Mean value theorem for double integral
Let D be an elementary region and f : D → IR be continuous, then for some pt.
( x0 , y 0 ) in D we have
∫∫ f ( x, y)dA
= f ( x0 , y 0 ) • A( D)
D
where A(D) denotes the area of D.
Pf. Since f is continuous on D, it has a maximum value M and a minimum value m.
Thus
m ≤ f ( x, y ) ≤ M for all ( x, y ) ∈ D
∴ mA( D) = ∫∫ mdA ≤ ∫∫ f ( x, y )dA ≤ ∫∫ MdA = M • A( D)
D
i.e. m ≤
D
D
1
f ( x, y )dA ≤ M
A( D ) ∫∫
D
Since a continuous function on D takes on every value between its maximum and
minimum values (Intermediate value theorem)., there must be a pt. ( x0 , y 0 ) ∈ D with
f ( x0 , y 0 ) =
1
f ( x, y )dA
A( D ) ∫∫
D
Triple Integral
Given a continuous function f : C → IR , where C is some rectangular box in IR 3 ,
we can define the integral of f over C as a limit of sums just as we did for a function
of two variables. Briefly, we partition the three sides of C into n equal parts and form
n −1 n −1 n −1
the sum S n = ∑∑∑ f (Cijk )∆V
i =0 j =0 k =0
Where cijk ∈ C ijk , the ijk th rectangular box in the partition of c and ∆V is the
volume of C ijk .
Def. Let f : C → IR be continuous. If lim S n exists and the limit is independent
n→∞
of the pts c ijk , we call the limit of S n the triple integral of f over C and denote it by
∫∫∫ f ( x, y, z)dV
C
Theorem Let f : C → IR be a bounded real-valued function on the rectangular box
C and suppose that f is continuous except for a finite set of pts, then f is integrable
over C.
Fubini’s theorem
Let f be a continuous function with domain
C = [a, b] × [c, d ] × [u , v ] . Then
b d v
∫∫∫ f ( x, y, z )dV
=
∫ ∫ ∫ f ( x, y, z)dzdydx
a c u
C
v b d
=
∫ ∫ ∫ f ( x, y, z)dydxdz
u a c
c v b
= .... =
∫ ∫ ∫ f ( x, y, z)dxdzdy
d u a
A region W is of type I if
W = {( x, y, z ) : a ≤ x ≤ b, φ 2 ( x) ≤ y ≤ φ1 ( x), γ 2 ( x, y ) ≤ z ≤ γ 1 ( x, y )}
where φi , γ i , i = 1,2 are continuous functions.
or W = {( x, y, z ) : c ≤ y ≤ d ,ψ 2 ( y ) ≤ x ≤ ψ 1 ( y ), γ 2 ( x, y ) ≤ z ≤ γ 1 ( x, y )}
where ψ i , γ i , i = 1,2 are continuous functions.
For more general bounded sets W ⊆ IR 3 and f : W → IR a given
Definition
continuous function. Choose a box B that contains W and define
f * ( x, y , z ) =
{
f ( x , y , z ),( x , y , z )∈W
0,( x , y , z )∈B|W
then f * is integrable over B and we can define
