I am working on the following exercise:
Calculate the surface integral
$$\iint_{\partial B} (x^2-xy) \ dy \wedge dz + (1-3y+z) \ dz \wedge dx + (e^x-xz) \ dx \wedge dy$$
where $\partial B$ is the surface of the paraboloid $z=x^2+y^2, z \le 4$.
I do not understand that $\wedge$ means the wedge product, but I do not get what $dx,dy$ and $dz$ should mean in this context. Could you help me?
Your orientable surface $\partial B$ consists of small orientable patches $d\partial B$. Each patch can be seen as a small flat area. The value $dx\wedge dy$ of this area can be thought as an area of projection to the plane $xy$ (with negative sign if the patch and axis $z$ look in opposite direction).
In practice instead of integrating bivector $\pmb v=a\,dy\wedge dz+b\,dz\wedge dx+c\,dx\wedge dy$ over the surface $\partial B$, one is usually integrating the flow of dual vector $\pmb u=a\,dx+b\,dy+c\,dz$ through the surface $\partial B$ by calculating the normal $\pmb n=(n_x,n_y,n_z)$:
$$ \iint_{\partial B}\pmb u\cdot\pmb n\, dA $$
But you can also calculate the integral directly. Let's say that the orientation of the paraboloid coincides with orientation of $dx\wedge dy$. Then, since $z=x^2+y^2$, $dz=2x\,dx+2y\,dy$, so $$ dy\wedge dz = 2x\,dy\wedge dx+2y\,dy\wedge dy=-2x\,dx\wedge dy,\\ dz\wedge dx = 2x\,dx\wedge dx+2y\,dy\wedge dx=-2y\,dx\wedge dy. $$ Here we used the antisymmetric property of wedge product: $dx\wedge dy=-dy\wedge dx$ and $dx\wedge dx=0$.
The integral becomes: $$ \iint_{\partial B}(x^2-xy)(-2x)\,dx\wedge dy + (1-3y+z)(-2y)\,dx\wedge dy + (e^x-xz)\,dx\wedge dy = \\ \iint_{\partial B}\left(-2x^3+2x^2y-2y+6y^2-2yz+e^x-xz\right)\,dx\wedge dy $$
By substituting $z=x^2+y^2$, we finally transform the integral over the surface $\partial B$ to the integral over the projection on the plane $xy$ which is the circle $x^2+y^2\le 4$: $$ \iint_{x^2+y^2\le 4} \left(-2x^3+2x^2y-2y+6y^2-2y(x^2+y^2)+e^x-x(x^2+y^2)\right)\,dx\,dy $$