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Assignment 7

Problems

Review

This assignment does not have many problems, and instead I want you to set aside the time you would usually be reading and working on the homework for review. We just finished Euclidean geometry! Go back through the book, and re-look over the sections which defined all of our foundational concepts.

  • Isometries
  • Lines
  • Shapes
  • Angles
  • Area

Make sure you know the definitions and understand the theorems. Bring questions to me in office hours after the break! We will be starting the geometry of the sphere when we return, and we will be rebuilding all of this for a new geometry - but at a much quicker rate!

Trigonometric Identities:

The following exercise has you compute with some trigonometric identities, which we needed to find the volume of spheres:

Exercise 1 (Integrating \(\cos^4(\theta)\)) Start with the angle sum-identity we derived in class some lectures back \[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]

Use this to derive an identity relating \(\cos(2\theta)\) to \(\cos^2(\theta)\) (we did most of this in class - but repeat it for yourselves). Now use this identity twice to show

\[\cos^4(\theta)= \frac{3}{8}+\frac{\cos 2\theta}{2}+\frac{\cos 4\theta}{8}\]

by writing \(\cos^4\theta = (\cos^2\theta)^2\). Then use this to confirm that

\[\int_0^{\frac\tau 4}\cos^4\theta = \frac{3}{8}\frac{\tau}{4}\]

This result was crucial to our calculuation of the volume of the four dimensional sphere in class, where we showed (via a trigonometric substitution)

\[\mathrm{vol}=\frac{8\pi}{3}\int_0^{\tau/4}\cos^4(\theta)d\theta\]

Using this result gave us the final answer:

\[\begin{align*} \mathrm{vol}&=\frac{8\pi}{3}\frac{3}{8}\frac{\tau}{4}\\ &= \frac{\pi\tau}{4}\\ &=\frac{\pi^2}{2} \end{align*}\]

The 5-Dimensional Sphere

The unit sphere in five dimensions is given by the set of points \((x,y,z,w,u)\) satisfying \[x^2+y^2+z^2+w^2+u^2=1\]

Exercise 2 Calculuate the volume of this space by slicing along the \(u\) direction. Show that the slices are four dimensional spheres: what’s the radius? Use the volume formula in four dimensions we derived in class \[\mathrm{vol}(r)=\frac{\pi^2}{2}r^4\] to write down the volume of each slice, and then perform the integral to confirm that the 5-dimensional volume is \[\frac{8\pi^2}{15}\]

(This will not need any trigonometric substitution) Once you have this, find the “surface area” of this 5-dimensional sphere by differentiation.

High Dimensional Spheres and Cylinders

Exercise 3 The discovery Archimedes was most proud of was that the surface area of the sphere in 3-dimensions was the same as the area of the smallest cylinder surrounding it.

Is the same true in four dimensions? (A four-dimensional cylinder has a sphere’s surface as its “base”, just like a three dimensional cylinder has a circle’s length as its base!)

Solutions

Trigonometric Identites: Freya

The 5-Dimensional Sphere: Katie

We are asked to calculate the volume of the 5D unit sphere. The surface of the unit sphere in 5-Dimensions is represented by the set of points where \(x^2+y^2+z^2+w^2+u^2=1\). We can calculate the volume or our sphere by integrating cross sections across \(u\). To do this we must find the radii of these cross sections.

The radius of a slice at a given height.s

Taking cross section at height \(u\), we can find its radius utilizing the Pythagorean theorem. We see that \(r = \sqrt{1-u^2}\). Therefore, \(x^2+y^2+z^2+w^2=\sqrt{1-u^2}\) represent our cross sections at height u. But this is the formula for a 4D sphere. So cross sections of our 5D sphere are 4D spheres! Recall that \(\frac{\pi^2}{2}r^4\) to be the volume of the 4D unit sphere, so the volume of our cross section at height \(u\) is: \[\frac{\pi^2}{2}r^4= \frac{\pi^2}{2}(\sqrt{1-u^2})^4=\frac{\pi^2}{2}(1-u^2)^2\].

Therefore, the volume of our 5D unit sphere is given by:

\[\begin{align*} \int_{-1}^{1}\frac{\pi^2}{2}(1-u^2)^2du &= \frac{\pi^2}{2}\int_{-1}^{1}1-2u^2+u^4du\\ &=\frac{\pi^2}{2}[u-\frac{2}{3}u^3+\frac{1}{5}u^5]^{1}_{-1}\\ &=\frac{\pi^2}{2}[1-\frac{2}{3}+\frac{1}{5}-(1+\frac{2}{3}-\frac{1}{5})]\\ &=\frac{\pi^2}{2}(2-\frac{4}{3}+\frac{2}{5})\\ &=\frac{\pi^2}{2}(\frac{30}{15}-\frac{20}{15}+\frac{6}{15})\\ &=\frac{\pi^2}{2}\cdot\frac{16}{15}\\ &=\frac{8\pi^2}{15} \end{align*}\]

Therefore, we have shown that the volume of the 5D unit sphere is \(\frac{8\pi^2}{15}\)!