∫∫∫ f ( x, y, z )dV = ∫∫∫ f * ( x, y, z )dV
W
B
Suppose W is of type I. Then either
b φ1 ( x ) γ 1 ( x , y )
∫∫∫ f ( x, y, z )dV = ∫
W
a
f ( x, y, z )dzdydz
∫
∫
φ
γ
2 ( x)
2 ( x, y )
γ1 ( x, y )
=
∫∫ γ ∫ f ( x, y, z )dz )dydx
D
2 ( x, y )
e.g. Let W be the region bounded by the planes x=0, y=0, z=2 and the surface
z = x 2 + y 2 , x ≥ 0, y ≥ 0 . Compute
∫∫∫ xdxdydz
W
Sol. Let γ 1 ( x, y ) = 2, γ 2 ( x, y ) = x 2 + y 2
φ1 ( x) = 2 − x 2 , φ 2 ( x) = 0
a=0 and b = 2 . Then W is a region of type I.
1
2
2 ( 2− x ) 2
∫∫∫ xdxdydz
∫[ ∫
=
W
0
2
∫ xdz)dy]dx
(
x2 + y2
0
1
2
2 ( 2− x ) 2
∫ [ ∫ ( x(2 − x
=
0
− y 2 )dydx
2
0
3
3
2
2
(2 − x 2 ) 2
= ∫ x[(2 − x 2 ) −
]dx
3
0
3
2
2
= ∫ x(2 − x 2 ) 2 dx
3
0
3
2
= −
1
(2 − x 2 ) 2 d (2 − x 2 )
3 ∫0
− 2(2 − x )
15
2
=
= 2(
5
2
2
0
5
2
2
8 2
)=
15
15
2
2
2
∫∫∫ ( x + y + z )dV , where W = {( x, y, z ) :
e.g. Evaluate I =
W
x2 y2 z2
+
+
≤ 1}
a2 b2 c2
Sol : Let D z0 be the region obtained by the intersection of W with a plane // to x-y
plane
that
D z0 = {( x, y ) :
passes
x
∫∫∫ x
2
W
2
0
2
z
a (1 − )
c
2
I=
2
+
through
y
2
z2
b (1 − 02 )
c
dV + ∫∫∫ y 2 dV + ∫∫∫ z 2 dV = I 1 + I 2 + I 3
w
c
W
c
I 3 = ∫ ( ∫∫ z dxdy )dz = ∫ z 2πab(1 −
2
− c Dz
≤ 1}
2
−c
z2
4
)dz = πabc 3
2
15
c
(0,0, z 0 )
.
Then
(Q ∫∫ dxdy = Area of D z = πab(1 −
Dz
Similarly, I 1 =
z2
))
c2
4 3
4
πa bc, I 2 = πab 3 c
15
15
b
∴I =
4
πabc (a 2 + b 2 + c 2 ) = ∫ ( ∫∫ y 2 dxdz )dy
15
−b D y
Maps from IR 2 to IR 2
π
Let D* = [a, b] × [0, ] where 0<a<b
2
Define T ( r , θ ) = ( r cos θ , r sin θ ) on D*, then
e.g.1
T ( D)* = D = {( x, y ) : a 2 ≤ x 2 + y 2 ≤ b 2 , x ≥ 0, y ≥ 0}
e.g.2
T (u , v ) = (−u 2 + 4u , v) maps the square D* onto the rectangle D
= (x(u,v),y(u,v))
Change of variable theorem
Def.Let T : D* ⊆ IR 2 → IR 2 be a C’ transformation given by x=x(u, v) and y-y(u, v).
∂x
∂ ( x, y ) ∂u
=
The Jacobian of T is
∂ (u, v) ∂y
∂u
∂x
∂v = det[ DT (u , v)]
∂y
∂v
e.g. The function from IR 2 → IR 2 that transforms polar co-ordinate into Cartesian
co-ordinate is given by x = r cos θ , y = r sin θ
∂ ( x, y ) cos θ − r sin θ
and its Jacobian is
=
=r
∂ (u , v) sin θ r cos θ
Theorem Let D and D* be elementary regions in IR 2 and let T : D* → D be C’,
suppose that T is
onto. Then for any integrable function f : D → IR , we have
∂ ( x, y )
∫∫ f ( x, y)dxdy = ∫∫ f ( x(u, v), y(u, v) ∂(u, v) dudv
D
D*
Pf.
 ∂x

