Find the value of the 2-form $dxdy+3dxdz$ on the oriented triangle with $(0,0,0)$ $(1,2,3)$ $(1,4,0)$ in that order.
I have tried various subtractions and plugging in values and have been unable to figure out how to solve this type of problem. As far as I can tell the method of solution is not described in the book thus far. I am trying to work through, Edwards, 'Advanced Calculus: A Differential Forms Approach'. The answer is in the back of the book and is a negative rational number, assuming no mistakes.
If someone could please show me how to get started on this problem I would appreciate it since my efforts are in vain.
There is no need to integrate here your differential forms are constant and your region is a paralelotope, so what you need to do is to evaluate $$ \mathrm{d}x\wedge \mathrm{d}y(\mathbf{e}_1,\mathbf{e}_2)+3\mathrm{d}x\wedge\mathrm{d}z(\mathbf{e}_1,\mathbf{e}_2), $$ where $$ \mathbf{e}_1=(1,2,3),\ \mathbf{e}_2=(1,4,0). $$
By the property of simple 2-forms, this is $$ \begin{vmatrix} \mathrm{d}x(\mathbf{e}_1) & \mathrm{d}x(\mathbf{e}_2) \\ \mathrm{d}y(\mathbf{e}_1) & \mathrm{d}y(\mathbf{e}_2)\end{vmatrix} +3\begin{vmatrix} \mathrm{d}x(\mathbf{e}_1) & \mathrm{d}x(\mathbf{e}_2) \\ \mathrm{d}z(\mathbf{e}_1) & \mathrm{d}z(\mathbf{e}_2)\end{vmatrix}=1\cdot4-1\cdot2+3(0-3)=-7, $$ but this is the value of your form on the paralelogram, not the triangle, so we divide this by $2$, and the answer is $-7/2$.
I also integrated your 2-form and it got me the same result, but this is much simpler this way.
EDIT: I just checked your book a bit, it even says that it will describe how to evaluate two-forms one section later. You should read on and do the excercises later imo.
But, basically, the definition of the $\wedge$ "wedge" or "exterior" product is if $\omega$ is a $k$-form and $\mu$ is an $l$ form then $$ \omega\wedge\mu=\frac{(k+l)!}{k!l!}\mathrm{Alt}(\omega\otimes\mu). $$ Couple of notes: There are other conventions for the combinatorical factors, but this is the "most geometrical" one. By $\mathrm{Alt}$ I mean the following map: If $T$ is a purely covariant $k$-th order tensor, then $$ \mathrm{Alt}T(\mathbf{x}_1,...,\mathbf{x}_k)=\frac{1}{k!}\sum_{\pi\in S_k}\mathrm{sgn}(\pi)T(\mathbf{x}_{\pi(1)},...,\mathbf{x}_{\pi(k)}), $$ where the sum happens over all permutations. It canbe shown through laborous manipulation of permutations, that the exterior product is associative and my above definition holds for any number of factors, and that if $\omega_1,...,\omega_k$ are 1-forms, and $\mathbf{x}_1,...,x_k$ are vectors, then because all of my forms are 1-forms, all combinatorical factors disappear, and $$ \omega_1\wedge...\wedge\omega_k(\mathbf{x}_1,...,\mathbf{x}_k)=\sum_{\pi\in S_k}\mathrm{sgn}(\pi)\omega_1(\mathbf{x}_{\pi(1)})...\omega_k(\mathbf{x}_{\pi(k)}), $$ which is exactly the definition of a determinant, whose $i,j$-th element is $\omega_i(\mathbf{x}_j)$.
The reason I ignored the $(0,0,0)$ vector is that 2-forms operate on paralelograms subtended by vectors, half of which are triangles. In this case, since one of the vertices are the zero vector, the sides of the triangle are $(1,2,3)$ and $(1,4,0)$, so this is what we plug in to the 2-form.