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MODULE-4
Multivariable Calculus- I
(Multiple Integration)
Syllabus: Multiple integral: Double integral, Triple integral, Change of order of integration, Change of
variables, Applications: Area, Volume, Centre of Mass & Centre of Gravity (constant and variable
densities)
...


TABLE OF CONTENTS
S
...


Topic

Page No
...
1

Course Objectives of Module 4

1

4
...
3

Multiple Integration

2

4
...
1

Double Integral

3

4
...
2

Triple Integral

12

4
...
3

Change of order of integration

15

4
...
4

Change of variables

19

4
...
4
...
4
...
4
...
5

Related Links

46

1

4
...


4
...
1 APPLICATIONS

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Double and triple integrals are useful in finding Area
...

Double and triple integrals are useful in finding Mass
...

In finding Average value of a function
...

Useful in calculating Kinetic energy and Improper Integrals
...

The most important application of Multiple Integrals involves finding areas bounded by a curve and
coordinate axes and area between two curves
...

In mechanics, the moment of inertia is calculated as the volume integral (triple integral) of
the density weighed with the square of the distance from the axis
...
The gravitational potential associated with a mass
distribution given by a mass measure on three-dimensional Euclidean space R3 is calculated by triple
integration
...

We can determine the probability of an event if we know the probability density function using double
integration
...
3 MULTIPLE INTEGRAL
4
...
1 Double integral
A double integral is its counterpart in two dimensions
...

Then double integral of f(x, y) over the region R is denoted by,

Also we can express as

or

2

4
...
1
...
Let the region
be bounded by the curves

...
)
(ii) When

are functions of x and

variable x [

are constants: If we have functional limits of y in terms of dependent

and constant limits of variable x then we will first integrate with respect to variable y

in case of double integral, as follows:

=

3

(Here we have drawn the strip parallel to y axis, because variable limits [

(iii) When

are provided
...
)

From case no
...
r
...
the variable
limits first and then w
...
t
...


4
...
1
...


=

(Here we have constant limits for both x and y variables,
so we may integrate w
...
t
...
)
For the ellipse we may write

or

The region of integration R can be expressed as
-a

,

where we have chosen variable limits of y and constant limits of x
...
r
...
y,

5

=

=

=

+0

[using the property of even and odd functions
...


intersect each other at two distinct points O (0,0) and A (3,3)
...

=

6

=

dx

=
=
=
Example 4: Evaluate
Solution: Solving

-

] dx

-

] dx
=

(Answer)

over the region R bounded by the parabolas
, we have

x = 0, 4
When x = 4, y = 4
Co-ordinates of A (intersection point of parabolas) are (4, 4)
The region R can be expressed as
0

x

4,

7


...


Solution: Let OAB be the triangle formed by given vertices (0, 0), (10, 1) and (1, 1) as shown in the figure through shaded
area
...

Hence the region of integration can be expressed as

8

dxdy =

=

= 18

dxdy

dy =

= 18

(Answer)

Example 6: Let D be the region in the first quadrant bounded by the curves xy = 16, x = y, y = 0 and x = 8
...

Solution: In this question we have to integrate the given function within the region bounded by the straight line x = y,
hyperbola xy = 16, y = 0 and x = 8, so first we will draw the figure for clarity finding all intersection points of curves
provided in question
...

Now we are to decide with respect to which variable we should first integrate, we construct strips in such a manner that
the complete area may be covered
...

Because area is changing from dotted lines, if we plot strip parallel to y-axis and also area is changing from lines drawn, if
we plot the strip parallel to x-axis
...

Here we are splitting the area OABNO in two parts by AM as shown in figure and plotted strips parallel to y- axis from x
= 0 to x = 4 and from x= 4 to x = 8
Then,

=
=

10

=
=

=

= 64 +8(64 -16) = 64 +384 = 448 (Answer)

4
...
1
...
Evaluate

2 x dydx
1 0
2
2

Ans :

x +y

2
...
8( log8-2)+e


...
Evaluate:
(i)

Ans
...
41/210

(iii)

Ans
...
3/ 35

4
...
Evaluate
6
...


...

Ans
...


Ans
...
3
...
5 Evaluation of Double Integrals in Polar Coordinates
In polar coordinates we know that, x = r cos

and y = r sin

Sometimes integration can be easier by converting Cartesian form to polar form
...
Here we draw radial strip to decide the
limit in order to cover the whole area
...
3
...
6 Solved Examples
Example 1: Evaluate

, over the area bounded between the circles r = 2 cos

11

and r = 4 cos
...
Here r varies from r = 2 cos
from

to r = 4 cos


...


