Course:  Accelerated  Engineering  Calculus  II   Instructor:  Michael  Medvinsky      

69    

5.7 Triple Integrals (12.7)

We  now  define  a  triple  integral  in  a  similar  way  that  we  defined  a  single  and  double  integrals.  We  first  start  with  the  rectangular-­‐box  domains   B = x, y, z( ) : x1 ≤ x ≤ x2, y1 ≤ y ≤ y2, z1 ≤ z ≤ z2{ } .  We  divide  the  domain  into  sub-­‐boxes   Bijk = xi−1, xi[ ]× yj−1, yj⎡⎣ ⎤⎦ × zk−1, zk[ ]    with  volume  ΔV = ΔxΔyΔz  and  choose  sample  points  to  define  triple  Riemann  sum:  

f xijk* , yijk

* , zijk*( )ΔV

k=1

K

∑j=1

J

∑i=1

I

∑  

Finally  we  take  the  limit  of  the  triple  Riemann  sum  to  define  triple  integral:  

f x, y, z( )dVB∫∫∫ = lim

I ,J ,K→∞f xijk

* , yijk* , zijk

*( )ΔVk=1

K

∑j=1

J

∑i=1

I

∑  

Fubini’s  Thm:  If  f  is  continuous  on  rectangular  box B = x1, x2[ ]× y1, y2[ ]× z1, z2[ ] ,  then  f x, y, z( )dV

B∫∫∫ = f x, y, z( )dzdydx

z1

z2

∫y1

y2

∫x1

x2

∫  

The  above  is  the  triple  iterated  integral,  using  this  specific  notations  we  first  integrate  in  z,  then  in  y  and  finally  in  x.  However,  the  order  can  be  changed;  it  has  five  different  orders.      To  define  the  triple  integral  for  more  general  shaped  domain  E  we  distinct  between  3  different  domains:  

Type  1  E = x, y, z( ) : x, y( )∈D,u1 x, y( ) ≤ z ≤ u2 x, y( ){ } :   f x, y, z( )dVB∫∫∫ = f x, y, z( )dz

u1 x,y( )

u2 x,y( )

∫⎛

⎝⎜

⎞

⎠⎟ dA

D∫∫    

Type  2  E = x, y, z( ) : y, z( )∈D,u1 y, z( ) ≤ x ≤ u2 y, z( ){ } :   f x, y, z( )dVB∫∫∫ = f x, y, z( )dx

u1 y,z( )

u2 y,z( )

∫⎛

⎝⎜

⎞

⎠⎟ dA

D∫∫  

Type  3  E = x, y, z( ) : x, z( )∈D,u1 x, z( ) ≤ y ≤ u2 x, z( ){ } :   f x, y, z( )dVB∫∫∫ = f x, y, z( )dy

u1 x,z( )

u2 x,z( )

∫⎛

⎝⎜

⎞

⎠⎟ dA

D∫∫  

The domain of the double integral D above can be of either type I or II. Note that D lies on different planes for types 1,2,3 above, we will consider type 1, i.e. D is on xy plane, for the types 2 and 3 it is analogically. Thus, if D is of type I we have

E = x, y, z( ) : x1 ≤ x,≤ x2,g1 x( ) ≤ y ≤ g2 x( ),u1 x, y( ) ≤ z ≤ u2 x, y( ){ }  and  

f x, y, z( )dVB∫∫∫ = f x, y, z( )dzdydx

u1 x,y( )

u2 x,y( )

∫g1 x( )

g2 x( )

∫x1

x2

∫ .  

Similarly,  for  D  of  type  II  we  have  E = x, y, z( ) : y1 ≤ y,≤ y2,g1 y( ) ≤ x ≤ g2 y( ),u1 x, y( ) ≤ z ≤ u2 x, y( ){ }  and  

f x, y, z( )dVB∫∫∫ = f x, y, z( )dzdxdy

u1 x,y( )

u2 x,y( )

∫g1 y( )

g2 y( )

∫y1

y2

∫  

 

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Ex  1. Compute

dV

1+ x + y + z( )3G∫∫∫ where G is a pyramid 0,0,0( ); 1,0,0( ); 0,1,0( ); 0,0,1( ) : We

first we sketch the domain and find that 0 ≤ x ≤1, y + x = 1, z + x + y = 1, thus  

dV

1+ x + y + z( )3G∫∫∫ = dzdydx

1+ x + y + z( )3z=0

1−x− y

∫y=0

1−x

∫x=0

1

∫ = − 12

dydx

1+ x + y + z( )2

0

1−x− y

0

1−x

∫0

1

∫ =

− 12

1

1+ x + y +1− x − y( )2 −1

1+ x + y( )2 dy dx0

1−x

∫0

1

∫ = − 12

14− 1

1+ x + y( )2 dy dx0

1−x

∫0

1

∫

= − 12

14

y + 11+ x + y( )

