Problems

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Solution:

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*} 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*}