I need help in the following problem (exercise 3.13 Do Carmo- Differential Forms and Applications):
Let $\omega$ be differential 1-form given by $\omega=xdx+ydy+zdz$, and let $P$ be the field of planes in $R^3-{0}$ determined by $\omega$. Show that the integral surface of $P$ passing through $p=(x, y, z)$ is the sphere with center in the origin and passing through $p$.
Where differential 1-form is defined as follows:
Since the answer is given in the question, you can do the converse (which is easier) and check that the sphere is indeed an integral surface of the plane field. You can do this using implicit differentiation on the Cartesian equation of the sphere.
One way you could have seen this coming is to introduce a metric, which using duality, allows you to identify a $1-$form and a vector, which would point normal to the hyperplane. In this case, the dual vector field is simply the position vector, so you're trying to find a surface whose tangent planes are normal to the position vector from the origin.
In general, if you're trying to find the equation of the surface $z=f(x,y)$, this amounts to solving this system of partial differential equations: $$ f\partial_xf= -x $$ $$ f\partial_yf= -y $$
Hope this helps