06 Double Integrals over Rectangles.pdf

Double Integrals over Rectangles

Lucky Galvez

Institute of Mathematics University of the Philippines

Recall

Iff(x) is defined for a≤x≤b, we subdivide the interval [a, b] into subintervals of length ∆x. Choosex∗_i on each subinterval.

The definite integral off froma tobis

ˆ b

a

f(x)dx= lim

n→∞ n

X

i=1

The Volume Problem

Letf(x, y)≥0 and continuous on the rectangular region

The Volume Problem

PartitionR by dividing the interval [a, b] intom subintervals [xi−1, xi] of equal width ∆x and [c, d] inton subintervals

[yj−1, yj] of equal width ∆y.

LetRij = [xi−1, xi]×[yj−1, yj] and let ∆Aij be the area of Rij,

The Volume Problem

PartitionR by dividing the interval [a, b] intom subintervals [xi−1, xi] of equal width ∆x and [c, d] inton subintervals

[yj−1, yj] of equal width ∆y.

LetRij = [xi−1, xi]×[yj−1, yj] and let ∆Aij be the area of Rij,

The Volume Problem

Take an arbitrary point (x∗_i, y∗_j)∈Rij and construct rectangular

prisms with basesRij and heightf(x∗i, y∗j) and denote the

volume of the prism by ∆Vij =f(x∗i, y∗j)∆A

The volume of the solid is approximately

m

X

i=1

n

X

j=1

f(x∗_i, y_j∗)∆A

As we increase the number of partitions,

V = lim

m,n→∞ m

X

i=1

n

X

j=1

The Volume Problem

Take an arbitrary point (x∗_i, y∗_j)∈Rij and construct rectangular

prisms with basesRij and heightf(x∗i, y∗j) and denote the

volume of the prism by ∆Vij =f(x∗i, y∗j)∆A

The volume of the solid is approximately

m

X

i=1

n

X

j=1

f(x∗_i, y_j∗)∆A

As we increase the number of partitions,

V = lim

m,n→∞ m

X

i=1

n

X

j=1

The Volume Problem

Take an arbitrary point (x∗_i, y∗_j)∈Rij and construct rectangular

prisms with basesRij and heightf(x∗i, y∗j) and denote the

volume of the prism by ∆Vij =f(x∗i, y∗j)∆A

The volume of the solid is approximately

m

X

i=1

n

X

j=1

f(x∗_i, y_j∗)∆A

As we increase the number of partitions,

V = lim

m,n→∞ m

X

n

X

Double Integrals over Rectangles

Definition

Iff(x, y) is continuous on a rectangular regionR, then the

double integraloff overR is

¨

R

f(x, y)dA= lim

m,n→∞ m

X

i=1

n

X

j=1

f(x∗_i, y∗_i)∆A

provided this limit exists.

Double Integrals over Rectangles

Definition

Iff(x, y) is continuous on a rectangular regionR, then the

double integraloff overR is

¨

R

f(x, y)dA= lim

m,n→∞ m

X

i=1

n

X

j=1

f(x∗_i, y∗_i)∆A

provided this limit exists.

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d].

By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy wherex is held fixed, called thepartial integral with respect toy. Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy. Hence,

ˆ b

a

I(x)dx=

ˆ b

a

ˆ d c

f(x, y)dy

dx

called aniterated (double) integral

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d]. By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy where x is held fixed,

called the partial integral with respect toy. Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy. Hence,

ˆ b

a

I(x)dx=

ˆ b

a

ˆ d c

f(x, y)dy

dx

called aniterated (double) integral

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d]. By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy where x is held fixed, called thepartial integral with respect toy.

Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy. Hence,

ˆ b

a

I(x)dx=

ˆ b

a

ˆ d c

f(x, y)dy

dx

called aniterated (double) integral

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d]. By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy where x is held fixed, called thepartial integral with respect toy. Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy.