Solution: I =

=

12

(Answer)

while

varies

=

=

=

(Answer)

4
...
1
...
Evaluate





2

0



a cos 

0

r sin  drd

Ans:

2
...
Evaluate

Ans:

, over the area bounded between the circles r = 2 sin and r = 4 sin
...
3
...

The triple integral of f(x, y, z) over the region and is denoted by
For purpose of evaluation, it can be expressed as the repeated integral

………………………
...

Working rule:
Let
Let


...
t
...
x (keeping y and x constant)between the limits
13

and


...
r
...
y (keeping
z constant) between the limits

and


...
r
...
z between the limits

and


...


Limits involving two variables are to be kept innermost, then the limits involving
one variable and finally the constant limits
...
Thus
=

=

4
...
2
...


Solution: Let I =
=
=
=
=
=
=

Example 2: Evaluate

=

=

(Answer)


...


Example 3: Evaluate

Solution: Here limits are constant but different for x,y,z
...
r
...
x,y or z, but limits are
fixed for x,y,z [0 ≤ x ≤ 1, 1 ≤ y ≤ 2, 2 ≤ z ≤ 3]
=
=
=
=

dx

=
=

= ½+4 – 0 = 9/2(Answer)

=

4
...
2
...
Evaluate the integral:
2
...
1/48
, where S is the region bounded by the surfaces
and z = x+3
...
Evaluate

Ans
...


,
Ans
...
Evaluate

Ans
...
3
...
Thus,

But if the limits of integration are variable, then in order to change the order of limits of integration we have to construct
the rough figure of given region of integration and re construct the strip parallel to that axis with respect to which we want
to first integrate
...


4
...
3
...


Solution: The given limits show that the region of integration is bounded by the curves y =

, y = e, x = 0, x = 1
...


In given problem we had variable limits of y in terms of x, so we had to integrate w
...
t
...
But we are instructed to
solve this problem changing the order of integration
...
r
...
x we have to find variable limits of x in terms of y
...

From the strip we can see lower limit lies on x = 0 and y =

in between the constant limits of y from y = 1

to y = e
...


Example 2: Change the order of integration in I =

and hence evaluate the same

Solution: From the variable limits of integration, it is clear that we have to integrate first w
...
to y which varies from y
=
to y = 2-x and then with respect to x which varies from x = 0
y
B

to x = 1
...

For changing the order of integration, we divide the region of

X=0

integration into horizontal strips
...

M

OY
...

x

For the region OAM, x varies from 0 to 2-y and y varies from 1 to 2
...

Solution: From the limit of integral, it is clear that we have to first integrate with respect to y, having variable limits of y
in terms of x (y =

to y =

)
...


In above figure we drawn both the parabolas intersecting at A(4a,4a)
...

=

=

=

=

=

=

Example 4: Change the order of integration and hence evaluate

18

=

(Answer)

to Q’ on

Solution: The given limits shows that the area
of integration lies between
, y = a, x =
0, x = a
...
Hence the
given integral,

=

=
=
=

=

=

(Answer)

=

Example5: Evaluate the following integral by changing the order of integration:

Solution: The given limits shows that the area of
integration lies between y = x, y = ∞, x = 0 and x = ∞
...

Hence the given integral,

=

19

=
=
=
= 1-0 = 1 (Answer)

4
...
3
...


Evaluate the integrals by changing the order of integration:
(i)

Ans: 1-

(ii)

Ans:

(iii)

Ans: 3/8

(iv)

Ans: ½(1- cos 1)

(v)

Ans:

(vi)

Ans:

(vii)

Ans:

(viii)

Ans:

(ix)

Ans: 27

(x)

Ans:

4
...
4 CHANGE OF VARIABLES
In simple integration, we use substitution to make our integration simpler than before
...

In general, there are following four types of transformation:
i
...


ii
...


iii
...


iv
...


4
...
4
...
We are to change these variables to some
variables (u , v) under the transformation x =  (u , v), y =  (u , v)
...
Also R ' is the region in the uv-plane corresponding to the region R in xy-plane
...
3
...
2 To change Cartesian co-ordinates ( x, y ) to polar co-ordinates ( r ,  ) :
We know that x = r cos  , y = r sin  and x 2 + y 2 = r 2 ,so

x x
cos  − r sin 
 (x, y ) r 
J=
=
=
= r cos 2  + sin 2  = r
...


