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Green's theorem states, for a positively oriented simple boundary $C=\partial R$,The defining polynomial becomesso the region is the cardioid

The parametrisation $r(x,y)=(x,y,F(x,y))$ has area factorso $\mathrm{area}(S)={\iint }_D\sqrt{1+F_x^2+F_y^2}dxdy$.

For the unbounded saddle $F=(x^2-y^2)/2$, this factor is $\sqrt{1+x^2+y^2}$. Thus the requested integral iswhich converges because its radial tail is $O(r^{-2})$.

The centre isEvery subgroup $H\le Z(G)$ is normal, since $gh{g}^{-1}=h$ for all $g\in G$ and $h\in H$.

Write $D_{2n}=⟨r,s:r^n=s^2=1,srs=r^{-1}⟩$. A central rotation must satisfy $r^k=r^{-k}$, and no reflection is central for $n\ge 3$. Hence

The product rule givesIntegrating and applying the divergence theorem proves the identity.

Here $\mathrm{div}T=4(x,y,z)$ and $v=x(x,y,z)$, soOn the boundary, $(Tn)\cdot v=x(x\cdot n)(x^2+y^2+z^2-1)$. The nonzero contributions from the faces $x=1,y=1,z=1$ are respectively $2/3,5/12,5/12$, totaling $3/2$.

Finally, direct contraction giveswhose integral is $-5/6$. Thus the right side is $3/2-(-5/6)=7/3$, equal to the left side.

For coordinates $x=x(q)$, the Jacobian is $J_{ij}=\partial x_i/\partial q_j$. Expanding after separating $x_1=r\cos {\theta }_1$ from the remaining coordinates givesConsequentlyand on $r=R$ the surface element is obtained by omitting $dr$ and replacing $r$ by $R$.