# Verifying Stokes Theorem

The first part of the question states prove,

$$\oint_{c} r \wedge dr= 2 \iint_{s} ds$$

This I have done using stokes theorem, however stuck on the next part which states, verify this forumla is also applicable when s has a mild singularity, by evaluating both sides of the equation for the case when s is the cone,

${(x,y,z): (1-z)^{2}=x^{2}+y^{2}, x^{2}+y^{2} \leq 1, z\geq0}$ and c is the circle $x^{2}+y^{2}=1$ in the xy plane.

Consider the vector field ${\bf F}(x,y,z):=(-y, x, 0)$, a force field, in threespace. Then $$\int_\gamma {\bf r}\wedge d{\bf r}=\int_\gamma (x,y)\wedge (dx,dy)=\int_\gamma (x\,dy -y\,dx)=\int_\gamma {\bf F}(x,y,0)\cdot{\rm d}{\bf r}$$ for any curve $\gamma$ in the $(x,y)$-plane; whence the left side of your equation can be interpreted as work done along $\gamma$. If $\gamma=\partial S$ for a correctly oriented surface in threespace then by Stokes theorem $$\int_\gamma {\bf F}(x,y,0)\cdot{\rm d}{\bf r}=\int_S {\bf rot F}(x,y,z)\cdot {\rm d}{\pmb\omega}=2\int_S (0,0,1)\cdot {\rm d}{\pmb\omega}\ ,$$ where ${\rm d}{\pmb\omega}$ denotes the oriented surface element on $S$.
In your first example $S$ is the unit disk in the $(x,y)$-plane and $\gamma=\partial S$ the unit circle. Therefore we have ${\rm d}{\pmb\omega}=(0,0,1){\rm d}\omega$ and obtain $$\int_\gamma {\bf r}\wedge d{\bf r}=\ldots =2\int_S {\rm d}\omega=2\pi\ .$$ For the second example we have to parametrize the given cone, e.g., by $$(u,\phi)\mapsto {\bf r}=(x,y,z):=(u\cos\phi,u\sin\phi, 1-u)$$ with parameter domain $B$ given by $0\leq u\leq 1$ and $0\leq\phi\leq 2\pi$. To check Stokes' formula for this case you have to compute $${\rm d}{\pmb\omega}= {\bf r}_u\wedge {\bf r}_\phi\ \ {\rm d(u,\phi)}$$ and to verify that the third component of ${\rm d}{\pmb\omega}$ is indeed $>0$. If you now compute $$2\int_S (0,0,1)\cdot {\rm d}{\pmb\omega}=\int_B \ldots\ {\rm d}(u,\phi)$$ you should again obtain $2\pi$.