ouble integrls on regions (Set. 5.) Review: Fubini s on retngulr domins. Fubini s on non-retngulr domins. Tpe I: omin funtions (). Tpe II: omin funtions (). Finding the limits of integrtion. Review: Fubini s on retngulr domins If f : R R R is ontinuous in R = [, b] [, d], then R f (, ) d d = = b d d b f (, ) d d, f (, ) d d. Remrk: Fubini result ss tht double integrls n be omputed doing two one-vrible integrls. Remrk: On retngle is simple to swith the order of integrtion in double integrls of ontinuous funtions. z R f(,)
ouble integrls on regions (Set. 5.) Review: Fubini s on retngulr domins. Fubini s on non-retngulr domins. Tpe I: omin funtions (). Tpe II: omin funtions (). Finding the limits of integrtion. Fubini s on Tpe I domins, () If f : R R is ontinuous in, then hold (Tpe I): If = { (, ) R : [, b], [g (), g ()] }, with g, g ontinuous funtions on [, b], then f (, ) d d = b g () g () f (, ) d d. = g () z f(,) f(,g ()) f(,g ()) = g () b b g () g ()
ouble integrls on regions (Set. 5.) Review: Fubini s on retngulr domins. Fubini s on non-retngulr domins. Tpe I: omin funtions (). Tpe II: omin funtions (). Finding the limits of integrtion. Fubini s on Tpe II domins, () If f : R R is ontinuous in, then hold (Tpe II): If = { (, ) R : [h (), h ()], [, d] }, with h, h ontinuous funtions on [, d], then f (, ) d d = d h () h () f (, ) d d. d = h () = h () z f(,) f(h (),) h () f(h (),) d h ()
Summr: Fubini s on non-retngulr domins If f : R R is ontinuous in, then hold: () (Tpe I) If = { (, ) R : [, b], [g (), g ()] }, with g, g ontinuous funtions on [, b], then f (, ) d d = b g () g () f (, ) d d. (b) (Tpe II) If = { (, ) R : [h (), h ()], [, d] }, with h, h ontinuous funtions on [, d], then f (, ) d d = d h () h () f (, ) d d. A double integrl on Tpe I domin Emple Find the integrl of f (, ) = +, on the domin = {(, ) R :, }. Solution: This is Tpe I domin, with lower boundr = g () =, = g () = nd upper boundr = g () =. = g () =
A double integrl on Tpe I domin. Emple Find the integrl of f (, ) = +, on the domin = {(, ) R :, }. b g () Solution: f (, ) d d = f (, ) d d g () with g () = nd g () =, we obtin f (, ) d d = ( + )d d, [ ( ) ( + )] d. [ ( ) + ( 6)] d. A double integrl on Tpe I domin Emple Find the integrl of f (, ) = +, on the domin = {(, ) R :, }. Solution: Rell: [ ( ) + ( 6)] d. [ 4 + 6] [ 4 d = 4 5 5 + 4 7 ] We onlude: 5 = 9 ()(5)(7). f (, ) d d = 5.
Summr: Fubini s on non-retngulr domins If f : R R is ontinuous in, then hold: () (Tpe I) If = { (, ) R : [, b], [g (), g ()] }, with g, g ontinuous funtions on [, b], then f (, ) d d = b g () g () f (, ) d d. (b) (Tpe II) If = { (, ) R : [h (), h ()], [, d] }, with h, h ontinuous funtions on [, d], then f (, ) d d = d h () h () f (, ) d d. A double integrl on Tpe II domin Emple Find the integrl of f (, ) = + on the domin = {(, ) R :, }. Solution: This is Tpe II domin, with left boundr = h () =, = h () = nd right boundr = h () =. = h () = Remrk: This domin is both Tpe I nd Tpe II: = =.
A double integrl on Tpe I domin Emple Find the integrl of f (, ) = +, on the domin = {(, ) R :, }. Solution: f (, ) d d = d h () h () with h () = nd h () =, we obtin [( ( + ) d d, ) + ( f (, ) d d )] d, [ ( / ) + ( / )] d. A double integrl on Tpe I domin Emple Find the integrl of f (, ) = +, on the domin = {(, ) R :, }. Solution: [ ( / ) + ( / )] d. [ [ / + 5/ ] d, 5 5/ 4 4 + 7 7/ 4 ] 4, 5 + 7 4 = 9 ()(5)(7). We onlude f (, ) d d = 5.
omins Tpe I nd Tpe II Summr: We hve shown tht double integrl of funtion f on the domin given in the pitures below holds, f (, )d d = f (, )d d = f (, )d d. = g () = = h () = = g () = = h () = ouble integrls on regions (Set. 5.) Review: Fubini s on retngulr domins. Fubini s on non-retngulr domins. Tpe I: omin funtions (). Tpe II: omin funtions (). Finding the limits of integrtion.
omins Tpe I nd Tpe II Emple Find the limits of integrtion of f (, ) dd in the domin = { (, ) R : 9 + 4 } when is onsidered first s Tpe I nd then s Tpe II. Solution: The boundr is the ellipse 9 + 4 =. So, the boundr s Tpe I is given b = 9 = g (), = 9 = g (). The boundr s Tpe II is given b = 4 = h (), = 4 = h (). omins Tpe I nd Tpe II Emple Reverse the order of integrtion in e d d. Solution: This integrl is written s Tpe I, sine we first integrte on vertil intervls [, e ], with boundries = e, =, while [, ]. e = e = = ln() = Invert the first eqution nd from the figure we get the left nd right boundries: = ln(), =, with [, e]. Therefore, we onlude tht e d d = e ln() d d.