Problem with bounds on surface integral.

Question

Let $$S=\{(x,y,z)\in\mathbb{R}^3:z=xy\},$$ and consider the $1$-form of $\mathbb{R}^3$ given by $$\omega=(y^3+xz)dx-x^2dz.$$ I'm trying to compute the integral $\int_C \omega,$ where $C=S\cap \{x^2+y^2=1\}.$ Since $S=f^{-1}(\{0\})$, where $f(x,y,z)=xy-z$, it's a regular submanifold of $\mathbb{R}^3$ so we can just consider the restriction of $\omega$ to $S$ (with pullback of the inmersion). I first computed the differential of the form, getting $$d\omega=3y^2dy\wedge dx+xdz\wedge dx-2x dx\wedge dz=-3y^2dx\wedge dy-3x dx\wedge dz,$$ and since $z=xy$, $dz=ydx+xdy$, so $$d\omega=-3y^2dx\wedge dy-3x^2dx\wedge dy=-3dx\wedge dy,$$ because $x^2+y^2=1$. Now by Stoke's theorem we have $$\int_C\omega=\int_M d\omega=-3\cdot \operatorname{Surface}(M),$$ where $M$ is some surface such as $C=\partial M$. Now i'm having troubles to visualize which manifold is exactly $M$ and how to compute its surface to apply Stoke's Theorem.

Answer 1

A correct candidate of $M$ would be the surface parametrised by: $$ (r,\phi)\in [0,1)\times[0,2\pi)\to\begin{pmatrix} r\cos\phi \\ r\sin\phi \\ r^2\cos\phi\sin\phi \end{pmatrix} $$ (Bruno B only gave a parametrisation of $C$). You cannot use $x^2+y^2=1$ in calculating $d\omega$ since you'll be integrating in $M$ where $x^2+y^2<1$. If your calculation of $d\omega$ were correct, then yes the RHS would have been the area of its projection in the $xy$ plane, i.e. a unit disk. Starting from your penultimate expression for $d\omega$: $$ d\omega=-3(x^2+y^2)dx\wedge dy $$ I get: $$ \int_M d\omega = \int_{x^2+y^2<1} -3(x^2+y^2)dxdy $$ To actually compute it, the easiest way is to go to polar coordinates where I use: $$ \begin{align} x^2+y^2 &=r^2 & dx\wedge dy&= rdr\wedge d\phi \end{align} $$ (last equation being the Jacobian of polar coordinates), I get: $$ \begin{align} \int_M d\omega &= \int_0^{2\pi}\int_0^1 -3r^3drd\phi \\ &= -\frac{3\pi}{2} \end{align} $$

You can check the consistency with Stokes' formula. Using Bruno B's parametrisation of $C$: $$ \gamma:\phi\in[0,2\pi)\to\begin{pmatrix} \cos\phi \\ \sin\phi \\ \cos\phi\sin\phi \end{pmatrix} $$ and the expression of $\omega$ on $C$: $$ \omega = y^3dx-x^3dy $$ you need to calculate: $$ \int_C \omega = -\int_0^{2\pi}(\sin^4\phi+\cos^4\phi)d\phi $$ which is straightforward using: $$ \begin{align} \cos^4\phi &= \frac{\cos(4\phi)+4\cos(2\phi)+3}{8} \\ \sin^4\phi &= \frac{\cos(4\phi)-4\cos(2\phi)+3}{8} \end{align} $$ so: $$ \begin{align} \int_C \omega &= -\left[\frac{\sin(4\phi)+4\sin(2\phi)+3\phi}{8}+\frac{\sin(4\phi)-4\sin(2\phi)+3\phi}{8}\right]_0^{2\pi} \\ &= -\frac{3\pi}{2} \end{align} $$

Hope this helps.

Answer 2

Just a small remark if permitted, since I was puzzled by LPZ's elegant way of solving this.

The correct calculation of $d\omega$ is $$\tag{1} d\omega=3y^2\,dy\wedge dx+x\,dz\wedge dx-2x\,dx\wedge dz=-3y^2\,dx\wedge dy+3x\,dz\wedge dx\,. $$ Using $z=xy$ it is correct to write $dz=x\,dy+y\,dx$ and therefore $$\tag{2} d\omega=-3y^2\,dx\wedge dy-3x^2\,dx\wedge dy=-3(x^2+y^2)\,dx\wedge dy\,. $$ It is however $\color{red}{incorrect}$ to use here the constraint for the curve $x^2+y^2=1$ to write the form on the surface as $\color{red}{d\omega=-3\,dx\wedge dy}\,.$

To integrate $d\omega$ over $M$ with LPZ's parametrization we use \begin{align} dx&=\cos\phi\,dr-r\sin\phi\,d\phi\,,\\ dy&=\sin\phi\,dr+r\cos\phi\,d\phi\,,\\ dz&=2r\cos\phi\sin\phi\,dr-r^2\sin^2\phi\,d\phi+r^2\cos\phi\,d\phi \end{align} which gives \begin{align}\require{cancel} d\omega&=3r^3\sin^4\phi\,d\phi\wedge dr-\cancel{3r^3\sin^2\phi\cos^2\phi\,dr\wedge d\phi}\\ &~~~-6r^3\cos^2\phi\sin^2\phi\,dr\wedge d\phi-\cancel{3r^3\cos^2\phi\sin^2\phi\,d\phi\wedge dr}\\[2mm] &~~~+3r^3\cos^4\phi\,d\phi\wedge dr\\[2mm] &=-3r^3(\sin^2\phi+\cos^2\phi)^2\,dr\wedge d\phi\\[2mm] &=-3r^3\,dr\wedge d\phi\, \end{align} when we use (1). When we use (2) we see this much more quickly from $$ dx\wedge dy=r(\cos^2\phi+\sin^2\phi)\,dr\wedge d\phi=r\,dr\wedge d\phi\,. $$