0.1 Practical Guide - Double & Triple Integrals

0.1.1 Double Integrals

Note: in some gures "sqrt" stands for "p", because some graphical properties are not activated.Notation A domain D R2 , f : D ! R , a continuous function on D .ZZ

D

f(x; y)dxdy

for f(x; y) = 1 for all (x; y) 2 D , we get the "area" of D =ZZD

1dxdy

We present two basic situations when the double integral reduces to computing two successive simple integrals.- a rectangle- a domain which is projectable onto the Ox or Oy axis, or may be "decomposed" into several such domains.

Case I. D is a rectangle, parallel to the Ox and Oy axis.

D = [a; b] [c; d] = f(x; y) 2 R2 , x 2 [a; b] and y 2 [c; d]g == f(x; y) 2 R2 , a x b and c y d g

x

y

a b

d

c

x

y

d

c

ba

1) 2)

ZZD

f(x; y)dxdy =

bZa

0@ dZc

f(x; y)dy

1A dx| {z }

1)

=

dZc

0@ bZa

f(x; y)dx

1A dy| {z }

2)

The "order" of integration makes no dierence. However we may think about (1) as "covering" the rectangle withvertical lines y 2 [c; d] , for each x 2 [a; b] , and about (2) as "covering" the rectangle with horizontal lines x 2 [a; b], for each y 2 [c; d] .

1

Example. D = [1; 2] [1; 0] , f(x; y) = xy x+ y + 2

i)ZZD

f(x; y)dxdy =

2Z1

0@ 0Z1(xy x+ y + 2)dy

1A dx = 0Z1

0@ 2Z1

(xy x+ y + 2)dx1A dy

2Z1

0@ 0Z1(xy x+ y + 2)dy

1A dx = 2Z1

xy2

2 xy + y

2

2+ 2y

01

!dx =

=

2Z1

x12 x(1) + 1

2 2dx = x

2

4+x2

2+1

2x 2x

21

= 2 + 54= 3

4

ii)

0Z1

0@ 2Z1

(xy x+ y + 2)dx1A dy = 0Z

1(x2

2y x

2

2+ xy + 2x

21

)dy =

=

0Z1(4y + 2 3

2y 3

2)dy =

0Z1(5

2y +

1

2)dy =

5

4y2 +

1

2y

01= 5

4+1

2= 3

4

Comment. You clearly dont need to produce both computations. We did so here, just to show we get thesame value in both cases.

Case II. D is "projectable" on the Ox axis, '; : [a; b]! R continuous functions. Which actually meansthe projections are "one to one".

D = f(x; y) 2 R2 , a x b and '(x) y (x) g

a b x

y

x

y

j(x)

y(x)

ZZD

f(x; y)dxdy =

bZa

0B@ (x)Z'(x)

f(x; y)dy

1CA dxComment. You may think about "covering" the domain by vertical lines, in this case of dierent lenght. For

each x 2 [a; b] we have the corresponding vertical line y 2 ['(x); (x)]

2

Example.Compute the area of the domain D R2 , which is bounded by the parabola y = x2 1 and the line y = x+1.First nd the intersection points between the line and the parabola

y = x2 1 , y = x+ 1 ) x2 1 = x+ 1 , x2 x 2 = 0

) x1 = 1 , y1 = 0 and x2 = 2 , y2 = 3As for the line, we have x = 0 , y = 1 and y = 0 , x = 1Consequently we may "describe" the domain as

D = f(x; y) 2 R2 , 1 x 2 and x2 1 y x+ 1 g

which proves it is projectable onto the Ox axis.

x

y

y = x-1

y = x+1

y = x-1

y

x-1 1 2

3

area(D) =

ZZD

1dxdy =

2Z1

0@ x+1Zx21

1dy

1A dx = 2Z1

x+ 1 (x2 1) dx =

=

2Z1

x2 + x+ 2) dx = x33+x2

2+ 2x

21= 6 8

3 13 12+ 2 = 5 1

2

Case III. D is "projectable" on the Oy axis, ; : [c; d]! R continuous functions. Which actually meansthe projections are "one to one".

D = f(x; y) 2 R2 , c y d and (y) x (y) g

3

xy

y

x

c

d

a(y)b(y)

ZZD

f(x; y)dxdy =

dZc

0B@(y)Z(y)

f(x; y)dx

1CA dyComment. You may think about "covering" the domain by horizontal lines, in this case of dierent lenght.

