Part 4: Fubini's Theorem and Additional Properties

The definition of the double integral implies many other properties. For example, if f( x,y) £ g( x,y) on R, then

R   f( x,y) dA £ R  g( x,y) dA

and likewise, if f( x,y) ³ 0 on R and S Ì R, then

S   f( x,y) dA £ R  f( x,y) dA

Moreover, suppose that R and S are non-overlapping regions-i.e., that R and S do not intersect except possibly on the boundary:

Then as will be shown in the exercises, we must have

RÈS  f( x,y) dA = R  f( x,y)dA + S  f( x,y) dA

(8)

Finally, properties of the double integral also follow from their relationship to iterated integrals.. For example, since the rectangle [ a,b] ×[ c,d] is both a type I and a type II region, we must have

[ a,b] ×[ c,d]  f( x,y)dA = ó õ b a  ó õ d c  f( x,y) dydx        and       [ a,b] ×[ c,d] f( x,y)dA = ó õ d c  ó õ b a  f( x,y) dxdy

As a result, the two iterated integrals are the same. This result is known as Fubini's theorem, which says that if a,b,c and d are constant and if the double integral of f( x,y) exists, then

ó õ b a  ó õ d c  f( x,y)dydx = ó õ d c  ó õ b a  f( x,y) dxdy

That is, the order of integration may be switched if the limits of integration are constant.       

EXAMPLE 6    Use Fubini's theorem to evaluate

ó õ p 0  ó õ 1 0  cos( x) sin( y2)dydx

Solution: Fubini's theorem implies that

ó õ p 0  ó õ 1 0  cos( x) sin( y2)dydx = ó õ 1 0  ó õ p 0  sin( y2) cos(x) dxdy

As a result, we integrate cos( x) to obtain

ó õ p 0  ó õ 1 0  cos( x) sin( y2)dydx = ó õ 1 0  sin( y2) sin( x)| 0p  dy = ó õ 1 0  sin( y2) ( 0-0) dy = 0