v
DT (∇ui ) =  ∂u
 ∂y

 ∂u
∂x 
 ∂x 
 ∆u 
 
∂v   = ∆u  ∂u 
∂y  0 
 ∂y 

 
∂v 
 ∂u 
 ∂x

v
DT (∇vj ) =  ∂u
 ∂y

 ∂u
∂x 
 ∂x 
 0 
 
∂v   = ∆v ∂v 
∂y  ∆v 
 ∂y 

 
∂v 
 ∂v 
Area of T(R*) is approximately equal to the absolute value of
∂x
∆u
∂u
∂y
∆u
∂u
∂x
∂x
∆v
∂v
= ∂u
∂y
∂y
∆v
∂v
∂u
∂x
∂v ∆u∆v = ∂ ( x, y ) ∆u∆v =
∂y
∂ (u, v)
∂v
e.g Let P be the //gram bounded by y=2x,
∫∫ ( x, y)dxdy
y=2x-2,
y=x and
by making the change of variables : x=u-v,
y=x +1. Evaluate
y=2u-v
D
Solution :
T(u, v) = (u-v, 2u-v)
∴
Hence
 1 − 1
∂ ( x, y )
 = 1
= det
2
1
−
∂ (u, v)


∫∫ xydxdy
=
P
∫∫ (u − v)(2u − v)dudv
P*
0 1
∫ ∫ (2u
=
2
− 3uv + v 2 )dudv
−2 0
0
2 3 3u 2 v
= ∫[ u −
+ v 2 u ]10 dv
3
2
−2
2
3
v3
= [ v − v 2 + ]0−2
3
4
3
2
8
= − [ (−2) − 3 − ] = 7
3
3
e.g. Evaluate
∫∫ ln( x
2
+ y 2 )dxdy , where
D
D = {( x, y ) : a 2 ≤ x 2 + y 2 ≤ b 2 , x ≥ 0, y ≥ 0}
(0 < a < b )
Solution Let x = r cos θ , y = r sin θ
i.e. T ( r , θ ) = ( x, y )
Then T : D* → D is 1-1 onto, where D* = {( r , θ ) : a ≤ r ≤ b,0 ≤ θ ≤
π
}
2
∴ ∫∫ ln( x 2 + y 2 )dxdy = ∫∫ lnr 2
D*
D
π
b 2
= ∫ ∫ rlnr 2 drdθ =
a 0
=
∂ ( x, y )
drdθ
∂ (r ,θ )
πb
2rlnrdr
2 ∫a
1
π 2
[b lnb − a 2 lna − (b 2 − a 2 )]
2
2
x2
x2
(Using integration by parts : ∫ xlnxdx = lnx − )
2
4
e.g. Evaluate
∫∫
x 2 + y 2 dxdy
, where R = [0,1] × [0,1]
R
Solution Let D1 * = {( r , ϑ ) : 0 ≤ θ ≤
π
and 0 ≤ r ≤ sec θ },
4
π
π
≤ θ ≤ and 0 ≤ r ≤ csc θ }
4
2
T ( r , θ ) = ( r cos θ , r sin θ ) , then
T ( D1 *) = T1 , T ( D2 *) = T2 , R = T1 ∪ T2 = T ( D*)
D2 * = {( r , ϑ ) :
Define
We have to find the volume of the region under z = x 2 + y 2 and over R.
From symmetry of z = x 2 + y 2 on R, we can see that
∫∫
x 2 + y 2 dxdy
= 2∫∫ x 2 + y 2 dxdy
π
R
= 2 ∫∫ ( r 2 )rdrdθ
D1 *
π
4 sec θ
= 2 ∫ ( ∫ r 2 dr )dθ
0
0
π
4
=
π
4
∫ sec
3
θ dθ = [
0
2
sec 3 θ dθ
3 ∫0
π
4
π
4
0
sec θ tan θ
1
] secθdθ + ∫ sec θdθ
2
20
π
=
∴ ∫∫
R
2 1
2 1
+ [ln sec θ + tan θ ]04 =
+ ln(1 + 2 )
2
2
2
2
1
x 2 + y 2 dxdy = [ 2 + ln(1 + 2 )]
3
Def. Let T : W ⊆ IR 3 → IR 3 be a C 1 function defined by x=x(u, v, w), y=y(u, v, w),
z=z(u, v ,w) . Then the Jacobian of T is
∂x ∂x ∂x
∂u ∂v ∂w
∂ ( x, y, z ) ∂y ∂y ∂y
=
∂ (u , v, w) ∂u ∂v ∂w
∂z ∂z ∂z
∂u ∂v ∂w
Change of variable formula
∂ ( x, y , z )
∫∫∫ f ( x, y, z )dxdydz = ∫∫∫ f ( x(u, v, w), y(u, v, w), z (u, v, w) ∂(u, v, w) dudvdw
D
D*
where D* is an elementary region in uvw-space correspond to D in xyz-space, under a
