Let $\omega = x^2 dz \wedge dy + (y+z)(dx \wedge dy + dx \wedge dz)$ be a 2-form on $\mathbb R^3$. Let $S$ be the surface of rotation around the x axis of the function $f(t)=1+t^2$, $0 < t < 1$. Calculate $\int_S\omega$.
The surface $S$ is parametrized by $\varphi: [0, 1]\times[0, 2\pi] \to \mathbb R^3$ $\\ \varphi(t, \theta) = (t, (1+t^2)\cos\theta,(1+t^2)\sin\theta)\\$
The Jacobian is \begin{bmatrix}1&2t\cos\theta&2t\sin\theta\\0&-(1+t^2)\sin\theta&(1+t^2)\cos\theta \end{bmatrix}
$dx\wedge dz \to (1+t^2)\cos\theta dtd\theta\\ dx\wedge dy \to -(1+t^2)\sin\theta dtd\theta \\dy\wedge dz = 2t(1+t^2)dtd\theta$
So I simply transformed $\int_S\omega =\\ \int_0^1\int_0^{2\pi} -t^2(1+t^2)2t dtd\theta \\+\int_0^1\int_0^{2\pi}(1+t^2)^2(\cos^2\theta - \sin^2\theta)dtd\theta$
The last term goes away as $\int_0^{2\pi} \cos^\theta - \sin^2\theta d\theta = 0$. The firs term simply gives
$-4\pi\int_0^1t^3+t^5dt = -4\pi(1/4+1/6) = -\pi(1+2/3) = -\pi\frac{5}{3}$