(

)

4
...
4
...

0

  f ( x, y, z )dxdydz =  f (r sin  cos  , r sin  sin  , r cos  ) r 2 sin  drdd
...


•

For first (positive) octant of the sphere x 2 + y 2 + z 2 = a 2 ;0  r  a,0    

2

,0    
...
3
...
4 To change Cartesian co-ordinates ( x, y, z ) to cylindrical co-ordinates (r ,  , z ) :
In cylindrical coordinates,

x = r cos  ,
y = r sin  ,
z=z
and x 2 + y 2 = r 2 , so
22

x
r
 ( x, y, z ) y
J=
=
 (r ,  , z ) r
z
r

x

y

z


x
z cos 
y
= sin 
z
0
z
z

− r sin 
r cos 
0

0
0 = r cos 2  + sin 2  = r
...

V

V

•

For full volume of the cylinder x 2 + y 2 = a 2 & z = b to z = c;0  r  a, b  z  c,0    2
...


4
...
4
...


Solution: In the given integral,

x varies from 0 to 2
y varies from 0 to

2x − x2

Now, y = 2 x − x 2  y 2 = 2 x − x 2 or x 2 + y 2 = 2 x
...


For this region of integration r varies from 0 to 2 cos  and  varies from 0 to



...

= ( Answer )
 =0
3 3

Example 2: Evaluate  z ( x 2 + y 2 )dxdydz over the volume of the cylinder x 2 + y 2 = 1 intercepted by the planes z = 2
and z = 3
...
r
...
2   = ( 9 − 4 ) = ( Answer )
4 z =2
4
4
 2 2 4
0

24

Example 3: Evaluate the integral:  z 2 dxdydz over the volume of the sphere x 2 + y 2 + z 2 = 1 by changing into
spherical polar coordinates
...
r 2 sin  drd d

=0 =0 r =0

2



2



1

 r5 
I =     cos 2  sin  d d
 =0  =0  5 0
I=

1

  cos


  5

2

 sin  d d

=0 =0

2  /2

I=

2
cos 2  sin  d d


5  =0  =0

I=

2
(1 − sin 2  ) sin  d d
5 =0  =0

2  /2

2  /2

2
I =   ( sin  − sin 3  ) d d
5  =0  =0
2


2 
2
I =  ( − cos  )0 2 − d
5  =0 
3

(using Walli's formula)

2 1
4
I =
...

5 3
15
Example 4: Evaluate  xyzdxdydz over the positive octant of the sphere x 2 + y 2 + z 2 = b 2 by transforming into
spherical polar coordinates
...

6 =0  =0
48
Example 5: Evaluate  (x + y )2 dx dy, where R is the parallelogram in the xy-plane with vertices (1, 0), (3, 1), (2, 2),
R

(0, 1) using the transformation u = x + y and v = x − 2 y
...
e
...
e
...


Solving the given equations for x and y , we get x =

1
(2u + v ), y = 1 (u − v )
...

1
3
−
3
1 4

1

1
  ( x + y ) dx dy =  ' u J du dv =   u du dv =  7dv = 21( Answer )
...

Solution: Here, region R is a triangle OAB in xy-plane having sides x = 0, y = 0 and x + y = 1
...

v 1 1
y

Using given transformation, we get
If x = 0, y = 0 then u = −v, u = v
...

Thus corresponding region R ' in uv-plane is a triangle OPQ bounded by u = −v, u = v, v = 1
...


Example 7: Using the transformation u = x − y and v = x + y , evaluate the integral  R (x − y )e

x 2− y

2

dx dy where R is

the region bounded by the lines x + y = 1and x + y = 3 and the curves x 2 − y 2 = −1and x 2 − y 2 = 1
...
Also the curves x 2 − y 2 = −1; x 2 − y 2 = 1 become

uv = 1; uv = −1
...


1
 ( x, y ) 2
Also J ( x, y ) =
=
(u, v ) 1
2

1
2 =1
1
2
2

−

Therefore, under the given transformation
3

1/ v

( x − y )e x

2− y 2

dx dy =

1
ueuv du dv


2 v =1 u =−1/ v

x
R ( x − y )e

2− y 2

dx dy =

2
( Answer )
...

