Double Integrals over General Regions
Lucky Galvez
Institute of Mathematics University of the Philippines
Diliman
Recall
Iff(x, y) is integrable over the rectangular region
R= [a, b]×[c, d], then ¨
R
f(x, y)dA = ˆ b
a
ˆ d
c
f(x, y) dy dx
= ˆ d
c
ˆ b
a
Type I Region
A plane regionDis said to be of Type Iif it lies between two continuous functions ofx,
that is,
D={(x, y) :a≤x≤b, g1(x)≤y≤g2(x)}
whereg1 and g2 are continuous on [a, b].
a b
y=g2(x)
y=g1(x)
Type I Region
A plane regionDis said to be of Type Iif it lies between two continuous functions ofx, that is,
D={(x, y) :a≤x≤b, g1(x)≤y≤g2(x)}
whereg1 and g2 are continuous on [a, b].
a b
y=g2(x)
Type I Region
A plane regionDis said to be of Type Iif it lies between two continuous functions ofx, that is,
D={(x, y) :a≤x≤b, g1(x)≤y≤g2(x)}
whereg1 and g2 are continuous on [a, b].
a b
y=g2(x)
y=g1(x)
Type I Region
A plane regionDis said to be of Type Iif it lies between two continuous functions ofx, that is,
D={(x, y) :a≤x≤b, g1(x)≤y≤g2(x)}
whereg1 and g2 are continuous on [a, b].
a b
y=g2(x)
Integral over Type I Region
Iff is continuous on a Type I region Dsuch that
D={(x, y) :a≤x≤b, g1(x)≤y≤g2(x)},
then
¨
D
f(x, y)dA= ˆ b
a
ˆ g2(x)
g1(x)
f(x, y) dy dx
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution.
SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1 1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution.
SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1 1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
= 4x 5 2 5 − x2 4 − x4 2 + x5 10 ! 1 0 = 3 20
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region,
i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1 1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
= 4x 5 2 5 − x2 4 − x4 2 + x5 10 ! 1 0 = 3 20
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1 1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
= 4x 5 2 5 − x2 4 − x4 2 + x5 10 ! 1 0 = 3 20
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
2x32 −x
−
2x3−x
4
2
dx
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1 1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
= 4x 5 2 5 − x2 4 − x4 2 + x5 10 ! 1 0 = 3 20
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
2x32 −x
−
2x3−x
4
dx
Integral over Type I Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. SinceRis a Type I region, i.e.,
R=
(x, y) : 0≤x≤1, x2≤y≤√x ,
1 1
y=x2
y=√x
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
x
x2
(2x−y)dy dx
=
ˆ 1
0
2xy−y
2
2
y=√x
y=x2
dx
=
ˆ 1
0
2x32 −x 2
−
2x3−x
4
2
dx
Type II Region
A plane regionDis said to be of Type IIif it lies between two continuous functions ofy,
that is,
D={(x, y) :h1(y)≤x≤h2(y), c≤y≤d}
whereh1 and h2 are continuous.
x=h2(y) x=h1(y)
Type II Region
A plane regionDis said to be of Type IIif it lies between two continuous functions ofy, that is,
D={(x, y) :h1(y)≤x≤h2(y), c≤y≤d}
whereh1 and h2 are continuous.
x=h2(y) x=h1(y)
c d
Type II Region
A plane regionDis said to be of Type IIif it lies between two continuous functions ofy, that is,
D={(x, y) :h1(y)≤x≤h2(y), c≤y≤d}
whereh1 and h2 are continuous.
x=h2(y) x=h1(y)
Type II Region
A plane regionDis said to be of Type IIif it lies between two continuous functions ofy, that is,
D={(x, y) :h1(y)≤x≤h2(y), c≤y≤d}
whereh1 and h2 are continuous.
x=h2(y) x=h1(y)
c d
Integral over Type II Region
Iff is continuous on a Type II region Dsuch that
D={(x, y) :h1(y)≤x≤h2(y), c≤y≤d},
then
¨
D
f(x, y)dA= ˆ d
c
ˆ h2(y)
h1(y)
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution.