Hence,

ˆ b

a

I(x)dx=

ˆ b

a

ˆ d c

f(x, y)dy

dx

called aniterated (double) integral

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d]. By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy where x is held fixed, called thepartial integral with respect toy. Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy. Hence,

ˆ b

a

I(x)dx

=

ˆ b

a

ˆ d c

f(x, y)dy

dx

=

ˆ b

a

ˆ d

c

f(x, y)dy dx

called aniterated (double) integral

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d]. By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy where x is held fixed, called thepartial integral with respect toy. Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy. Hence,

ˆ b

a

I(x)dx=

ˆ b

a

ˆ d c

f(x, y)dy

dx

=

ˆ b

a

ˆ d

c

f(x, y)dy dx

called aniterated (double) integral

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d]. By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy where x is held fixed, called thepartial integral with respect toy. Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy. Hence,

ˆ b

a

I(x)dx=

ˆ b

a

ˆ d c

f(x, y)dy

dx=

ˆ b

a

ˆ d

c

f(x, y)dy dx

Iterated Integrals

Supposef(x, y) is integrable on the rectangular region

R= [a, b]×[c, d]. By

ˆ d

c

f(x, y)dy

we mean the definite integral fromctodof f(x, y) with respect toy where x is held fixed, called thepartial integral with respect toy. Note that the result is a function ofx, so let

I(x) =

ˆ d

c

f(x, y)dy. Hence,

ˆ b

a

I(x)dx=

ˆ b

a

ˆ d c

f(x, y)dy

dx=

ˆ b

a ˆ d

c

f(x, y)dy dx

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1

2x dx=x2

2 1 = 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1

2x dx=x2

2 1 = 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1

2x dx=x2

2 1 = 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1 2x dx

=x2

2 1 = 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2

2x dx=x2

2

= 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1

2x dx=x2

2 1 = 4−1

= 3 b. ˆ 1 0 ˆ 2 1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2

2x dx=x2

2 b. ˆ 1 0 ˆ 2 1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1

2x dx=x2

2 1 = 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2

2x dx=x2

2 b. ˆ 1 0 ˆ 2 1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1

2x dx=x2

2 1 = 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0 9y2dy

= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2

2x dx=x2

2 b. ˆ 1 0 ˆ 2 1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1

9y2dy= 3y3

Iterated Integrals

Example

Evaluate the iterated integrals:

a. ˆ 2

1 ˆ 1

0

6xy2dy dx b. ˆ 1

0 ˆ 2

1

6xy2dx dy

Solution. a. ˆ 2 1 ˆ 1 0

6xy2dy dx

=

ˆ 2

1

ˆ 1

0

6xy2dy

dx

=

ˆ 2

1 2xy3

y=1 y=0 dx = ˆ 2 1

2x dx=x2

2 1 = 4−1 = 3

b.

ˆ 1

0 ˆ 2

1

6xy2dx dy

=

ˆ 1

0

ˆ 2

1

6xy2dx

dy

=

ˆ 1

0 3x2y2

x=2 x=1 dy = ˆ 1 0

9y2dy= 3y3

Double Integrals over Rectangles

Theorem (Fubini’s Theorem)

If f is continuous on 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

Double Integrals over Rectangles

Example

Evaluate

¨

R

ysin(xy)dA, whereR = [0,2]×[0, π].

Solution. We choose to integrate first with respect to x: ¨

R

ysin(xy)dA = ˆ π

0 ˆ 2

0

ysin(xy)dx dy

= ˆ π

0

−cos(xy)

x=2

x=0

dy

= ˆ π

0

(−cos 2y+ 1)dy

= −sin 2y 2 +y

π

Double Integrals over Rectangles

Example

Evaluate

¨

R

ysin(xy)dA, whereR = [0,2]×[0, π].

Solution. We choose to integrate first with respect to x: ¨

R

ysin(xy)dA = ˆ π

0 ˆ 2

0

ysin(xy)dx dy

= ˆ π

0

−cos(xy)

x=2

x=0

dy

= ˆ π

0

(−cos 2y+ 1)dy

= −sin 2y 2 +y

π

Double Integrals over Rectangles

Example

Evaluate

¨

R

ysin(xy)dA, whereR = [0,2]×[0, π].

