 2 \qquad \text{as this is just a geometric series} \\
	\end{align*}
	\[ M = \sum m_i = \frac{1}{1-1/4} = 4/3 \]
	Then
	\[ x_{CM} = \frac{\sum m_i x_i}{M} = 3/2 \]
\end{solution}

\vfill

\problem{Bonus}
Try to actually find $h$ from Problem \ref{3D soda}. Good luck.

