$f, g : \mathbb{R}^2 \to \mathbb{R}$ are functions of class $C^1$. We describe a differential 2-form:
$$\eta = f(y, z) dy \wedge dz \ + \ g(x, z) dz \wedge dx \ + \ z^2 (x^2 + y^2) dx \wedge dy$$
Calculate $\int_M \eta$, where M is a part of paraboloid:
$$M = \{(x, y, z) \in \mathbb{R}^3 \ : \ z = x^2 + y^2 < 1\}$$
oriented in such way that in every point $p \in M$ the third component of the vector normal to M consistent with the orientation is negative.
Solution:
I can see that the set $M$ defined by $x^2 + y^2 < 1$, $0 < z < 1$, is kind of a bowl as presented below:
Image may be NSFW.
Clik here to view.
Red colour means: $z = x^2 + y^2$
Mint colour means: $z = 1$
I get that:
$$d \eta = d \left( f(y, z) dy \wedge dz \ + \ g(x, z) dz \wedge dx \ + \ z^2 (x^2 + y^2) dx \wedge dy \right)$$
$$d \eta = d \left( f(y, z) dy \wedge dz \right) \ + \ d \left( \ g(x, z) dz \wedge dx \right) \ + \ d \left( z^2 (x^2 + y^2) dx \wedge dy \right)$$
$$d \eta = \frac{z}{2} (x^2 + y^2) dz \wedge dx \wedge dy = \frac{z}{2} (x^2 + y^2) dx \wedge dy \wedge dz$$
We look for a set for which it is true that: $\partial G = M$. It looks like that set is equal to:
$$G = \{(x, y, z) \in \mathbb{R}^3 \ : \ x^2 + y^2 \leq z, \ 0 < z < 1 \}$$
Therefore, form Stokes theorem, we can say that:
$$\int_{M} \omega = \int_{G} d \omega = \int_{G} \frac{z}{2} (x^2 + y^2) \ dx dy dz$$
To solve such integral we can use such parametrization:
$$x = r cos \alpha, y = r sin \alpha, z = h $$
using those, our set is described by: $\{ (r, \alpha, h) : r \in [0,1], \alpha \in [−π, π], h \in (0, 1) \}$
Jacobian of such parametrization is equal to $r$, so we can write that:
$$\int_{G} \frac{z}{2} (x^2 + y^2) \ dx dy dz = \int_0^1 \int_0^1 \int_{- \pi}^{\pi} \frac{h}{2} \left( (r cos \alpha)^2 + (r sin \alpha)^2 \right) \ r \ d \alpha \ dr \ dh = $$
$$= \frac{1}{2} \int_0^1 \int_0^1 \int_{- \pi}^{\pi} h \ r^3 \ d \alpha \ dr \ dh = \frac{1}{2} \int_0^1 \int_0^1 2 \pi \ h \ r^3 \ dr \ dh $$
$$= \frac{1}{2} \int_0^1 2 \pi \ h \ \frac{1}{4} \ dh = \frac{1}{2} 2 \pi \ \frac{1}{2} \ \frac{1}{4} = \frac{\pi}{8}$$
Is that correct now?