Solution: Changing into cylindrical coordinates,



D



D



D

e

x2 + y 2

 /4 2 4 − r 2

dV =

  


e r r dz dr d
2

=0 r =0 z =0

ex

2

+ y2

 /4 2

dV =



r
2
 e r ( 4 − r ) dr d
2

0 0

e

x2 + y 2

 /4

dV =


0

e4 − 5) 
(
 4
e4 1 
4
( Answer )
...
3
...
6 Practice problems
1
...

2
...
Use cylindrical coordinates to evaluate  x 2 + y 2 dx dy dz taken over the region V bounded by the paraboloid
V

z = 9 − x 2 − y 2 and the plane z = 0
...
Evaluate  

0 0
0

Ans:

dz dy dx
1 − x2 − y2 − z 2

, by changing to spherical polar coordinates
...
Let D be the region in the first quadrant bounded by x = 0, y = 0 and x + y = 1
...


Ans: 16/945

6
...

3
a
7
...


Ans:

 ( 5e3 − 2 )
4

Hint: 0  r  3;0     / 4;0     / 2

4
...
4
...
4
...
1 Cartesian Co-ordinates: The area A of the region bounded by two curves y = f1 (x), y = f 2 (x) and the
b f 2 (x )
 dy dx
...
4
...
2 Polar Co-ordinates: The area A of the region bounded by two curves r = f1 ( ), r = f 2 ( ) and the lines
 f 2 ( )

 =  , =  is given by A = 

 rdr d
...
4
...
3 Solved Examples
Example 1: Find the area lying between the parabola y = 4 x − x 2 and the line y = x
...


Selecting the vertical strip, the required area lies between x = 0, x = 3 and y = x, y = 4 x − x 2
...

3 0 2
2
 2

Example 2: Determine the area of region bounded by the curves xy = 2,4 y = x 2 , y = 4
...
e
...


30

Therefore, required area
4

A=

2 y

 

dx dy

y =1 x = 2/ y
4

(

)

A =  2 y − 2 / y dy
1

4

2

A = 2  y 3/2 − log y 
3
1

 16
 2  28
A = 2  − 2 log 2  −  =
− 4 log 2 ( Answer )
 3 3
 3
Example 3: Find, by double integration, the area lying inside the cardioid r = a(1+ cos ) and outside the circle r = a
...


31

Required area


2 r (cardioid)
 rd dr
 = 0 r (circle)

A=2 


2 a (1+ cos  )
 rd dr
 =0
a
a (1+ cos  )

2 r2 
A=2   
d
 = 0  2  a

A=2 

A=a

2

A=a

2











2
2
 (1 + cos  ) − 1 d
 =0



2
2
 cos  + 2 cos  d
 =0
2

1 
 a
( + 8)(using Walli' s formula
A = a2 
...


Solution: Since bounded region is symmetric about initial line,

Therefore, required area=2area above the initial line

32

 / 2 r = cardioid

 r dr d

A=2 

 = 0 r = parabola

1+ cos 

 / 2 r2 

A=2   
 
0  2 

d
1
1+ cos 

 /2

1
A =  (1 + cos  )2 −
d
2
(
)
0 
1 + cos  
 /2
 /2
1
A =  (1 + cos  )2 d − 
d
2
0
0 (1 + cos  )
 /2

(

)

A =  1 + cos 2  + 2 cos  d −
0

 /2

(

1 /2 4 
d
 sec
4 0
2

)

I1 =  1 + cos 2  + 2 cos  d

Let

0

and I 2 =
 /2

(

1 /2 4 
d , then
 sec
4 0
2

)

I1 =  1 + cos 2  + 2 cos  d
0

1 
3
+
...
2d
4 0

let


2

=

 d = 2d

I2 =

I2 =

(

)

1 /4
2
2
 1 + tan  sec  d
2 0

(

let t = tan 
 dt = sec 2 d

)

11
2
 1 + t dt
20
1

1  t 3 
2
I2 = t +
=


2
3
3
0

Hence required area=

3
2 3 4
+2− =
+ ( Answer )
...

Solution: Clearly, dotted portion is the required area
...
4
...
4 Practice problems
1
...


Ans:1/12

2
...


Ans:

Hint: − 2  x 

16 2
3

2; x 2 − 3  y  1 − x 2

x2

3
...


Ans:  ab

4
...


Ans:

1
6

5
...