0

1−x

dx0

1

∫ = − 12

1− x4

+ 11+ x +1− x( ) −

11+ x

dx0

1

∫ =

− 12

34− x

4− 1

1+ xdx

0

1

∫ = − 12

34

x − x2

8− ln(1+ x)

⎛⎝⎜

⎞⎠⎟

0

1

= − 12

34− 1

8− ln(2)

⎛⎝⎜

⎞⎠⎟= ln(2)

2− 5

16

 

5.8 Triple Integrals in Cylindrical and Spherical Coordinates (12.8)

Cylindrical  Coordinates:    Recall:  1)  Cylindrical  Coodrinates:   x, y, z( ) = r cosθ ,r sinθ , z( )      

2)  Let  D = r,θ( ) :α ≤θ ≤ β,h1 θ( ) ≤ r ≤ h2 θ( ){ } ,  then   f x, y( )dAD∫∫ = f r cosθ ,r sinθ( )r dr dθ

h1 θ( )

h2 θ( )

∫α

β

∫  

Consider  E = x, y, z( ) : x, y( )∈D,u1 x, y( ) ≤ z ≤ u2 x, y( ){ }  where    D = r,θ( ) :α ≤θ ≤ β,h1 θ( ) ≤ r ≤ h2 θ( ){ } ,    

then   f x, y, z( )dVE∫∫∫ = f x, y, z( )dz

u1 x,y( )

u2 x,y( )

∫⎛

⎝⎜

⎞

⎠⎟ dA

D∫∫ = f r cosθ ,r sinθ , z( )r dzdr dθ

u1 r cosθ ,r sinθ( )

u2 r cosθ ,r sinθ( )

∫h1 θ( )

h2 θ( )

∫α

θ

∫  

Ex 2. Evaluate xzdVG∫∫∫ where

G = x, y, z( ) :0 ≤ x ≤1,− 1− x2 ≤ y ≤ 1− x2 ,0 ≤ z ≤ y{ }We first

sketch the domain, then

Course:  Accelerated  Engineering  Calculus  II   Instructor:  Michael  Medvinsky      

71    

 

xzdVG∫∫∫ = xzdzdydx

z=0

y

∫ =y=− 1−x2

1−x2

∫x=0

1

∫ r cosθ ⋅ z ⋅r dzdr dθz=0

r sinθ

∫r=0

1

∫θ=0

π

∫ =

= r2zcosθ dzdr dθz=0

r sinθ

∫r=0

1

∫θ=0

π

∫ = r2 z2

2cosθ

0

r sinθ

dr dθ0

1

∫0

π

∫ = 12

r4 sin2θ cosθ dr dθ0

1

∫0

π

∫ =

= 110

r5 sin2θ cosθ0

1dθ =

0

π

∫110

sin2θ cosθ dθ0

π

∫ =t=sinθdt=cosθdθ

sinθ=0⇒t=0sinθ=π⇒t=0

110

t 2 dt =0

0

∫ 0

 

Ex 3. Find volume of intersection between balls x2 + y2 + z2 = 9 and x2 + y2 + z − 2( )2 = 9 .

We  need  to  evaluate  V = dVG∫∫∫ ,  but  first  we  have  to  define  G.  We  first  find  the  

intersection  coordinates 0 = z2 − z − 2( )2 = z2 − z2 − 4z + 4( )⇒ 4 = 4z⇒ z = 1 ,  

therefore  on  a  yz  plane  we  have  an  intersection  of  circles   z2 = 9 − r2  on  top  and   z − 2( )2 = 9 − r2 pn  bottom;  and  a  circular  projection  of  radius 8  on  xy  plane.  

V = dVG∫∫∫ = r dzdr dθ

2− 9−r2

9−r2

∫0

8

∫0

2π

∫ = r dzdr dθ2− 9−r2

9−r2

∫0

8

∫0

2π

∫ = dθ 2r 9 − r2 −1( )dr0

8

∫0

2π

∫ =

=t=9−r2dt=−2r

r=0⇒t=9,r= 8⇒t=1

t dt dθ1

9

∫0

2π

∫ − 2r dr dθ0

8

∫0

2π

∫ = t 3/2

3 / 2 1

9

dθ0

2π

∫ − 8dθ0

2π

∫ = 2393/2 −13/2( )− 8⎛

⎝⎜⎞⎠⎟ 2π = 56

3π

 

