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CALCULUS 
C being the boundary of an arbitrary region 2 lying in S, then 
dP = dQ 
dy ox 
at every point of S. 
Suppose the theorem false. Then the continuous function 
dP_dQ 
dy dx 
must be different from 0 at some interior point A of S, and hence 
must be either positive at every point of a suitably chosen neighbor 
hood of A (say, throughout the interior of a small circle about A, 
lying wholly in S) or else negative throughout such a region. But 
then the double integral of ((7), § 4, could not vanish when extended 
over this region; and since the line integral which forms the right- 
hand side of (0) vanishes by hypothesis, we are led to a contradic 
tion. Hence the theorem is established. 
EXERCISES 
1. Prove by an example that the following theorem is false : Let 
P and Q be two functions which satisfy the conditions of Theorem 1 ; 
and let C be a simple closed curve lying in S. Then 
Pdx 4- Qcly — 0. 
2. Let S be a ring-shaped region bounded by the curves C x and 
C 2 , and let P and Q satisfy the conditions of Theorem 1 in S. Then 
j Pdx + Q dy 
Pdx -f Q dy, 
where each integral is extended in the clock-wise sense over C x 
or 0 2 . 
6. Simply and Multiply Connected Regions. By a simply connected 
region is meant a region such that no closed curve drawn in the 
region contains in its interior a boundary point of the region. All 
other regions are called multiply connected. 
Thus a square or an ellipse is simply connected; more generally, 
the interior of any simple closed curve, together with the boundary, 
forms a simply connected region. It is not necessary that the 
region be finite. The whole plane, or a half-plane, or the region 
bounded by two rays which emanate from a point, or the (x, y)-plane