Ans:

6
...

3

7
...


Ans: 3

8
...


Ans:

a2
2

9
...


Ans:

3a 2
2

4
...
2 VOLUME
4
...
2(a) VOLUME BY DOUBLE INTEGRATION
4
...
2
...

R

4
...
2
...
The volume V is given by

V =  z rdr d
...
4
...

If the region is bounded by z = f1 (x, y ), z = f 2 (x, y ); y = 1 (x), y = 2 (x)and x = a, x = b , then the volume of the
b

bounded region is V =

2 ( x )

 

f2 ( x, y )



dz dy dx
...

In spherical polar coordinates: V = V r 2 sin  dr d d
...
4
...
3 Solved examples
Example 1: Find the volume bounded by the cylinder x 2 + y 2 = 4 and the planes y + z = 4 and z = 0
...

35

Therefore, required volume
4− y 2

2



V=



z dx dy

y =−2 x =− 4 − y 2
2

4− y 2

−2

0

 ( 4 − y )dx dy

V = 2
2

V = 2  ( 4 − y ) 4 − y 2 dy
−2
2

(

)

2

V = 2  4 4 − y dy − 2  y 4 − y 2 dy
−2

2

using the property of odd function

−2

2

2

−2

0

V = 8  4 − y 2 dy = 16  4 − y 2 dy
 y 4 − y2
V = 16 
+ 2sin −1
2

V = 16  2sin −1 1 = 32 


2

y

2 

2

0

= 16 ( Answer )

Example2: A triangular prism is formed by planes whose equations are ay = bx, y = 0 and x = a
...

Solution: The volume of the bounded region
a

bx
a c + xy

  

V=

dz dy dx

x =0 y =0 z =0
bx
a a

V =   ( c + xy ) dy dx
0 0
a

bx

V =  ( cx + xy 2 / 2 ) a dx
0

0

a
 cxb b 2 3 
V = 
+ 2 x dx
a
2a


0
a

a

cb  x 2 
b2  x4 
V=   + 2 
a  2  0 2a  4  0
V=

abc a 2b 2
+
( Answer )
...

Solution: Since the column is standing on xy − plane , therefore at the base of the cylinder z = 0
...

4 7
10 5 140
Example 4: Find, by triple integration, the volume of the region bounded by the paraboloid az = x 2 + y 2 and
the cylinder R 2 = x 2 + y 2
...

a

For cylinder : r 2 = R 2 or r = R
...
r

2

0 0



a

dr d

R

 r4 
1
V = 4    d =
4a  0
a
0 
2



2



R 4 d =

0

 R4
2a

( Answer )
...

Solution: It is clear from given equations that the base of one cylinder is on xy − plane and of other is on

xz − plane
...

3

Example 6: Find the volume bounded above by the sphere x 2 + y 2 + z 2 = a 2 and below by the cone

x2 + y2 = z 2
...


So required volume is

V = V dz dy dx
...

2

Using polar coordinates,
a

2

V=

2

  

=0 r =0

2

V=

a2 −r 2

2

r dzdr d

z =r

  ( z )r

a2 −r 2

rdr d

0 0

  (r

2

V=

2

0 0

1
V=
3

2


0

)

a 2 − r 2 − r 2 dr d 

  a2 32
3 
 −   + a 3 − a d
  2 
2 2



(

)

1 
1 
a3
V = a 3 1 −

...


3 
3
2
4
...
2
...
Calculate the volume of the solid bounded by the surface x = 0, y = 0, x + y + z = 1and z = 0
...
Find by triple integration, the volume of the paraboloid of revolution x 2 + y 2 = 4 z cut-off by the plane
Ans: 32

z = 4
...
The sphere x 2 + y 2 + z 2 = a 2 is pierced by the cylinder x 2 + y 2 = a 2 x 2 − y 2
...

3  4 3
3 

4
...


Ans:


4

x y z
5
...

a

b

c

Ans:

a 3bc
60

x y z
8
...

a

b

c

Ans:
9
...

39

a3b 2c 2
2520

Ans: 8 2

Find the volume bounded by the elliptic paraboloids z = x 2 + 9 y 2 and z = 18 − x 2 − 9 y 2
...


Hint: V =

Ans: 27

1
9− x 2
18 − x 2 − 9 y 2
3 3


 dz dy dx
2
2
− 3 −1
9− x 2 x +9 y
3

11
...

