1
1.1
Double Integral
Double Integral over Rectangular Domain
As the definite integral of a positive function of one variable represents the area
of the region between the graph and the x-asis, the double integral of a positive
function of two variables represents the volume of the region between the surface
defined by the function z = f (x, y) and the xy-plane which contains its domain.
We start with rectangular domain D = {[x, y] ∈ R2 : x ∈ ha, bi, y ∈ hc, di} on
the xy−plane according to Fig 1.
We devide interval ha, bi, (resp. hc, di) by sequences of points a = x0 < x1 <
Fig. 1. Double integral on rectangular domain
x2 < . . . < xm = b, (resp. c = y0 < y1 < y2 < . . . < yn = d) to intervals
hxi−1 , xi i, i = 1, 2, . . . , m, (resp. hyj−1 , yj i, j = 1, 2, . . . , n). We denote sizes
of each component ∆xi = xi − xi−1 , ∆yj = yj − yj−1 . This way is the whole
rectangular domain divided into m · n small rectangles with area ∆Dij = ∆xi ·
∆yj . Now we can choose an arbitrary point [ξi , ηj ] in each rectangle Dij and we
can evaluate the volume of a prism with basis Dij and and height z = f (ξi , ηj ).
m P
n
P
The sum of the volumes
f (ξi , ηj ) · ∆xi · ∆yj represents the volume of the
i=1 j=1
body consisted of such prisms over all rectangles Dij if f (x, y) ≥ 0 on D.
Definition 1.1.
m P
n
P
If there exists lim
f (ξi , ηj ) · ∆xi · ∆yj for m → ∞, n → ∞, ∆xi →
i=1 j=1
0, ∆yj → 0 for all i = 1, 2, . . . , m, j = 1, 2, . . . , n, we call itRR
double integral of
function f (x, y) over the rectangular domain D and denote it
f (x, y)dxdy.
D
1
Theorem 1.1. (Dirichlet’s)
Let D = {[x, y] ∈ R2 : x ∈ ha, bi, y ∈ hc, di}. If function f (x, y) is continuous
on rectangle D, then




Zd Zb
Zb Zd
ZZ
(1)
f (x, y)dxdy =  f (x, y)dy  dx =  f (x, y)dx dy.
c
a
D
a
c
In fact there are two ways of computing the double integral. If the inner differential is dy then the limits on the inner integral must have y limits of integration
and outer integral must have x limits of integration. We compute the integral
Rd
f (x, y)dy by holding x constant and integrating with respect to y as if this were
c
a single integral (similar approach is used for partial derivatives of function of
more than one variable). This will result as a function of a single variable x which
we can integrate once again. We use similar approach for the second way of computing of the doubleintegral. Rb Rd
Rb Rd
Rd Rb
We usually write
f (x, y)dy dx = dx f (x, y)dy and
f (x, y)dx dy =
a
Rd
dy
c
Rb
c
a
c
c
a
f (x, y)dx.
a
Theorem 1.2. (Properties of the double integral over a rectangular domain)
RR
RR
1.
cf (x, y)dxdy = c f (x, y)dxdy,
D
2.
RR
D
(f (x, y) + g(x, y))dxdy =
D
3.
RR
RR
f (x, y)dxdy +
D
f (x, y)dxdy =
D
RR
RR
g(x, y)dxdy,
D
f (x, y)dxdy +
D1
RR
f (x, y)dxdy,
D2
where f (x, y), g(x, y) are continuous functions on D, c ∈ R and D1 , D2 are rectangles that fullfil D = D1 ∪ D2 .
Example RR
1.1.
Compute
6xy 2 dxdy over the domain D = {[x, y] : 2 ≤ x ≤ 4, 1 ≤ y ≤ 2}.
D
Solution: We will show both ways of the computing.
RR
R4 R2
R4
R4
R4
2
a)
6xy 2 dxdy = dx 6xy 2 dy = [2xy 3 ]1 dx = (16x−2x)dx = 14xdx =
D
=
4
[7x2 ]2
2
1
2
2
= 112 − 28 = 84.
2
2
b)
RR
=
R2
6xy 2 dxdy =
R2
dy
R2
4
[3x2 y 2 ]2 dy = (48y 2 − 12y 2 )dy =
R2
6xy 2 dx =
1
1
2
1
D
R4
2
36y 2 dy = [12y 3 ]1 = 96 − 12 = 84.
1
If the integrand f (x, y) can be writen as a multiplication of two functions of one
variable f (x, y) = f1 (x) · f2 (y), it holds:
Z
ZZ
b
Z
d
f1 (x)dx ·
f (x, y)dxdy =
a
D
f2 (y)dy.
(2)
c
Example 1.2.
Compute the integral from Example 1.1. by using equation (2).
Solution:
RR
R4
R2
2
4
6xy 2 dxdy = 6 x dx · y 2 dy = [x2 ]2 · [y 3 ]1 = 12 · 7 = 84.
2
D
1
If the decomposition is not possible, we can always use Dirichlet’s Theorem and
relation (1).