We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1 1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
= 3 20
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution.
We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region,
i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1 1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
= 3 20
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1 1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
= 3 20
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1 1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
= 3 20
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1 1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1
0 h
y−y32
− y4−y3i
dy
= y
2
2 − 2y52
5 − y5
5 + y4
4
!
1
0
= 3 20
Integral over Type II Region
Example
Evaluate ¨
R
(2x−y) dAwhere R is the region enclosed by
y=x2 and x=y2.
Solution. We considerRas a Type II region, i.e.,
R=
(x, y) :y2≤x≤√y, 0≤y≤1 ,
1
x=√y
x=y2
¨
R
(2x−y)dA =
ˆ 1
0 ˆ √
y
y2
(2x−y)dx dy
=
ˆ 1
0
x2−xy
x=√y
x=y2
dy
=
ˆ 1h
y−y32
− y4−y3i
Double Integrals over General Regions
Some properties
1 If cis a constant,
¨
R
cf(x, y) dA=c
¨
R
f(x, y) dA
2
¨
R
[f(x, y)±g(x, y)] dA= ¨
R
f(x, y)dA± ¨
R
g(x, y) dA
3 If R=R1∪R2 such thatR1∩R2 =∅, then
¨
R
f(x, y) dA= ¨
R1
f(x, y) dA+ ¨
R2
f(x, y) dA
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤2−x
2 , hence, ¨
R
xy dA=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx+
ˆ 2
0 ˆ 2−x
2
0
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y ≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤ 2−x
2 , hence, ¨
R
xy dA=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx+
ˆ 2
0 ˆ 2−x
2
0
xy dy dx
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y ≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤ 2−x
2 , hence, ¨
R
xy dA=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx+
ˆ 2
0 ˆ 2−x
2
0
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y ≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤ 2−x
2 , hence, ¨
R
xy dA=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx+
ˆ 2
0 ˆ 2−x
2
0
xy dy dx
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y ≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤ 2−x
2 , hence, ¨
R
xy dA=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx+
ˆ 2
0 ˆ 2−x
2
0
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤ 2−x
2 , hence,
¨
R
xy dA=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx+
ˆ 2
0 ˆ 2−x
2
0
xy dy dx
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤ 2−x , hence,
=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx+
ˆ 2
0 ˆ 2−x
2
0
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y≤√x+ 1 ∪
(x, y) : 0≤x≤2, 0≤y≤ 2−x
2 , hence, ¨
R
xy dA=
ˆ 0
−1 ˆ √
x+1
0
xy dy dx
+
ˆ 2
0 ˆ 2−x
2
0
xy dy dx
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
y= 2−2x
a. R=
(x, y) :−1≤x≤0, 0≤y≤√x+ 1 ∪
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
x=y2−1 x= 2−2y
b. R=
(x, y) :y2−1≤x≤2−2y, 0≤y ≤1 , hence, ¨
R
xy dA= ˆ 1
0
ˆ 2−2y
y2−1
xy dx dy
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
y=√x+ 1
x+ 2y= 2
x=y2−1 x= 2−2y
b. R=
(x, y) :y2−1≤x≤2−2y, 0≤y ≤1 , hence, ¨
R
xy dA= ˆ 1
0
ˆ 2−2y
y2−1
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
x+ 2y= 2
x=y2−1
x= 2−2y
b. R=
(x, y) :y2−1≤x≤2−2y, 0≤y ≤1 , hence, ¨
R
xy dA= ˆ 1
0
ˆ 2−2y
y2−1
xy dx dy
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
x=y2−1 x= 2−2y
b. R=
(x, y) :y2−1≤x≤2−2y, 0≤y ≤1 , hence, ¨
R
xy dA= ˆ 1
0
ˆ 2−2y
y2−1
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
x=y2−1 x= 2−2y
b. R=
(x, y) :y2−1≤x≤2−2y, 0≤y≤1 , hence,
¨
R
xy dA= ˆ 1
0
ˆ 2−2y
y2−1
xy dx dy
Double Integrals over General Regions
Example
Set up the iterated double integral equal to ¨
R
xy dA whereR
is the region bounded byy=√x+ 1,x+ 2y= 2 and the x-axis by consideringR as a
a. Type I region b. Type II region
Solution.
