Let \[ T=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1,\ x,y,z\ge 0\} \] be the spherical triangle, and let \(\alpha=z\,dx\wedge dz\).
Find a smooth \(1\)-form \(\beta\) on \(\mathbb R^3\) such that \(\alpha=d\beta\).
Define consistent orientations for \(T\) and \(\partial T\).
Using those orientations, compute \(\int_T\alpha\) and \(\int_{\partial T}\beta\) directly and check that they agree. Explain why they should agree.
Solution
1. The smooth \(1\)-form\[ \beta=xz\,dz \]satisfies \(d\beta=\alpha\) on \(\mathbb R^3\).
Since \(d(dz)=0\), \[ d\beta=d(xz)\wedge dz=(z\,dx+x\,dz)\wedge dz=z\,dx\wedge dz=\alpha. \]
2. Orient \(T\) by the outward unit normal\[ n=(x,y,z). \]Give \(\partial T\) the induced boundary orientation: if \(\nu\) is the outward-pointing unit conormal to an edge inside \(T\), then a tangent vector \(\tau\) is positive when \((\nu,\tau)\) is a positive basis of \(TT\).
Equivalently, along an edge we require \(\nu\times\tau=n\).
Writing \[ e_1=(1,0,0),\qquad e_2=(0,1,0),\qquad e_3=(0,0,1), \] the three positively oriented edges are therefore \[ e_1\longrightarrow e_2\quad(z=0),\qquad e_2\longrightarrow e_3\quad(x=0),\qquad e_3\longrightarrow e_1\quad(y=0). \] Indeed, on these edges the outward conormals are respectively \(-e_z,-e_x,-e_y\), and the indicated tangent directions satisfy \(\nu\times\tau=n\).
3. With the orientation in 2,\[ \int_T\alpha=-\frac13. \]
On the interior of \(T\), write it as the graph \[ \Phi(x,z)=\bigl(x,\sqrt{1-x^2-z^2},z\bigr) \] over \[ D=\{(x,z):x\ge0,\ z\ge0,\ x^2+z^2\le1\}. \] If \(y=\sqrt{1-x^2-z^2}\), then \[ \Phi_x\times\Phi_z =\left(-\frac{x}{y},-1,-\frac{z}{y}\right) =-\frac1y\,n, \] so the \((x,z)\)-orientation is opposite to the chosen outward orientation. Hence \[ \int_T\alpha =-\int_D z\,dx\,dz =-\int_0^1\int_0^{\sqrt{1-x^2}}z\,dz\,dx =-\frac12\int_0^1(1-x^2)\,dx =-\frac13. \]
4. With the induced boundary orientation in 2,\[ \int_{\partial T}\beta=-\frac13. \]
On the edges \(z=0\) and \(x=0\), the form \(\beta=xz\,dz\) vanishes identically.
The remaining edge is \(y=0\), oriented from \(e_3\) to \(e_1\). Parametrize it by \[ \gamma(t)=(\sin t,0,\cos t),\qquad 0\le t\le\frac\pi2. \] This has precisely the required orientation. Along \(\gamma\), \[ x=\sin t,\qquad z=\cos t,\qquad dz=-\sin t\,dt, \] and therefore \[ \gamma^*\beta =-\sin^2 t\cos t\,dt. \] Thus \[ \int_{\partial T}\beta =-\int_0^{\pi/2}\sin^2t\cos t\,dt =-\frac13. \]
5. The two direct computations agree, as required by Stokes’ theorem.
By 1, \(\alpha=d\beta\), and by 2 the orientations on \(T\) and \(\partial T\) are compatible. Hence Stokes’ theorem gives \[ \int_T\alpha=\int_Td\beta=\int_{\partial T}\beta. \] Steps 3 and 4 compute both sides directly as \(-1/3\).