Solution. We choose to integrate first with respect to x: ¨

R

ysin(xy)dA = ˆ π

0 ˆ 2

0

ysin(xy)dx dy

= ˆ π

0

−cos(xy)

x=2

x=0

dy

= ˆ π

0

(−cos 2y+ 1)dy

= −sin 2y 2 +y

π

Double Integrals over Rectangles

Example

Evaluate

¨

R

ysin(xy)dA, whereR = [0,2]×[0, π].

Solution. We choose to integrate first with respect to x: ¨

R

ysin(xy)dA = ˆ π

0 ˆ 2

0

ysin(xy)dx dy

= ˆ π

0

−cos(xy)

x=2

x=0

dy

= ˆ π

0

(−cos 2y+ 1)dy

= −sin 2y 2 +y

π

Double Integrals over Rectangles

Example

Evaluate

¨

R

ysin(xy)dA, whereR = [0,2]×[0, π].

Solution. We choose to integrate first with respect to x: ¨

R

ysin(xy)dA = ˆ π

0 ˆ 2

0

ysin(xy)dx dy

= ˆ π

0

−cos(xy)

x=2

x=0

dy

= ˆ π

0

(−cos 2y+ 1)dy

= −sin 2y 2 +y

π

0

Double Integrals over Rectangles

Example

Evaluate

¨

R

ysin(xy)dA, whereR = [0,2]×[0, π].

Solution. We choose to integrate first with respect to x: ¨

R

ysin(xy)dA = ˆ π

0 ˆ 2

0

ysin(xy)dx dy

= ˆ π

0

−cos(xy)

x=2

x=0

dy

= ˆ π

0

(−cos 2y+ 1)dy

= −sin 2y 2 +y

π

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2].

The volume of S is ¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2].

The volume of S is ¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2]. The volume ofS is

¨

R

(20−4x3−3y2)dA

= ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2]. The volume ofS is

¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2]. The volume ofS is

¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2]. The volume ofS is

¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2]. The volume ofS is

¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2]. The volume ofS is

¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

2

Double Integrals over Rectangles

Example

Find the volume of the solidS bounded by the surface

4x3+ 3y2+z= 20, the planes x= 1,y= 2 and the coordinate planes.

Solution. S is the solid under the surfacez= 20−4x3_−3y2_above the regionR= [0,1]×[0,2]. The volume ofS is

¨

R

(20−4x3−3y2)dA = ˆ 2

0 ˆ 1

0

(20−4x3−3y2)dx dy

= ˆ 2

0

20x−x4−3y2x

x=1

x=0

dy

= ˆ 2

0

19−3y2

dy

= 19y−y3

2

0

Double Integrals over Rectangles

Supposef(x, y) can be written as a product of a function of x

and a function ofy, i.e., f(x, y) =g(x)h(y).

Iff is continous on

R= [a, b]×[c, d], then by Fubini’s Theorem,

¨

R

f(x, y)dA =

ˆ b

a

ˆ d

c

f(x, y)dy dx

=

ˆ b

a

ˆ d

c

g(x)h(y)dy dx

=

ˆ b

a

ˆ d c

g(x)h(y)dy

dx

=

ˆ b

a

g(x)

ˆ d c

h(y)dy

dx

=

ˆ b

a

g(x)dx

ˆ d

c

Double Integrals over Rectangles

Supposef(x, y) can be written as a product of a function of x

and a function ofy, i.e., f(x, y) =g(x)h(y). Iff is continous on

R= [a, b]×[c, d], then by Fubini’s Theorem,

¨

R

f(x, y)dA =

ˆ b

a

ˆ d

c

f(x, y)dy dx

=

ˆ b

a

ˆ d

c

g(x)h(y)dy dx

=

ˆ b

a

ˆ d c

g(x)h(y)dy

dx

=

ˆ b

a

g(x)

