Surface Integrals of Vector Fields — Question 8

Question 8 – Solution

Strategy. Follow a normal direction once around the band and test whether it returns consistently at the identified seam.

Step 1: Identify the seam Direct substitution gives $$\mathbf r(0,v)=\left\langle 1+\frac v2,0,0\right\rangle =\mathbf r(2\pi,-v).$$ Thus the left edge at height $v$ is glued to the right edge at height $-v$.

Step 2: Compare transverse tangents Along the centerline $v=0$, $$\mathbf r_v(0,0)=\left\langle\frac 12,0,0\right\rangle, \qquad \mathbf r_v(2\pi,0)=\left\langle-\frac 12,0,0\right\rangle.$$ Also $\mathbf r_u(0,0)=\mathbf r_u(2\pi,0)=\langle 0,1,0\rangle$. Hence $$(\mathbf r_u\times\mathbf r_v)(0,0)=\left\langle 0,0,-\frac 12\right\rangle, \quad (\mathbf r_u\times\mathbf r_v)(2\pi,0)=\left\langle 0,0,\frac 12\right\rangle.$$

Step 3: Conclude The two parameter points represent the same surface point, yet continuation around the strip reverses the normal. Therefore no continuous global unit normal exists, and $$\boxed{\text{a global oriented flux integral over the M\"obius band is not defined}.}$$

Verification Cutting the band at the seam permits a local choice of normal, but gluing the seam back forces opposite choices, exposing the obstruction.