co-ordinate change T : (u , v, w) → ( x(u , v, w), y (u , v, w), z (u , v, w))
(i)
Jacobian for the map defining the change to cylindrical co-ordinates
x = r cos θ , y = r sin θ , z = z
cos θ
∂ ( x, y , z )
= sin θ
∴
∂ (r ,θ , z )
0
− r sin θ
r cos θ
0
0
0 =r
1
∴ ∫∫∫ f ( x, y, z )dxdydz = ∫∫∫ f (r cosθ , r sin θ )rdrdθdz
D
(ii)
D*
Consider the spherical co-ordinate system given by
x = ρ sin φ cos θ , y = ρ sin φ sin θ , z = ρ cos φ
sin φ cos θ
∂ ( x, y , z )
= sin φ sin θ
∂( ρ ,θ , φ )
cos φ
= cos φ
− ρ sin φ sin θ
ρ sin φ cos θ
− ρ sin φ sin θ
ρ sin φ cos θ
0
ρ cos φ cos θ
ρ cos φ cos θ
− ρ sin φ
sin φ cos θ
ρ cos φ cos θ
− ρ sin φ
sin φ sin θ
ρ cos φ sin θ
− ρ sin φ sin θ
ρ sin φ cos θ
(expand along the last row)
= − ρ 2 cos 2 φ sin φ sin 2 θ − ρ 2 cos 2 φ sin φ cos 2 θ − ρ 2 sin 3 φ cos 2 θ − ρ 2 sin 3 φ sin 2 θ
= − ρ 2 cos 2 φ sin φ − ρ 2 sin 3 φ
= − ρ 2 sin φ
∴ ∫∫∫ f ( x, y, z )dxdydz
D
=
∫∫∫ f ( ρ sin φ cosθ , ρ sin φ sin θ , ρ cos φ ) ρ
D*
2
sin φdρdθdφ
3
e.g. Evaluate ∴ ∫∫∫ exp( x 2 + y 2 + z 2 ) 2 dV , where D = {( x, y, z ) : x 2 + y 2 + x 2 ≤ 1}
D
Sol. Use spherical co-ordinates, then D* = {( ρ , θ , φ ) : 0 ≤ ρ ≤ 1,0 ≤ θ ≤ 2π ,0 ≤ φ ≤ π }
3
∴ ∫∫∫ exp( x 2 + y 2 + z 2 ) 2 dv = ∫∫∫ ρ 2 e ρ sin φdρdθdφ
3
D
D*
1 π 2π
=
2 ρ
∫ ∫ ∫ ρ e sin φdθdφdρ
3
0 0 0
1π
= 2π ∫ ∫ e ρ ρ 2 sin φdφdρ
3
0 0
1
= − 2π ∫ ρ 2 e ρ [cos φ ]π0 dρ
3
0
1
= − 4π ∫ ρ 2 e ρ dρ =
0
=
3
4π ρ 3 1
[e ] 0
3
4π
(e − 1)
3
e.g. Let D = {( x, y, z ) : x 2 + y 2 + z 2 ≤ R 2 } be the ball of radius R and center
(0,0,0) ∈ IR 3
then Vol(D) = ∫∫∫ dxdydz =
D
=
π 2π R
∫ ∫∫ρ
2
sin φdρdθdφ
0 0 0
R3
3
π 2π
∫ ∫ sin φdθdφ
0 0
π
2πR 3
sin φdφ
=
3 ∫0
=
2πR 3
[− cos φ ]π0
3
=
4πR 3
(vol. of solid sphere)
3
Integrals over paths
Def. The path integral or the integral of f(x, y, z) along the path σ : I = [a, b] → IR 3
is defined by
b
fds ≡ ∫ f ( x(t ), y (t ), z (t )) σ ' (t ) dt
∫
σ
a
where σ ' (t ) = x' (t ) 2 + y ' (t ) 2 + z ' (t ) 2 ,
ds = elt. of arc length
e.g. Let σ : [0,2π ] → IR 3 be the helix given by t → (cos t , sin t , t ) and let
f ( x, y, z ) = x 2 + y 2 + z 2 , then
σ ' (t )
d
d
dt
cos t ] 2 + [ sin t ] 2 +( ) 2
dt
dt
dt
=
[
=
sin 2 t + cos 2 t + 1 = 2
f (σ (t ) = cos 2 t + sin 2 t + t 2 = 1 + t 2
∴ ∫ f ( x, y, z )ds =
σ
2π
2
∫ (1 + t ) 2dt = 2[t +
0
t 3 2π
]0
3
2 2π
(3 + 4π 2 )
3