ˆ d c

h(y)dy

dx

=

ˆ b

a

g(x)dx

ˆ d

c

Double Integrals over Rectangles

Supposef(x, y) can be written as a product of a function of x

and a function ofy, i.e., f(x, y) =g(x)h(y). Iff is continous on

R= [a, b]×[c, d], then by Fubini’s Theorem,

¨

R

f(x, y)dA =

ˆ b

a

ˆ d

c

f(x, y)dy dx

=

ˆ b

a

ˆ d

c

g(x)h(y)dy dx

=

ˆ b

a

ˆ d c

g(x)h(y)dy

dx

=

ˆ b

a

g(x)

ˆ d c

h(y)dy

dx

=

ˆ b

a

g(x)dx

ˆ d

c

Double Integrals over Rectangles

Supposef(x, y) can be written as a product of a function of x

and a function ofy, i.e., f(x, y) =g(x)h(y). Iff is continous on

R= [a, b]×[c, d], then by Fubini’s Theorem,

¨

R

f(x, y)dA =

ˆ b

a

ˆ d

c

f(x, y)dy dx

=

ˆ b

a

ˆ d

c

g(x)h(y)dy dx

=

ˆ b

a

ˆ d c

g(x)h(y)dy

dx

=

ˆ b

a

g(x)

ˆ d c

h(y)dy

dx

=

ˆ b

a

g(x)dx

ˆ d

c

Double Integrals over Rectangles

Supposef(x, y) can be written as a product of a function of x

and a function ofy, i.e., f(x, y) =g(x)h(y). Iff is continous on

R= [a, b]×[c, d], then by Fubini’s Theorem,

¨

R

f(x, y)dA =

ˆ b

a

ˆ d

c

f(x, y)dy dx

=

ˆ b

a

ˆ d

c

g(x)h(y)dy dx

=

ˆ b

a

ˆ d c

g(x)h(y)dy

dx

=

ˆ b

a

g(x)

ˆ d c

h(y)dy

dx

=

ˆ b

a

g(x)dx

ˆ d

c

Double Integrals over Rectangles

Supposef(x, y) can be written as a product of a function of x

and a function ofy, i.e., f(x, y) =g(x)h(y). Iff is continous on

R= [a, b]×[c, d], then by Fubini’s Theorem,

¨

R

f(x, y)dA =

ˆ b

a

ˆ d

c

f(x, y)dy dx

=

ˆ b

a

ˆ d

c

g(x)h(y)dy dx

=

ˆ b

a

ˆ d c

g(x)h(y)dy

dx

=

ˆ b

a

g(x)

ˆ d c

h(y)dy

dx

=

ˆ b

a

g(x)dx

ˆ d

c

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

=

₈

3+ 1 3

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny,

by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

=

₈

3+ 1 3

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

=

₈

3+ 1 3

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

=

₈

3+ 1 3

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

=

₈

3+ 1 3

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

=

₈

3+ 1 3

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

_{8 1}

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

=

₈

3+ 1 3

(0 + 1)

Double Integrals over Rectangles

Example

Evaluate

¨

R

x2siny dA whereR= [−1,2]×[0,π₂].

Solution. Since x2siny is a product ofg(x) =x2 and

h(y) = siny, by the prevoius theorem,

¨

R

x2siny = ˆ 2

−1 ˆ π

2

0

x2siny dy dx

= ˆ 2

−1

x2dx

ˆ π

2

0

siny dy

= x 3

3

2

−1

!

−cosy

π

2

0

!

Exercises

1 Calculate the iterated integral.

a.

ˆ 2

0 ˆ 1

0

(2x+y)8dx dy

b.

ˆ 4

1 ˆ 2

1

_x

y + y x

dy dx

2 _{Calculate the double integral.}

a.

¨

R

yexydA,R= [0,2]×[0,1]

b.