−1 1 2 1
x=y2−1 x= 2−2y
b. R=
(x, y) :y2−1≤x≤2−2y, 0≤y≤1 , hence,
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2
has no elementary antiderivative with respect tox.
Consider the regionR={(x, y) :y≤x≤2, 0≤y≤2}.
1 2
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
2
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2
has no elementary antiderivative with respect tox.
Consider the regionR={(x, y) :y≤x≤2, 0≤y≤2}.
1 2
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2
has no elementary antiderivative with respect tox. Consider the regionR={(x, y) :y≤x≤2, 0≤y≤2}.
1 2
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
2
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2
has no elementary antiderivative with respect tox. Consider the regionR={(x, y) :y≤x≤2, 0≤y≤2}.
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2 has no elementary antiderivative with respect tox. WriteR asR={(x, y) : 0≤x≤2, 0≤y≤x}.
1 2
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
2
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2 has no elementary antiderivative with respect tox. WriteR asR={(x, y) : 0≤x≤2, 0≤y≤x}.
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA
=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2 has no elementary antiderivative with respect tox. WriteR asR={(x, y) : 0≤x≤2, 0≤y≤x}.
1 2
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
2
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2 has no elementary antiderivative with respect tox. WriteR asR={(x, y) : 0≤x≤2, 0≤y≤x}.
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx
=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2 has no elementary antiderivative with respect tox. WriteR asR={(x, y) : 0≤x≤2, 0≤y≤x}.
1 2
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
2
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2 has no elementary antiderivative with respect tox. WriteR asR={(x, y) : 0≤x≤2, 0≤y≤x}.
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
= e 4−1
Double Integrals over General Regions
Example
Evaluate ˆ 2
0
ˆ 2
y
ex2 dx dy.
Solution. f(x, y) =ex2 has no elementary antiderivative with respect tox. WriteR asR={(x, y) : 0≤x≤2, 0≤y≤x}.
1 2
1 2
x=y
ˆ 2
0 ˆ 2
y
ex2 dx dy =
¨
R
ex2 dA=
ˆ 2
0 ˆ x
0
ex2 dy dx
=
ˆ 2
0
yex2
y=x
y=0
dx=
ˆ 2
0
xex2dx
= e x2
2
2
0
= e 4−1
2
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
sketch the regionDand express the double integral as an iterated integral with reversed order of integration.
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
sketch the regionDand express the double integral as an iterated integral with reversed order of integration.
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
sketch the regionDand express the double integral as an iterated integral with reversed order of integration.
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
Exercises
1 Evaluate the following iterated double integrals:
a.
ˆ 2
0
ˆ 2y
y
xy dx dy
b.
ˆ √ π
0
ˆ √ π
y
cosx2 dx dy
c.
ˆ 4
0
ˆ 2
√ x
1
1 +y3 dy dx
d.
ˆ 1
0
ˆ 1
x
exy dy dx
2 Evaluate the following double integrals over the given regionR.
a.
¨
R
(x+y2)dA,Ris the triangle with vertices (0,0), (1,1), (0,2)
b.
¨
R
xsiny dA,Ris bounded byy=x2,x= 0 andy=π.
c.
¨
R
x2 dA,Ris bounded byy=x, 2x+y= 6 and they-axis.
3 Given the iterated integral ¨
D
f(x, y)dA=
ˆ 1
0
ˆ 2y
0
f(x, y)dx dy+
ˆ 3
1
ˆ 3−y
0
f(x, y)dx dy,
sketch the regionD and express the double integral as an iterated integral with reversed order of integration.
References
1 Stewart, J., Calculus, Early Transcendentals, 6 ed., Thomson Brooks/Cole, 2008
2 Leithold, L.,The Calculus 7, Harper Collins College Div., 1995
