18.10. EXERCISES 497

for some c a positive constant. (This is the cross product from calculus. Reviewthis.), use the area formula to describe the boundary integral from the divergencetheorem as an integral over [a,b]. We write this integral as

∫∂U Pdx+Qdy. This

yields a version of Green’s theorem,∫

U Qx−Pydm2 =∫

∂U Pdx+Qdy provided ∂Uis a Lipschitz curve oriented as just described. However, more generality is possible,although I am not sure how far this has been generalized. You really only need tohave the boundary of U be a rectifiable simple closed curve meaning it has finitelength. In this setting, the Jordan curve theorem makes it possible to make sense ofGreen’s theorem and in fact it holds. In what was just discussed, Ū was the Lipschitzimage of some rectangle. The general version only requires that the boundary of Ube the image of the unit circle. Nevertheless, this version in this problem is prettygood.

13. Let U be a bounded open set for which Green’s theorem holds. Let C be the orientedboundary consistent with Green’s theorem. Show: area of U =

∫C xdy .

14. Let U be an open set with which is on one side of its boundary as above with theboundary being the image of a Lipschitz map which is one to one. (Note this condi-tion eliminates the curve crossing itself.) Now suppose you have a closed polygonalcurve going from (x0,y0)→ (x1,y1)→ (x2,y2) · · ·(xp,yp) = (x0,y0) oriented suchthat the direction of motion is in the direction k×n where n is the unit outer normalfrom the divergence theorem. Show that if U is this enclosed polygon, then

Area of U =12

n

∑k=1

(xk + xk−1)(yk− yk−1)

This is a pretty remarkable result. Just draw a few such polygons and think how youwould find their area without it.

15. Orient the u,v axes just like the usual arrangement of the x and y axes, u axis like thex axis. Let r : U → R3 where U is an open subset of R2. Suppose r is C2 and letF be a C1 vector field defined in V, an open set containing r (U). Show, using theabove reduction identity of Problem 8 that

(ru×rv) · (∇×F )(r (u,v)) = ((F ◦r)u ·rv− (F ◦r)v ·ru)(u,v) . (18.15)

The left side is the dot product of the curl of the vector field F with a normal vectorto the surface r (U) , namely ru × rv. Show that the right side can be written as((F ◦r) ·rv)u− ((F ◦r) ·ru)v thanks to equality of mixed partial derivatives. Nowsuppose U is a region for which Green’s theorem holds, the curve C bounding Ubeing Lipschitz. Verify Stokes’ theorem∫

U(ru×rv) · (∇×F )(r (u,v))dudv =

∫R(F ◦r) ·rudu+(F ◦r) ·rvdv

whereR(t) = (u(t) ,v(t)) for t→ (u(t) ,v(t)) being a parametrization of C orientedso that (u′ (t) ,v′ (t) ,0)×k is an exterior normal to U . Show the left integral can bewritten as

∫r(U) ∇×F ·NdH 2 whereN is a unit normal to the surface r (U). Show

the integral on the right can be written as the differential form∫R F1dx+F2dy+F3dz

whereR(t) = r (u(t) ,v(t)).