¨

R

x

x2_+y2dA,R={(x, y)|1≤x≤2,0≤y≤1}

3 Find the volume of the solid in the first octant bounded by

the cylinder z= 16−x2 and the plane y= 5.

4 Sketch the solid bounded by the paraboloid

x2_+y2_{+z= 4, the planes 2x+ 2y+z= 6, x= 1,y= 1}

Exercises

1 Calculate the iterated integral.

a.

ˆ 2

0 ˆ 1

0

(2x+y)8dx dy b.

ˆ 4

1 ˆ 2

1

_x

y + y x

dy dx

2 _{Calculate the double integral.}

a.

¨

R

yexydA,R= [0,2]×[0,1]

b.

¨

R

x

x2_+y2dA,R={(x, y)|1≤x≤2,0≤y≤1}

3 Find the volume of the solid in the first octant bounded by

the cylinder z= 16−x2 and the plane y= 5.

4 Sketch the solid bounded by the paraboloid

x2_+y2_{+z= 4, the planes 2x+ 2y+z= 6, x= 1,y= 1}

Exercises

1 Calculate the iterated integral.

a.

ˆ 2

0 ˆ 1

0

(2x+y)8dx dy b.

ˆ 4

1 ˆ 2

1

_x

y + y x

dy dx

2 _{Calculate the double integral.}

a.

¨

R

yexydA,R= [0,2]×[0,1]

b.

¨

R

x

x2_+y2dA,R={(x, y)|1≤x≤2,0≤y≤1}

3 Find the volume of the solid in the first octant bounded by

the cylinder z= 16−x2 and the plane y= 5.

4 Sketch the solid bounded by the paraboloid

x2_+y2_{+z= 4, the planes 2x+ 2y+z= 6, x= 1,y= 1}

Exercises

1 Calculate the iterated integral.

a.

ˆ 2

0 ˆ 1

0

(2x+y)8dx dy b.

ˆ 4

1 ˆ 2

1

_x

y + y x

dy dx

2 _{Calculate the double integral.}

a.

¨

R

yexydA,R= [0,2]×[0,1]

b.

¨

R

x

x2_+y2dA,R={(x, y)|1≤x≤2,0≤y≤1}

3 Find the volume of the solid in the first octant bounded by

the cylinder z= 16−x2 and the plane y= 5.

4 Sketch the solid bounded by the paraboloid

x2_+y2_{+z= 4, the planes 2x+ 2y+z= 6, x= 1,y= 1}

Exercises

1 Calculate the iterated integral.

a.

ˆ 2

0 ˆ 1

0

(2x+y)8dx dy b.

ˆ 4

1 ˆ 2

1

_x

y + y x

dy dx

2 _{Calculate the double integral.}

a.

¨

R

yexydA,R= [0,2]×[0,1]

b.

¨

R

x

x2_+y2dA,R={(x, y)|1≤x≤2,0≤y≤1}

3 Find the volume of the solid in the first octant bounded by

the cylinder z= 16−x2 and the plane y= 5.

4 Sketch the solid bounded by the paraboloid

x2_+y2_{+z= 4, the planes 2x+ 2y+z= 6, x= 1,y= 1}

Exercises

1 Calculate the iterated integral.

a.

ˆ 2

0 ˆ 1

0

(2x+y)8dx dy b.

ˆ 4

1 ˆ 2

1

_x

y + y x

dy dx

2 _{Calculate the double integral.}

a.

¨

R

yexydA,R= [0,2]×[0,1]

b.

¨

R

x

x2_+y2dA,R={(x, y)|1≤x≤2,0≤y≤1}

3 Find the volume of the solid in the first octant bounded by

the cylinder z= 16−x2 and the plane y= 5.

4 Sketch the solid bounded by the paraboloid

References

1 _{Stewart, J., Calculus, Early Transcendentals, 6 ed., Thomson} Brooks/Cole, 2008

2 _{Leithold, L.,The Calculus 7, Harper Collins College Div., 1995}