Problems

A spherical loaf of bread is cut into parallel slices of equal thickness. Show that, after any number of the slices have been eaten, the area of crust remaining is proportional to the number of slices remaining. A European ruling decrees that a parallel-sliced spherical loaf can only be referred to as `crusty' if the ratio of volume \(V\) (in cubic metres) of bread remaining to area \(A\) (in square metres) of crust remaining after any number of slices have been eaten satisfies \(V/A<1\). Show that the radius of a crusty parallel-sliced spherical loaf must be less than \(2\frac23\) metres. [{\sl The area \(A\) and volume \(V\) formed by rotating a curve in the \(x\)--\(y\) plane round the \(x\)-axis from \(x=-a\) to \(x=-a+t\) are given by \[ A= 2\pi\int_{-a}^{-a+t} { y}\left( 1+ \Big(\frac{\d {y}}{\d x}\Big)^2\right)^{\frac12} \d x\;, \ \ \ \ \ \ \ \ \ \ \ V= \pi \int_{-a}^{-a+t} {y}^2 \d x \;. \ \ ] \] }

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.

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Solution:

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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*}

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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*}

A kingdom consists of a vast plane with a central parabolic hill. In a vertical cross-section through the centre of the hill, with the \(x\)-axis horizontal and the \(z\)-axis vertical, the surface of the plane and hill is given by \[ z=\begin{cases} \dfrac{1}{2a}(a^{2}-x^{2}) & \mbox{ for }\left|x\right|\leqslant a,\\ 0 & \mbox{ for }\left|x\right|>a. \end{cases} \] The whole surface is formed by rotating this cross-section about the \(z\)-axis. In the \((x,z)\) plane through the centre of the hill, the king has a summer residence at \((-R,0)\) and a winter residence at \((R,0)\), where \(R>a.\) He wishes to connect them by a road, consisting of the following segments: \begin{itemize} \item a path in the \((x,z)\) plane joining \((-R,0)\) to \((-b,(a^{2}-b^{2})/2a),\) where \(0\leqslant b\leqslant a.\) \item a horizontal semicircular path joining the two points \((\pm b,(a^{2}-b^{2})/2a),\) if \(b\neq0;\) \item a path in the \((x,z)\) plane joining \((b,(a^{2}-b^{2})/2a)\) to \((R,0).\) \end{itemiz} The king wants the road to be as short as possible. Advise him on his choice of \(b.\)