If we think of the helix as a wire and f ( x, y, z ) = x 2 + y 2 + z 2 as the mass density,
=
2 2π
(3 + 4π 2 ) .
3
then the total mass of the wire is
Limit definition
Subdivide the interval I = [ a, b] by a partition
a = t 0 < t1 < ... < t n = b
This leads to a decomposition of σ into paths σ i defined on [t i , t i +1 ] for
0 ≤ i ≤ n − 1 . Denote the arc length of σ i by ∆S i .
i.e. ∆S i = ∫
t i +1
ti
σ ' (t ) dt
When n is large, the arc length ∆S i is small and f(x, y, z) is approximately constant
for points on σ i . Then
∴ ∫ f ( x, y, z )ds
σ
n −1
= lim ∑ f ( x, y, z )∆S i
n→∞
i =0
n −1
= lim ∑ ∫
n→∞
b
=
i =0
ti +1
ti
f ( x(t ), y (t ), z (t ) γ ' (t ) dt
∫ f ( x, y, z) γ ' (t ) dt
a
when σ (t ) = ( x (t ), y (t )) is a plane curve and f ( x, y ) ≥ 0 . We can construct a
“fence” with base the image of I=[a, b] under σ and with altitude f(x, y) at (x, y).
Then
b
∫ f ( x, y)ds
σ
=
∫ f ( x(t ), y(t )
x' (t ) + y ' (t )dt
a
represents the area of a side of this fence.
e.g. Evaluate I = ∫ ( x + y )ds
σ
where σ is the line segments joining (0,0), (1,0) and (1,1)
I =(∫ +
Sol.
OA
∫
+
AB
∫
)( x + y )ds
BO
y=0 on OA and ds = dx,
1
∴ ∫ ( x + y )ds = ∫ xdx =
0
OA
1
2
x=0 on AB and ds = dy,
∴
∫
AB
1
( x + y )ds = ∫ (1 + y )dy =
0
3
2
y=x on BO and ds = − 2dx
∴
∫
BO
1
( x + y )ds = ∫ − 2 x 2dx = − 2
0
∴I = 2− 2
“The adventures of Tow Sawyer” Mark Twain
at 2
e.g. Find the total mass of a wire σ (t ) = (a, at ,
) 0 ≤ t ≤ 1, a > 0 with linear
2
density ρ ( x, y, z ) =
2z
a
Sol. ds = a 1 + t 2 dt
1
a
∴ m = ∫ ρ ( x, y, z )ds = ∫ at 1 + t 2 dt = (2 2 − 1)
3
0
σ
e.g. Tom Sawyer’s aunt has ask him to whitewash both sides of the old fence shown.
Tow estimated that for each 25 sq. ft of whitewashing he lets someone do for him, the
willing victim will pay 5 cents. How much can Tom hope to earn, assuming his aunt
will provide whitewash free of charge?
Solution σ ' (t ) = (−90 cos 2 t sin t ,90 sin 2 t cos t )
∴ σ ' (t ) = 90 sin t cos t
Area of one side of half of the fence
y
= ∫ (1 + )ds =
3
σ
π
2
30 sin 3 t
∫0 (1 + 3 )90 sin t cos tdt
π
2
= 90 ∫ ( sin t + 10 sin 4 t ) cos tdt
0
π
sin 2 t
= 90[
+ 2 sin 5 t ]02
2
1
= 90( + 2) = 225
2
Hence, area of one side of fence is 450 sq.ft.
Since both sides are to be whitewashed, the total area is 900sq.ft. Thus the amount
that Tom could realize for the job = $(
900 5
)(
) = $ 1 .8
25 100