Differential form surface integral

[vickyr]

Question:
Evaluate the surface integral

$$J = 2xzdy \land dz+2yzdz \land dx-{z}^{2}dx \land dy$$

where \(\displaystyle S \subset {\Bbb{R}}^{3}\) is the rectangle parametrised by:

$$x(u,v) = 1-u,\ y(u,v) = u,\ z(u,v) = v,\ \ 0\le u, v \le 1$$

so far I have:
\begin{array}{}x = u\cos v, &dx = \cos v\, du - u\sin v\, du \\
y = u\sin v, &dy = u\cos v\, dv + \sin v\, du \\
z = v, &dz = dv\end{array}

The next part requires me to find
\begin{array}{}\d{x}{u} du; &\d{x}{v} dv \\
\d{y}{u} du; &\d{y}{v} dv \\
\d{z}{u} du; &\d{z}{v} dv\end{array}

but I have no idea how to do this, please help!

 

[jvicens]

it is important to have a solid understanding of mathematical concepts and techniques. Let's break down the problem and approach it step by step.

First, we have the surface integral:

$$J = 2xzdy \land dz+2yzdz \land dx-{z}^{2}dx \land dy$$

This integral is a bit different from the usual ones we encounter, as it includes wedge products (represented by the symbol $\land$). These products represent the cross product of two vectors, which in this case are the differentials $dy$, $dz$, and $dx$. To evaluate this integral, we need to express the vectors in terms of the parametric variables $u$ and $v$.

Using the parametric equations given in the problem, we can express the differentials as:

\begin{array}{}\d{x}{u} du = -du, &\d{x}{v} dv = \cos v\, dv \\
\d{y}{u} du = du, &\d{y}{v} dv = \sin v\, dv \\
\d{z}{u} du = 0, &\d{z}{v} dv = dv\end{array}

Note that the partial derivatives with respect to $u$ and $v$ are simply the coefficients of $du$ and $dv$, respectively.

Now, let's substitute these expressions into the original integral:

$$J = 2(1-u)v(-du)\land dv+2(uv)(dv)\land (\cos v\, dv)-v^2(\cos v\, dv)\land (\sin v\, dv)$$

Simplifying this, we get:

$$J = -2vdu \land dv + 2uv\cos v\, dv \land dv - v^2\cos v\sin v\, dv \land dv$$

Next, we need to evaluate the wedge products. The wedge product of two differentials $du$ and $dv$ can be computed using the following formula:

$$du \land dv = \d{x}{u}\d{y}{v}\, du \land dv + \d{x}{v}\d{y}{u}\, dv \land du$$

Substituting the expressions we found earlier for the partial derivatives, we get:

$$du \land dv = (-1)(1)(-du