# solutions for problems on vector analysis and differential forms

Solutions

Exercise:
Assume that 4 vectors in $$\mathbb{R}^3$$ satisfy $$A+B+C+D$$=0.
Simplify
$$A\times B-B\times C+C \times D-D\times A$$

Solution: By the anti-symmetry $$-B\times C = C\times B$$
and $$-D\times A = A\times D$$ we can rewrite the expression as
$$A\times B+C\times B+C \times D+A\times D$$
Using the distributive law of the cross product, we have
$$(A+C)\times B+(C+A) \times D$$
and again
$$(A+C)\times (B+D) =**$$
However, as $$F=(A+C) = -(B+D)$$ we get
$$**= F\times (-F) = 0$$

Exercise: Integrate $$dx/y$$ over the circle (oriented counterclockwise)
$$C:=\{x^2+y^2=R^2\}$$.

Solution: utilize a counter clockwise parametrization
$$x(t)=R\cos(2\pi t),~~~~~y(t)=R\sin(2\pi t),~~~~ t\in [0,1)$$
The differential becomes
$$\frac{dx}{y} = \frac{ -2\pi R\sin(2\pi t) dt}{R\sin(2\pi t)}= -2\pi dt$$
(except two removable singularities at $$t=0,\frac{1}{2}$$).
The overall integral would then be
$$\int_{C} \frac{dx}{y} = \int_{0}^1 (-2\pi dt) = -2\pi$$

An alternative approach is to integrate over $$x$$ and use the explicit
form of $$y=\pm \sqrt{R^2-x^2}$$ (and account for integration direction):
$$\underbrace{\int_{R}^{-R} \frac{dx}{\sqrt{R^2-x^2}} dx}_{ \begin{array}{l} \text{Upper half integral} \\ \text{Note: right to left direction} \end{array}} + \underbrace{\int_{-R}^{R} \frac{dx}{(-\sqrt{R^2-x^2})} dx}_{ \text{Lowed half integral}} = -2\int_{-R}^{R} \frac{dx}{\sqrt{R^2-x^2}} =**$$
Then, by substitution $$x = -\cos(\pi t)$$ with end-points $$t=0,1$$ we get
$$** = -2\int_{0}^1 \frac{\pi R\sin(\pi t)}{R\sin(\pi t)} dt = -2\pi$$

Exercise: let
$$\omega = dx\wedge dy + ds \wedge dt + du\wedge dv$$. Find $$\omega^3$$.

Solution: This is a $$2$$-form on $$\mathbb{R}^6$$, so $$\omega^3$$
would be a $$6$$-form on $$\mathbb{R}^6$$, hence a multiple of the volume form:
$$\omega^3 = cdx \wedge dy \wedge du \wedge dv \wedge ds \wedge dt$$
So that we just need to find $$c$$.

First compute the $$4$$-form $$\omega^2$$:
\begin{align*} \omega\wedge\omega &= (dx\wedge dy + ds \wedge dt + du\wedge dv )\wedge (dx\wedge dy + ds \wedge dt + du\wedge dv ) \\ = &(dx\wedge dy)\wedge (dx\wedge dy) + (dx\wedge dy)\wedge (ds \wedge dt) + (dx\wedge dy)\wedge (du\wedge dv) \\ & +(ds \wedge dt)\wedge (dx\wedge dy) +(ds \wedge dt)\wedge (ds \wedge dt) +(ds \wedge dt)\wedge (du\wedge dv) \\ & +(du\wedge dv)\wedge (dx\wedge dy) +(du\wedge dv)\wedge (ds \wedge dt) +(du\wedge dv)\wedge (du\wedge dv) \\ \end{align*}
We can get rid of parenthesis since the wedge product is also associative
(in contrast to the cross product ). The diagonal elements vanish
since $$dx\wedge dx = ds\wedge ds = du\wedge du =0$$, and the
off diagonal elements are equal as $$2$$-forms have commutative wedge
$$\alpha^\wedge \beta = (-1)^4 \beta\wedge \alpha$$
so that
\begin{align*} \omega^2 &= 2\left[ (dx\wedge dy)\wedge (ds \wedge dt) + (dx\wedge dy)\wedge (du\wedge dv) + (ds \wedge dt)\wedge (du\wedge dv) \right] \end{align*}

Then by the same rules we get
\begin{align*} \omega^3 = \omega \wedge \omega^2 &= 2\left[ dx\wedge dy + ds \wedge dt + du\wedge dv\right] \wedge \left[(dx\wedge dy)\wedge (ds \wedge dt) + (dx\wedge dy)\wedge (du\wedge dv) + (ds \wedge dt)\wedge (du\wedge dv) \right] \\ &= 2 \left[(dx\wedge dy)\wedge (du \wedge dv) \wedge (ds\wedge dt) + (ds\wedge dt)\wedge(dx\wedge dy)\wedge (du \wedge dv) + (du \wedge dv) \wedge (dx\wedge dy)\wedge (ds\wedge dt) \right] \\ & = 6 dx\wedge dy\wedge du \wedge dv \wedge ds\wedge dt \end{align*}