For each y 2 [c; d] we have the corresponding horizontal line x 2 [(y); (y)]

Example. Compute the area of the domain D R2 , which is bounded by the parabola x = 4 y2 and theline x = 1.First nd the intersection between the line and the parabola.

x = 4 y2 , x = 1 ) y2 = 5 ) y = p5

4

xy

-1

2

-2

x = -1x = 4-y

x

4

We may "describe" the domain as

D = f(x; y) 2 R2 , p5 y

p5 and 1 x 4 y2 g

area(D) =

ZZD

1dxdy =

p5Z

p5

0B@4y2Z

11dx

1CA dy =p5Z

p5

4 y2 + 1 dy =

=

p5Z

p5

5 y2 dy = 5y y3

3

p5

p5= 5 2

p5 1

32 5p5 =

20

3

p5

Comment. In both examples, we may consider the "opposite" projection.

5

xy

-1

y = sqrt( 4-x)

y = -sqrt( 4-x)

x = sqrt( 1+y)x = -sqrt( 1+y)

x = y-1

4

-1

3

1) 2)

However the successive simple integrals we get seem to be more "complicated". We have

1)ZZD

1dxdy =

0Z1

0B@py+1Z

py+1

1dx

1CA dy + 3Z0

0B@py+1Zy1

1dx

1CA dy =

=

0Z12py + 1dy +

3Z0

py + 1 (y 1)dy = 4

3

py + 1

301+2

3

py + 1

330 y

2

2 y30

=4

3+16

3 23 92+ 3 =

9

2

2)ZZD

1dxdy =

4Z1

0B@p4xZ

p4x

1dy

1CA dx = 4Z12p4 xdx = 4

3

p4 x3

41=4

35p5 =

20

3

p5

Change of variable for double integrals.Consider two domains D;E R2Let h : E ! D bijective (one to one), of class C1 (u; v)! (x; y) ,x = x(u; v) , y = y(u; v) , h(u; v) = (x(u; v); y(u; v))and the jacobian nonzero at any point (u; v)

det Jh = det

@x@u

@x@v

@y@u

@y@v

6= 0

ZZD

f(x; y)dxdy =

ZZE

f(x(u; v); y(u; v)) jdet Jhj dudv

Comment. The main purpose to use a change of variable in a double integral is not to get a simpler function,but a "simpler" domain "E" instead of "D". We can imagine innitely many changes of variable. However for"school" problems, polar and elliptical cordinates would be frequently used, whenever the domain D is "round"shaped.

Example.1) Compute the area of the disc D = f(x; y) 2 R2 , x2 + y2 R2 g

6

xy

Consider polar coordinates x = x(r; t) = r cos t , y = y(r; t) = r sin t ,To cover the whole disc we need r 2 [0; R] and t 2 [0; 2] , with the previous notation E = [0; R] [0; 2]The jacobian for polar coodinates is

det J = det

@x@r

@x@t

@y@r

@y@t

= det

cos t r sin tsin t r cos t

= r cos2 t+ r sin2 t = r

area(D) =

ZZD

1dxdy =

ZZE

1 rdrdt =2Z0

0@ RZ0

rdr

1A dt = 2Z0

r2

2

R0

!dt =

2Z0

R2

2dt = R2

Example.2) Compute the area bounded by the ellipse x

2

a2 +y2

b2 = 1 , that is the domain D R2

D = f(x; y) 2 R2 , x2

a2+y2

b2 1 g

x

y

a

b

7

This is how we get "elliptical" (or "generalized" polar) coordinates

x2

a2+y2

b2= r2 1

and clearly r 2 [0; 1] , for r = 0 we get the origin (0; 0) and for r = 1 we get a point on the ellipse.x2

a2r2+

y2

b2r2= 1 ) x

2

a2r2= cos2 t ,

y2

b2r2= sin2 t

x = ar cos t , y = br sin t

and in order to cover the domain D we need r 2 [0; 1] , t 2 [0; 2] , so E = [0; 1] [0; 2]

