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The Theorem of George Green and its Proof

George Green (1793-1841) is somewhat of an anomaly in mathematics. He was the son of a baker/miller in a rural area. He had only one year of formal education. For the rest he was self-taught, yet he discovered major elements of mathematical physics. One of those elements was a theorem, now known as Green's Theorem.

Let P(x,y) and Q(x,y) be differentiable functions defined over a region S. Let C be the piecewise smooth boundary of S.

The curve C is traversed in a counterclockwise direction so that the region S is always to the left of the direction of travel.


C(Pdx + Qdy) = ∫S∫(∂Q/∂x − ∂P/∂y)dS

The proof will be carried out using successively more general condition on the boundary C. First it is assumed that the region S has no holes. Second, it is assumed that C is vertically simple and horizontally simple.

Vertically simple means that the lower edge of C can be expressed as a single valued function of x; i.e., y=f1(x). Call this the curve C1. Likewise the upper edge of C can be expressed a single valued function of x; i.e., y=f2(x). This curve will be called C2. The function-curve C1 runs from (x1 , y1) to (x2, y2), but C2 runs in the reverse direction from (x2, y2) to (x1, y1).


Similarly horizontally simple means that the left edge of C can be expressed x=g1(y) and the right edge as x=g2(y). These will be denoted as the curves C3 and C4, respectively. The function-curve C3 runs from (x3, y3) to (x4, y4) and C4 in the reverse direction (x4, y4) to (x3, y3).

Note the bit of asymmetry to this formulation. In traversing the curve C starting at (x1, y1) the lower (minimum) edge of S is first encountered. On the other hand, starting at (x3, y3) the right-most (maximum) edge of S is first encountered.


CPdx = ∫C1Pdx + ∫C2Pdx
= ∫x1x2P(x, f1(x))dx + ∫x2x1P(x, f2(x))dx

But a reversal of the direction of integration changes the sign of the integral; i.e.,

x2x1P(x, f2(x))dx = −∫x1x2P(x, f2(x))dx


CPdx = ∫x1x2 [P(x, f1(x))−P(x, f2(x))]dx

Now consider ∫S∫(∂P/∂y)dydx.

S∫(∂P/∂y)dydx = ∫x1x2f1(x)f2(x)(∂P/∂y)dydx

The inner integral with respect to y can be evaluated so

S∫(∂P/∂y)dydx = ∫x1x2[P(x,y)]f1(x)f2(x)dx
and hence
S∫(∂P/∂y)dydx = ∫x1x2 [P(x, f2(x))−P(x, f1(x))]dx

This latter expression is just the negative of the expression found above.


CPdx = −∫S∫(∂P/∂y)dydx

Similarly, working with the functions g1 and g2 it can and will be shown below that

CQdy = ∫S∫(∂Q/∂x)dxdy


C(Pdx + Qdy) = ∫S∫(∂Q/∂x − ∂P/∂y)dS

The proof that ∫CQdy = ∫S∫(∂Q/∂x)dxdy starts with

CQdy = ∫C3Qdy + ∫C4Qdy
= ∫y3y4Q(g2(y), y)dy + ∫y4y3Q(g1(y))dy

The integration from (x4, y4) to (x3, y3) is just the negative of the integration from (x3, y3) to (x4, y4).


y4y3Q(g1(y),y)dy = −∫y3y4Q(g1(y),y)dy


CQdy = ∫y3y4(Q(g2(y),y)−Q(g1(y),y)))dy

Now consider

S∫(∂Q/∂x)dxdy = ∫y3y4g1(y3)g2(y4)(∂Q/∂x)dxdy

The inner integral evaluates to

(Q(g2(y), y)−Q(g1(y), y))


S∫(∂Q/∂x)dxdy = ∫y3y4(Q(g2(y), y)−Q(g1(y), y))dy

This is exactly the expression found previously for ∫CQdy. Therefore

S∫(∂Q/∂x)dxdy = ∫CQdy

When the two propositions proved above are combined it can be concluded that

C(Pdx + Qdy) = ∫S∫(∂Q/∂x − ∂P/∂y)dS

Extention to Regions with Holes

Consider a region with a hole in it, such as shown below

The boundary of this region consists of the outer boundary curve C' and the boundary curve of the hole, C". Traversing the boundary of S now consists of a counterclockwise traversal of the outer boundary and a clockwise traversal of the boundary of the hole. This procedure keeps the region S always to the left of the direction of travel.

Green's Theorem can be applied to a region with holes by cutting lines from the outer boundary to each hole, such as shown below.

This creates a region without holes. But the cut lines are traversed twice into the hole and out of the holes, as two parts of the boundary of the created region, as shown by c1 and c2. The effects on the line integral of c1 and c2 cancel out. They enclose no area so there is no effect on the surface integral. Thus Green's Theorem applies to regions with holes.

The formal proof of this would be by induction. The theorem is true for zero holes. If the theorem is true for k holes then, as shown above, it is true for (k+1) holes because a cut from the outside to one hole, or from one hole to another, reduces the situation to a region of k holes.

(To be continued.)

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