$$\require{cancel}$$

# Exercise 1.1.1

Solution not available yet.

# Exercise 1.2.1

\begin{equation*} \begin{aligned} dx' &= \cos\theta dx - \sin\theta dy \\ dy' &= \sin\theta dx + \cos\theta dy \\ dz' &= dz \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} (dx')^2 &= \cos^2\theta dx^2 - 2\cos\theta\sin\theta dxdy + \sin^2\theta dy^2 \\ (dy')^2 &= \sin^2\theta dx^2 + 2\cos\theta\sin\theta dxdy + \cos^2\theta dy^2 \\ (dz')^2 &= dz^2 \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} (d\ell')^2 &= dx^2 + dy^2 + dz^2 \\ d\ell' &= d\ell \end{aligned} \end{equation*}

# Exercise 1.3.1

\begin{equation*} \begin{aligned} dx &= \cos\theta dr - r\sin\theta d\theta \\ dy &= \sin\theta dr + r\cos\theta d\theta \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} dx^2 &= \cos^2\theta dr^2 - 2r\cos\theta\sin\theta drd\theta + r^2sin^2\theta d\theta^2 \\ dy^2 &= \sin^2\theta dr^2 + 2r\cos\theta\sin\theta drd\theta + r^2\cos^2\theta d\theta^2 \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} d\ell^2 &= dx^2 + dy^2 \\ d\ell^2 &= dr^2 + r^2d\theta^2 \end{aligned} \end{equation*}

# Exercise 1.4.1

Using $$d\ell^2 = dx^2 + dy^2 + dz^2$$ we get

\begin{equation*} \begin{aligned} d\ell^2 = r_1^2sin^2\lambda d\lambda^2 + r_1^2cos^2\lambda d\lambda^2 \end{aligned} \end{equation*}

which simplifies to

\begin{equation*} \begin{aligned} d\ell^2 = r_1^2 d\lambda^2 \quad \rightarrow \quad d\ell = r_1 d\lambda \end{aligned} \end{equation*}

Now integrate both sides, with $$\lambda$$ from $$0$$ to $$2\pi$$ (one period)

\begin{equation*} \begin{aligned} \ell = \int_{0}^{2\pi} r_1 d\lambda \quad \rightarrow \quad \ell = 2\pi r_1 \end{aligned} \end{equation*}

This is the circles circumference.

# Exercise 1.4.2

To Cartesian coordinates to Spherical:

\begin{equation*} \begin{aligned} x &= rcos\phi sin\theta \\ y &= rsin\phi sin\theta \\ z &= rcos\theta \end{aligned} \end{equation*}

substituting in $$\phi = \lambda$$, $$\theta = \pi/2$$, and $$r = r_1$$, we get

\begin{equation*} \begin{aligned} x &= r_1cos\lambda \\ y &= r_1sin\lambda \\ z &= 0 \end{aligned} \end{equation*}

which is indeed the same path given in Exercise 1.3.1.

# Exercise 1.4.3

Substituting in the above to Equation 1.7 we get

\begin{equation*} \begin{aligned} d\ell^2 = dr^2 + r_1^2(d\theta^2 + sin^2(\frac{\pi}{2})d\lambda^2) \end{aligned} \end{equation*}

and since $$r$$ and $$\lambda$$ are constants the whole thing simplifies to

\begin{equation*} \begin{aligned} d\ell^2 = r_1^2 d\lambda^2 \quad \rightarrow \quad d\ell = r_1 d\lambda \end{aligned} \end{equation*}

Now integrate both sides, again with $$\lambda$$ from $$0$$ to $$2\pi$$

\begin{equation*} \begin{aligned} \ell = \int_{0}^{2\pi} r_1 d\lambda \quad \rightarrow \quad \ell = 2\pi r_1 \end{aligned} \end{equation*}

This is the same result as we got in Exercise 1.3.1.

# Exercise 1.4.4

Substituting in the above to Equation 1.7 we get

\begin{equation*} \begin{aligned} d\ell^2 = d\mu^2 + \mu^2(d\theta^2 + sin^2(\frac{\pi}{2})d\lambda^2) \end{aligned} \end{equation*}

and since $$\theta$$ and $$\lambda$$ are fixed the whole thing simplifies to

\begin{equation*} \begin{aligned} d\ell^2 = d\mu^2 \quad \rightarrow \quad d\ell = d\mu \end{aligned} \end{equation*}

It's now easy to see that $$\int_0^{r_1} (d\ell/d\mu) d\mu$$ does indeed equal $$r_1$$. This is the radius.