Ans: 16 / 9
2  1
4
12
...
Evaluate V
...
4
...
4
...
1 Calculation of Mass
(a) For a plane lamina: If the area of lamina is A and density at any point is  = f (x, y ) then the total mass

M =  A dx dy =M =  A f (x, y )dx dy

In polar co-ordinates,  = f (r, ) then the total mass

M =  A rdr d =M =  A f (r , )rdr d
...

4
...
3
...
r cos 
...
r sin 
...


4
...
3
...
Then find:
(i) the mass of the plate
(ii) the position of centre of gravity G
...

3 0 3

(ii) Centre of gravity

x=

  xdx dy =   xdx dy
M
  dx dy
2 2x

3
x=
(1 + x + y )xdx dy
44 0 0
2

x=

3
2 x + 4 x 2 )xdx
(

44 0
2

3 2
 16
x =  x3 + x 4  =
44  3
0 11

41

y=
y=

 

ydx dy   ydx dy
=
M
  dx dy

3 22x
  (1 + x + y )ydydx
44 0 0
2x

3 2
y 2 y3 
(
)
y=
1
+
x
+
 dx

44 0 
2
3 
0
y=

3 2
y3 
2
(
)
1
+
x
2
x
+
 dx

44 0 
3 
2

3  2 x 3 14 x 4 
18
y=
+
...

 11 11 

Therefore, the position of centre of gravity (C
...
)= 

Example 2: Find the centroid (centre of gravity) of a loop of the lemniscates r 2 = a 2 cos 2
...


So in this case y = 0
...

For one loop, −  / 4     / 4 and 0  r  a cos 2
...
 r dr d

cr 2 cos  dr d

0

 /4 a cos 2




− /4



c r dr d

0

a cos 2

 /4

 r3 
cos

3

 0
− /4

x=

 /4

d

a cos 2

 r2 
  2 
− /4 
0
 /4

2a
3

x=

4a
x=
3



d
3

cos  1 − sin 2   2 d

0

 /4

 sin 2 
 2 
0
 /4



cos  (1 − 2sin 2  ) 2 d
3

0

Let 2 sin  = sin t  2 cos  d = cos tdt;
x=
x=

4a
3

 /2


0

(1 − sin 2 t ) 2
...

=

...

8



Therefore, the position of centre of gravity (C
...
) = 

Example 3: A solid is in the form of the positive octant of the sphere x 2 + y 2 + z 2 = a 2
...
Find the co-ordinate of centre of gravity of the solid
...
If x, y, z are the coordinates of C
...

Then
 x dx dy dz
 y dx dy dz
 z dx dy dz
x= V
,y = V
, z == V

...
1
sin  cos 2 d

5

...

1
6a 0
24a 2 16a
x=
=

...


/
2
7
35 3 35
2
sin  cos d

0 4
...
Therefore, the position of centre of gravity (C
...
) = 
,
,
 (Answer)
...
4
...
4 Practice problems
1
...


Ans:
2

21
a 4
32

2

 x
 y
2
...
Find the

 8a 8b 
, 
 15 15 

Ans: 

coordinates of the centre of gravity of the plane
...
Find the mass of the plate in the form of the curve   +  
= 1 , density at any point is given by  = xy
...
Find the centre of gravity of the area of the circle x 2 + y 2 = a 2 lying in the first quadrant
...
Find the mass of the area bounded by the curves y 2 = x and y = x 3 , if  =  x 2 + y 2
...
5 RELATED LINKS
1
...
youtube
...
https://www
...
com/watch?time_continue=1&v=wtY5fx6VMGQ&feature=emb_logo
3
...
youtube
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/double-integralstopic/v/double-integral-1

5
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/double-integralstopic/v/double-integrals-3

7
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/double-integralstopic/v/double-integrals-5

9
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/double-integralsa/a/double-integrals

11
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/double-integralsa/a/double-integrals-beyond-volume

13
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/double-integralsa/a/double-integrals-in-polar-coordinates

45

15
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/triple-integralstopic/v/triple-integrals-2

17
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/triple-integralsa/a/triple-integrals

19
...
khanacademy
...
https://www
...
org/math/multivariable-calculus/integrating-multivariable-functions/triple-integralsa/a/triple-integrals-in-spherical-coordinates

21
...
org/multi/S-11-4-Double-Integrals-Applications
...
https://www
...
com/watch?v=FmhMUTmUjhM

46



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