det J = det

@x@r

@x@t

@y@r

@y@t

= det

a cos t ar sin tb sin t br cos t

= abr cos2 t+ abr sin2 t = abr

area(D) =

ZZD

1dxdy =

ZZE

ab rdrdt =2Z0

0@ 1Z0

abrdr

1A dt = 2Z0

ab

r2

2

10

!dt =

2Z0

ab

2dt = ab

Example.3) Compute the area of the domain D R2 bounded by the curve of equation

(x2 + y2)2 = a2(y2 x2) , a > 0

First remark that this curve has left-right symmetry, ( it is symmetric with respect to Oy axis) since the equationdoes not change by replacing x with x .Also up-down symmetry, (it is symmetric with respect to Ox axis) since the equation does not change by

replacing y with y .We use polar coordinates, and write the equation of the curve in polar coodinates x = r cos t , y = r sin t

(r2 cos2 t+ r2 sin2 t)2 = a2(r2 sin2 t r2 cos2 t)

r2 = a2(sin2 t cos2 t) = a2 cos 2tr = a

p cos 2tThis means cos 2t 0 , therefore 2t 2 [2 ; 32 ] , , t 2 [4 ; 34 ] also t 2 [ 54 ; 74 ] due to up-down symmetry.To cover the domain D we need t 2 [4 ; 34 ] , t 2 [ 54 ; 74 ] and r 2 [0;

p cos 2t]

E = f(r; t) , t 2 [4;3

4] [ [ 5

4;7

4] , r 2 [0;p cos 2t]g

8

xy

r = a sqrt(-cos2t)

Also due to up-down symmetry the area(D) is twice the area for y 0

area(D) =

ZZD

1dxdy =

ZZE

1 rdrdt = 234Z4

0B@ap cos 2tZ

0

rdr

1CA dt = 234Z4

r2

2

ap cos 2t

0

!dt =

= 2

34Z4

a2 cos 2t2

dt = 2a2 sin 2t4

344

= 12(sin

3

2 sin

2) = a2

Example.4) Compute the integral ZZ

D

xdxdy

where D R2 is the domain D = f(x; y) 2 R2 , x2 + y2 4xg:First we determine the domain D.

x2 + y2 4x, x2 + y2 4x+ 4, (x 2)2 + y2 4

which means D is the disc centered at (2; 0) with radius = 2.

9

xyr = 4cost

42

t

We use polar coordinates, x = r cos t , y = r sin t , the previous inequality becomes

x2 + y2 4x, r2 4r cos t , r 4 cos tTherefore cos t 0 and so t 2 [2 ; 2 ] , 0 r 4 cos t . By using this change of variable we get

ZZD

xdxdy =

2Z2

0@4 cos tZ0

r cos t rdr1A dt =

2Z2

0@4 cos tZ0

cos t r2dr1A dt =

=

2Z2

cos t r

3

3

4 cos t0

!dt =

2Z2

43

3cos4 tdt =

43

3

2Z2

1 + cos 2t

2

2dt =

=42

3

2Z2

1 + 2 cos 2t+ cos2 2t

dt =

42

3

2Z2

1 + 2 cos 2t+

1 + cos 4t

2

dt =

=16

3

3

2 + sin 2t+

sin 4t

8

=2=2

!= 8

0.1.2 Triple Integrals

Notation A domain M R3 f :M ! R , continuous function on M .ZZZM

f(x; y; z)dxdydz

for f(x; y; z) = 1 for all (x; y; z) 2M , we get the "volume" of M =ZZZM

1dxdydz

10

We present only two basic situations when a triple integral reduces to three successive simple integrals.- a parallelipiped- a domain which is projectable onto one of the planes xOy , xOz or yOz or may be "decomposed" into such

domains.

Case I. M is a parallelepiped M = [a; b] [c; d] [; ]

a

bdc

b

a

M

y

z

x

ZZZM

f(x; y; z)dxdydz =

bZa

0@ dZc

0@ Z

f(x; y; z)dz

1A dy1A dx

or any other order of integration, there are 6 of them.