Spherical  Coordinates:  x, y, z( ) = ρ sinϕ cosθ ,ρ sinϕ sinθ ,ρ cosϕ( )    In  spherical  coordinates  with  domain  E = ρ,θ ,ϕ( ) :ρ1 ≤ ρ ≤ ρ2,θ1 ≤θ ≤θ2,ϕ1 ≤ϕ ≤ϕ2{ }  we  “change”  the  boxes  into  spherical  wedges  Eijk  which  volume  is  approximated  by  rectangular  box  with  volumeΔρ × ρiΔϕ( )× ρi sinϕkΔθ( ) = ρi

2 sinϕkΔρΔθΔϕ .  The  quantities   ρiΔϕ  and   ρi sinϕkΔθ  are  appropriate  arcs  of  the  wedge.  Thus  we  get    

f x, y, z( )dVE∫∫∫ = f ρ sinϕ cosθ ,ρ sinϕ sinθ ,ρ cosϕ( )ρ2 sinϕ dρ dθ dϕ

ρ1

ρ2

∫θ1

θ2

∫ϕ1

ϕ2

∫  

Ex 4. Solve xydzdxdyx2+y2

2− x2+y2( )∫

0

1−x2

∫0

1

∫ z = y2 + x2 and below a sphere

x2 + y2 + z2 = 2 (an ice-cream cone) in first octant. The cone gives

ρ cosϕ = z = y2 + x2 = ρ sinϕ , therefore tanϕ = 1 thus 0 ≤ϕ ≤ π4

and

finally 0 ≤ ρ ≤ 2 . Thus

-­‐3  

1  

z  

y  

Course:  Accelerated  Engineering  Calculus  II   Instructor:  Michael  Medvinsky      

72    

xydzdxdyx2+y2

2− x2+y2( )∫

0

1−x2

∫0

1

∫ = ρ sinϕ cosθ ⋅ ρ sinϕ sinθ ⋅ ρ2 sinϕ dρ dθ dϕ0

2

∫0

π /2

∫0

π /4

∫ =

= ρ 4 sin3ϕ sinθ cosθ dρ dθ dϕ0

2

∫0

π /2

∫0

π /4

∫ = sin3ϕ dϕ0

π /4

∫ ⋅ sinθ cosθ dθ0

π /2

∫ ⋅ ρ 4 dρ0

2

∫ =

= 4 2 − 56 2

⋅ 12⋅ 2

5/2

5= 4 2 − 5

15

ρ 4 dρ0

2

∫ = ρ5

5 0

2

= 25/2

5, sinθ cosθ dθ

0

π /2

∫ = t dt0

1

∫ = t2

2 0

1

= 12,

sin3ϕ dϕ0

π /4

∫ = 1− cos2ϕ( )sinϕ dϕ0

π /4

∫ = − 1− t 2( )dt1

1/ 2

∫ = 1− t 2 dt1/ 2

1

∫ = t − t3

3⎛⎝⎜

⎞⎠⎟ 1/ 2

1

=

= 23− 1

2− 13⋅23/2

⎛⎝⎜

⎞⎠⎟= 23− 1

21− 16

⎛⎝⎜

⎞⎠⎟ =

23− 56 2

= 4 2 − 56 2

 

5.9 Change of Variables in Multiple Integrals (12.9)

We  already  discussed  in  the  class  that  the  changing  of  variables  in  integrals  of  multiply  variables  is   actually   a   change   of   coordinate   system.   We   also   already   saw   that   when   we   changed  coordinates   from   Cartesian   to   polar\cylindrical\spherical   we   were   needed   to   multiply   the  function  by  some  quantity.  We  now  give  a  name  to   this  quantity  and  generalize   the  change  of  coordinates  of  integration  to  an  arbitrary  coordinate  system.  Def,  Jacobian:  Given  a  coordinate  system   u x, y( ),v x, y( )( )  ,  the  Jacobian  is  given  by  determinant  of  matrix  of  partial  derivatives    

Ju,v( ) x, y( ) = ∂ u,v( )

∂ x, y( ) = detux uy

vx vy

⎛

⎝⎜⎜

⎞

⎠⎟⎟= uxvy − uyvx  

Similarly,  for  a  coordinate  system   u x, y, z( ),v x, y, z( ),w x, y, z( )( )  the  Jacobian  is  

Ju,v ,w( ) x, y, z( ) = ∂ u,v,w( )

∂ x, y, z( ) = det

ux uy uz

vx vy vz

wx wy wz

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

 

 

Thm: f u,v( )dudvA∫∫ = f u s,t( ),v s,t( )( ) ∂ u,v( )

∂ s,t( ) dsdtB∫∫  and  similarly    

f u,v,w( )dudvdwA∫∫∫ = f u s,t, p( ),v s,t, p( ),w s,t, p( )( ) ∂ u,v,w( )

