Problem source

Prove that the area of the zone of the surface of a sphere between two parallel planes cutting the sphere is given by 
\[
2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}).
\]
A tangent from the origin $O$ to the curve with cartesian equation
\[
(x-c)^{2}+y^{2}=a^{2},
\]
where $a$ and $c$ are positive constants with $c>a,$ touches the curve at $P$. The $x$-axis cuts the curve at $Q$ and $R$, the points lying in the order $OQR$ on the axis. The line $OP$ and the arc $PR$ are rotated through $2\pi$ radians about the line $OQR$ to form a surface. Find the area of this surface.

Solution source

\begin{center}
    \begin{tikzpicture}
        \coordinate (O) at (0,0);

        \draw (O) circle (2);
        \def\a{70};
        \def\b{120};

        \draw[domain = 0:180, samples=180, variable = \x]  plot ({2*cos(\x)},{-0.2*(sin(\x))});
        \draw[domain = 180:360, dashed, samples=180, variable = \x]  plot ({2*cos(\x)},{-0.2*(sin(\x))});

        \draw[domain = 0:180, samples=180, variable = \x]  plot ({2*sin(\a)*cos(\x)},{2*cos(\a)+-0.2*(sin(\x))});
        \draw[domain = 180:360, dashed, samples=180, variable = \x]  plot ({2*sin(\a)*cos(\x)},{2*cos(\a)+-0.2*(sin(\x))});

        \draw[domain = 0:180, samples=180, variable = \x]  plot ({2*sin(\b)*cos(\x)},{2*cos(\b)+-0.2*(sin(\x))});
        \draw[domain = 180:360, dashed, samples=180, variable = \x]  plot ({2*sin(\b)*cos(\x)},{2*cos(\b)+-0.2*(sin(\x))});
                
    \end{tikzpicture}
\end{center}


We can choose a coordinate frame where the parallel planes are parallel  to the $y-z$ axis. Then we can compute the surface area as an integral of the surface of revolution for $x^2 + y^2 = r^2$. Using $y = r \sin t, x = r \cos t$ we have:

\begin{align*}
S &= 2\pi\int_{\cos^{-1}a}^{\cos^{-1}b}y \sqrt{\left ( \frac{\d x}{\d t} \right)^2+\left ( \frac{\d y}{\d t} \right)^2} \d t \\
&=2\pi\int_{\cos^{-1}a}^{\cos^{-1}b} r^2 \sin t \d t \\
&= 2\pi \cdot r^2 \cdot (a - b) \\
&= 2 \pi \cdot r \cdot (ra-rb) \\
&= 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}).
\end{align*}

\begin{center}
    \begin{tikzpicture}
        \def\a{2};
        \def\c{3};
        \def\t{120};
        \coordinate (Q) at ({\c-\a},0);
        \coordinate (R) at ({\c+\a},0);
        \coordinate (C) at (\c,0);

        \coordinate (P) at ({\c+\a*cos(\t)},{\a*sin(\t)});
        \coordinate (X) at ({\c+\a*cos(\t)},{0});
        \coordinate (L) at ({\a*tan(\t)*sin(\t)},{-\a*tan(\t)*cos(\t)});
        
        \coordinate (O) at ($(P)+(L)$);
        

        \draw (P) -- (O);
        
        \draw (C) circle (\a);


        \node [left] at (O) {$O$};
        \node [above] at (P) {$P$};
        \node [above left] at (Q) {$Q$};
        \node [above right] at (R) {$R$};
        \node [above] at (C) {$C$};
        \node [above] at (X) {$X$};

        \draw[->] (O) -- ({\c+\a+1}, 0) node [right] {$x$};
        \draw[->] (O) -- ++(0,{\a+1}) node [above] {$y$};
        \draw (O) -- ++(0,{-\a-1});

        \draw[dashed] (P) -- ({\c+\a*cos(\t)},{-\a*sin(\t)}) -- (O);
        \draw[dashed] (P) -- (C);
    \end{tikzpicture}
\end{center}

We can view this surface as a sphere missing a cap of height $XQ$ and adding a cone of slant height $OP$ and radius $PX$

The centre of the circle is at $(c,0)$ and $OP^2 + a^2 = c^2 \Rightarrow OP = \sqrt{c^2-a^2}$

Since $OPC \sim OXP$ we must have that 

$\frac{OX}{OP} = \frac{OP}{OC} \Rightarrow OX = \frac{c^2-a^2}{c}$ and $\frac{PX}{OP} = \frac{CP}{OC} \Rightarrow PX =  \frac{a}{c}\sqrt{c^2-a^2}$

$QX = OX - OQ = \frac{c^2-a^2}{c}-(c-a) = \frac{ac-a^2}{c}$

Therefore the surface area is:

\begin{align*}
S &= 4 \pi a^2 - 2\pi \cdot a \cdot QX+ \pi PX \cdot OP \\
&= 4 \pi a^2 - 2\pi a \cdot \frac{ac-a^2}{c}+\pi \frac{a}{c}\sqrt{c^2-a^2}\cdot \sqrt{c^2-a^2} \\
&= 4\pi a^2 -2\pi \frac{a^2c-a^3}{c}+\pi \frac{ac^2-a^3}{c} \\
&= \pi a \frac{(a+c)^2}{c}

\end{align*}