Example. For M = [0; 1] [2; 3] [1; 0] we haveZZZM

(x+ y + z)dxdydz =

1Z0

0@ 3Z2

0@ 0Z1(x+ y + zdz

1A dy1A dx =

=

1Z0

0@ 3Z2

0@ 0Z1

xz + yz +z2

2

z=0z=1

1A dy1A dx = 1Z

0

0@ 3Z2

x+ y 1

2

dy

1A dx ==

1Z0

xy +

y2

2 12y

y=3y=2

!dx =

1Z0

(x+ 2) dx =x2

2+ 2x

10

=5

2

Case II M is a "projectable" on the "horizontal" plane ( the xOy plane). Which actually means theprojection is "one to one".Let D be the projection of M onto the xOy plane.We may say M is between two surfaces S1 and S2 , dened by z = (x; y) , z = (x; y)

11

S1

S2

D

x

y

z

O

z = a(x,y)

z = b(x,y)

z

M = f(x; y; z) 2 R3 ; (x; y) 2 D R2 , z 2 [(x; y); (x; y)]gZZZM

f(x; y; z)dxdydz =

ZZD

0B@(x;y)Z(x;y)

f(x; y; z)dz

1CA dxdyComment. We clearly identify R2 with the xOy plane by (x; y) ! (x; y; 0)We actually reduce the triple integral to a simple integral along the Oz axis and a double integral into the xOy

plane.

Example.Compute the volume of a sphere S = f(x; y; z) 2 R3 , x2 + y2 + z2 R2gWe can project the sphere into the horizontal plane xOy and we get the disk

D = f(x; y) 2 R2 , x2 + y2 R2 g

S = f(x; y; z) 2 R3 , z 2hpR2 x2 y2;

pR2 x2 y2

i, x2 + y2 R2g

12

xz

y

z = sqrt(R-x-y)

z = -sqrt(R-x-y)

vol (S) =ZZZS

1dxdydz =

ZZD

0BB@pR2x2y2Z

pR2x2y2

1dz

1CCA dxdy = ZZD

2pR2 x2 y2

dxdy =

then we use polar coordinates as we did for double integrals x = r cos t , y = r sin twe need r 2 [0; R] and t 2 [0; 2] to cover the whole disk D .

=

2Z0

0@ RZ0

2pR2 r2rdr

1A dt = 2Z0

23

pR2 r23

r=Rr=0

!dt =

2Z0

2

3R

dt =

4R3

3

Example.Compute the volume of the domain bounded by the surface S dened by z = x2 + y2 and the plane z = 4 .

Looks like a "cup".The equation z = x2 + y2 represents a paraboloid, and z = 4 a horizontal plane.First nd the intersection between the paraboloid and the plane

z = x2 + y2 , z = 4 ) x2 + y2 = 4

we get a circle centered at (0; 0; 4) with radius 2 .

13

xy

z

4

z = 4

z = x+y

D

Consequently we may project the domain onto the xOy plane into the disk D = f(x; y) 2 R2 , x2 + y2 4gTherefore we may "describe" the domain as

M = f(x; y; z) 2 R3 , x2 + y2 z 4 , x2 + y2 4 g

and compute the volume as

vol (M) =ZZZM

1dxdydz =

ZZD

0B@ 4Zx2+y2

1dz

1CA dxdy = ZZD

4 x2 y2 dxdy =

next use polar coordinates to compute the double integral, x = r cos t , y = r sin t , r 2 [0; 2] , t 2 [0; 2]

=

2Z0

0@ 2Z0

(4 r2)rdr1A dt = 2Z

0

0@2Z0

(4 r2)dt1A rdr = 2 2Z

0

(4 r2)rdr = 2 4r r

3

3

r=2r=0

!= 2(8 8

3)

For domains which are "projectable" on the plane xOz or the plane yOz the procedure is quite similar.

Change of variable for triple integralsLet h : N !M , bijective (one to one) of class C1 N 3 (u; v; t)! (x; y; z) 2Mx = x(u; v; t) , y = y(u; v; t) , z = z(u; v; t) , and the jacobian nonzero at every point (u; v; t)

Jh = det

0@ @x@u @x@v @x@t@y@u

@y@v

@y@t

@z@u

@z@v

@z@t

1A 6= 0vol(M) =

ZZZM

1dxdydz =

ZZZN

f(x(u; v; t); y(u; v; t); z(u; v; t)) jJhj dudvdt

Example. Compute the volume of a sphere S = f(x; y; z) 2 R3 , x2 + y2 + z2 R2g and we use sphericalcoordinates

x = r cos' sin , x = r sin' sin , x = r cos

14

were r 2 [0; R] , ' 2 [0; 2] , 2 [0; ] so N = [0; R] [0; 2] [0; ] and the Jacobian is