∂ s,t, p( ) dsdt dpB∫∫∫  

Course:  Accelerated  Engineering  Calculus  II   Instructor:  Michael  Medvinsky      

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1

Ex 5. J x,y( ) r,θ( ) = cosθ −r sinθsinθ r cosθ

= r cos2θ + r sin2θ = r > 0

Ex 6. J x,y,z( ) r,θ , z( ) =cosθ −r sinθ 0sinθ r cosθ 00 0 1

= cosθ r cosθ − 0( )− sinθ −r sinθ − 0( ) + 0 = r > 0

Ex 7. J x,y,z( ) r,θ ,ϕ( ) =sinϕ cosθ −ρ sinϕ sinθ ρ cosϕ cosθsinϕ sinθ ρ sinϕ cosθ ρ cosϕ sinθcosϕ 0 −ρ sinϕ

=

= cosϕ −ρ sinϕ sinθρ cosϕ sinθ − ρ sinϕ cosθρ cosϕ cosθ( ) +−ρ sinϕ( ) sinϕ cosθρ sinϕ cosθ − sinϕ sinθ −ρ sinϕ sinθ( )( )= ρ2 cosϕ sinϕ cosϕ −sin2θ − cos2θ( )− ρ2 sin3ϕ cos2θ + sin2θ( )= −ρ2 cos2ϕ sinϕ − ρ2 sin3ϕ = −ρ2 sinϕ cos2ϕ + sin2ϕ( ) = −ρ2 sinϕ

⇒ J x,y,z( ) r,θ ,ϕ( ) = −ρ2 sinϕ =0≤ϕ≤π⇒sinϕ≥0

ρ2 sinϕ

 

   

Ex 8.

xy 1− x4 − y4 dAR∫∫ , where R is a quart of the unit circle in the first

quadrant. Define x = r cosθ , y = r sinθ , compute Jacobian:

J =cosθ −r sinθ

2 cosθ

sinθ r cosθ2 sinθ

= cosθ r cosθ2 sinθ

+ sinθ r sinθ2 cosθ

= r(cos2θ + sin2θ )

2 cosθ sinθ= r

2 cosθ sinθ  

xy 1− x4 − y4 dAR∫∫ = r cosθ ⋅r sinθ ⋅ 1− r 4 r

2 cosθ sinθr=0

1

∫θ=0

π 2

∫ drdθ =

= 12

r3 ⋅ 1− r 4

r=0

1

∫θ=0

π 2

∫ drdθ =

t=1−r4

dt=−4r3dr

r=0⇒t=1,r=1⇒t=0

12

t40

1

∫0

π 2

∫ dtdθ = 18

t3 2

3 20

1

0

π 2

∫ dθ = 224

dθ0

π 2

∫ = π24

 

 The  following  theorem  could  be  helpful  in  some  cases.  

Thm:  If   J

u,v( ) x, y( ) ≠ 0  then  the  coordinate  system   x u,v( ), y u,v( )( )  exists  and J

x ,y( ) u,v( ) = 1J

u,v( ) x, y( ) .  

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Similarly,  if J

u,v ,w( ) x, y, z( ) ≠ 0 then  the  coordinate  system   x u,v,w( ), y u,v,w( ), z u,v,w( )( )  exists  and  

J

x ,y ,z( ) u,v,w( ) = 1J

u,v ,w( ) x, y, z( ) .  

Ex 9. Given ( ) ( ), , ,x s t y s t find ( ) ( ), , ,s x y t x y if possible

a. x = as+ bt, y = cs− dt : We require

a bc −d

= − ad + cb( ) ≠ 0 , then

x = as+ bty = cs− dt

⎧⎨⎩

⇒s = 1

a x − bt( )y = c

a x − bt( )− dt = cxa − cb

a + d( )t⎧⎨⎪

⎩⎪⇒

t =cxa − ycba + d

= cx − aycb+ ad

s = xa− b

a⋅ cx − aycb+ ad

= dx + bycb+ ad

⎧

⎨⎪⎪

⎩⎪⎪

b. x = s+ t( )n, y = a s+ t( )m : DNE due

n s+ t( )n−1n s+ t( )n−1

m s+ t( )m−1m s+ t( )m−1

= 0

x = s+ t; y = s2 − t2 :  Under  condition  

1 12s −2t

= −2 s+ t( ) ≠ 0 ,  i.e.   s ≠ −t  we  get  

x = s+ t

y = s2 − t2

⎧⎨⎩⎪

⇒s2 = x2 − 2xt + t2

y = s2 − t2 = x2 − 2xt

⎧⎨⎪

⎩⎪⇒

t = x2 − y2x

s = x − t = 2x2 − x2 + y2x

= x2 + y2x

⎧

⎨⎪⎪

⎩⎪⎪