596 CHAPTER 30. STOKES AND GREEN’S THEOREMS

=m

∑i=1

∫ bi

ai

(Q(xi (t) ,yi (t)) ,−P(xi (t) ,yi (t)))

· 1√(x′i)

2 +(y′i)2

(y′i,−x′i

) dS︷ ︸︸ ︷√(x′i)

2 +(y′i)2dt

=m

∑i=1

∫ bi

ai

(Q(xi (t) ,yi (t)) ,−P(xi (t) ,yi (t))) ·(y′i,−x′i

)dt

=m

∑i=1

∫ bi

ai

Q(xi (t) ,yi (t))y′i (t)+P(xi (t) ,yi (t))x′i (t)dt ≡∫

∂UPdx+Qdy

This proves Green’s theorem from the divergence theorem.

Proposition 30.1.2 Let U be an open set in R2 for which Green’s theorem holds. ThenArea of U =

∫∂U F ·dR where F (x,y) = 1

2 (−y,x) ,(0,x), or (−y,0).

Proof: This follows immediately from Green’s theorem.

Example 30.1.3 Use Proposition 30.1.2 to find the area of the ellipse x2

a2 +y2

b2 ≤ 1.

You can parameterize the boundary of this ellipse as x = acos t, y = bsin t, t ∈ [0,2π].Then from Proposition 30.1.2,

Area equals =12

∫ 2π

0(−bsin t,acos t) · (−asin t,bcos t)dt =

12

∫ 2π

0(ab)dt = πab.

Example 30.1.4 Find∫

∂U F ·dR where U is the set{(x,y) : x2 +3y2 ≤ 9

}and F (x,y) =

(y,−x).

One way to do this is to parameterize the boundary of U and then compute the lineintegral directly. It is easier to use Green’s theorem. The desired line integral equals∫

U ((−1)−1)dA = −2∫

U dA.Now U is an ellipse having area equal to 3√

3 and so theanswer is −6

√3.

Example 30.1.5 Find∫

∂U F ·dR where U is the set {(x,y) : 2 ≤ x ≤ 4,0 ≤ y ≤ 3} and

F (x,y) =(xsiny,y3 cosx

)From Green’s theorem this line integral equals∫ 4

2

∫ 3

0

(−y3 sinx− xcosy

)dydx =

814

cos4−6sin3− 814

cos2.

This is much easier than computing the line integral because you don’t have to break theboundary in pieces and consider each separately.

Example 30.1.6 Find∫

∂U F ·dR where U is the set {(x,y) : 2 ≤ x ≤ 4,x ≤ y ≤ 4} and

F (x,y) = (xsiny,ysinx)

From Green’s theorem, this line integral equals∫ 4

2∫ 4

x (ycosx− xcosy)dydx = 4cos2−8cos4−8sin2−4sin4.