Jh = det

0B@@x@r

@x@

@x@'

@y@r

@y@

@y@'

@z@r

@z@

@z@'

1CA =cos' sin r cos' cos r sin' sin sin' sin r sin' cos r cos' sin cos r sin 0

= ::: = r2 sin

vol(S) =

ZZZS

1dxdydz =

ZZZN

1 r2 sin drd'd =Z0

0@ RZ0

0@2Z0

r2 sin d'

1A dr1A d =

=

Z0

0@ RZ0

2r2 sin dr

1A d = Z0

0@ RZ0

2r3

3sin

R0

1A d = Z0

2R3

3sin d =

= 2R3

3cos

0

=2R3

3( cos + cos 0) = 4R

3

3

Example. Compute the volume of an ellipsoid (the "rugby ball")

M = f(x; y; z) 2 R3 , x2

a2+y2

b2+z2

c2 1g

We use 3D-elliptical coordinates ("generalized" spherical coordinates). This is how we get them

x2

a2+y2

b2+z2

c2= r2 1

and clearly r 2 [0; 1] , for r = 0 we get the origin (0; 0; 0) and for r = 1 we get a point on the ellipsoid.x2

a2r2+

y2

b2r2+

z2

c2r2= 1 ,

x2

a2r2+

y2

b2r2

+ zcr

2= 1

it is natural to let x2

a2r2+

y2

b2r2

= sin 2 ,

zcr

2= cos2 ) z = cr cos , 2 [0; ]

x2

a2r2 sin 2+

y2

b2r2 sin 2= 1 and again

x2

a2r2 sin 2= cos2 ' ,

y2

b2r2 sin 2= sin2 '

so nally we get r 2 [0; 1] , ' 2 [0; 2] , 2 [0; ]

x = ar cos' sin , y = br sin' sin , z = cr cos

just as before the jacobian is

Jh = det

0B@@x@r

@x@

@x@'

@y@r

@y@

@y@'

@z@r

@z@

@z@'

1CA =a cos' sin ar cos' cos ar sin' sin b sin' sin br sin' cos br cos' sin c cos cr sin 0

= ::: = abcr2 sin So we may compute the volume

vol(M) =

ZZZM

1dxdydz =

ZZZN

abcr2 sin drd'd =

=

Z0

0@ 1Z0

0@2Z0

r2 sin d'

1A dr1A d =

15

= abc

Z0

0@ 1Z0

2r2 sin

dr

1A d = abc Z0

0@ 1Z0

2r3

3sin

10

1A d = 2abc3

Z0

sin d =

=2abc

3( cos )j0 =

2abc

3( cos + cos 0) = 4abc

3

Example. Compute the volume of the domain M R3 located at the intersection of the following domains :

the ball B dened by x2+ y2+ z2 9 , exterior of the cone dened by z2 = x2+ y2 and the upper half space z 0.We use spherical coordinates and write the inequalities in spherical coordinates

x2 + y2 + z2 9 , r2 9 , r 2 [0; 3] , z 0) 2 [0; 2]

z2 x2 + y2 , r2 cos2 (r cos' sin )2 + (r sin' sin )2 = r2 sin2 ,, cos sin ) 2 [

4;

2] because 2 [0;

2]

x

y

z

j

q

3

3

First nd the intersection between these surfaces

x2 + y2 + z2 = 9 , z2 = x2 + y2 ) 2z2 = 9 ) z = 3p2

16

yz

q

O C

AB

Consequently sin =3p2

3 =1p2) = 4 . Or you may think "geometrically", cut the cone z2 = x2+ y2 with

the yOz plane, that is z = 0 , we get z2 = y2 which leads to z = y and z = y , two lines, the "main" bisectors,so clearly the angle = 4we nally get "N" = [0; 3] [4 ; 2 ] [0; 2] and compute

vol(M) =

ZZZM

1dxdydz =

ZZZN

r2 sin drd'd =

=

=2Z=4

0@ 3Z0

0@2Z0

r2 sin d'

1A dr1A d =

=

=2Z=4

0@ 3Z0

2r2 sin

dr

1A d = =2Z=4

0@ 3Z0

2r3

3sin

30

1A d =

= 18

=2Z=4

sin d = 18( cos 2+ cos

4) =

18p2

17