Example RR
1.3.
Compute (2x − 4y)dxdy over the domain D = {[x, y] : 1 ≤ x ≤ 3, −1 ≤ y ≤
D
1}.
Solution:
RR
(2x − 4y)dxdy =
=
dx
R1
(2x − 4y)dy =
−1
1
D
R3
R3
R3
1
[2xy − 2y 2 ]−1 dx =
1
3
4x dx = [2x2 ]1 = 16.
1
Although generally the order of integration doesn’t matter, in some cases the integral can be easily solved by using one way of integration while it can be rather
complicated using the other way. Everything depends on the integrand f (x, y)
itself and the limits of integral.
Example RR
1.4.
Compute
xy dxdy over the domain D = {[x, y] : 0 ≤ x ≤ 1, 1 ≤ y ≤ 2}.
D
Solution: We will use both ways of the computing to show the situation.
RR y
R2 R1
R2 h xy+1 i1
R2 1
R2 dy
0
a)
x dxdy = dy xy dx =
dy
=
−
dy
=
=
y+1
y+1
y+1
y+1
1
D
= [ln |y +
1|]21
0
0
1
= ln 3 − ln 2 = ln
3
.
2
3
1
1
b)
RR
xy dxdy =
R1
0
D
dx
R2
xy dy =
R1 xy 2
0
1
ln x
dx =
1
R1 x2
ln x
0
−
x
ln x
dx = . . .
The further computation of the integral is tough.
Exercises
Compute the following integrals over their domains D.
RR 2
1.
(x + y 2 ) dxdy, D = {−2 ≤ x ≤ 0, −1 ≤ y ≤ 2}.
[14]
D
2.
RR p
x x2 + y dxdy,
D = {0 ≤ x ≤ 1, 0 ≤ y ≤ 3}.
D
3.
RR
D = {0 ≤ x ≤ π,
sin(2x + y) dxdy,
D
4.
RR
D
1.2
1
(x+y+1)2
dxdy,
π
4
√
2
[ 15
(31 − 9 3)]
≤ y ≤ π}.
[0]
[ln 34 ]
D = {0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.
Double Integral over General Domain
There is no reason to limit our problem to rectangular regions. The integral domain can be of a general shape. We extend the Riemann’s definition of the double
integral over rectangular domain to a closed connected bounded domain Ω without any problem. The domain is connected if we can connect every two points
from it by curve that lies within the domain. We can always find a rectangle D
that fullfils Ω ⊆ D (see Fig. 2.) and we can define function f ∗ (x, y) by

 f (x, y) ∀ [x, y] ∈ Ω,
f ∗ (x, y) =
 0
∀ [x, y] ∈ D \ Ω.
Then it holds
RR
f (x, y) dxdy =
Ω
RR
f ∗ (x, y) dxdy.
D
The properties of the double integral over a general domain must correspond to
Theorem 1.2.
Theorem 1.3. (Properties of the double integral over a general domain)
RR
RR
1.
cf (x, y)dxdy = c f (x, y)dxdy,
Ω
2.
RR
Ω
(f (x, y) + g(x, y))dxdy =
Ω
3.
RR
Ω
RR
f (x, y)dxdy +
Ω
f (x, y)dxdy =
RR
RR
g(x, y)dxdy,
Ω
f (x, y)dxdy +
Ω1
RR
Ω2
4
f (x, y)dxdy,
Fig. 2. Extension to closed connected domain Ω.
where f (x, y), g(x, y) are continuous functions on Ω, c ∈ R and Ω1 , Ω2 are domains that fullfil Ω = Ω1 ∪ Ω2 .
There are two types of domains we need to look at.
Definition 1.2.
1. Normal domain with respect to the x−axis is bounded by lines x =
a, x = b, where a < b and continuous curves y = g1 (x), y = g2 (x),
where g1 (x) < g2 (x) for all x ∈ ha, bi.
2. Normal domain with respect to the y−axis is bounded by lines y =
c, y = d, where c < d and continuous curves x = h1 (y), x = h2 (y),
where h1 (y) < h2 (y) for all y ∈ hc, di.
Fig. 3. a) Normal domain with respect to the x−axis, b) Normal domain with
respect to the y−axis
5
Theorem 1.4. (Fubini’s)
1. If the function f (x, y) is continuous on a normal domain with respect to the
x−axis, then it holds
gZ2 (x)
Zb
ZZ
f (x, y) dxdy =
a
Ω
f (x, y) dy.
dx
(3)
g1 (x)
2. If the function f (x, y) is continuous on a normal domain with respect to the
y−axis, then it holds
hZ2 (y)
Zd
ZZ
f (x, y) dxdy =
dy
c
Ω
f (x, y) dx.
(4)
h1 (y)
Example 1.5.
RR
Determine integration limits for
f (x, y) dxdy over the domain Ω, which is
Ω
bounded by curves y = x2 and x = y 2 .
Solution: We need to find intersections of curves y = x2 and x = y 2 by solving
the system of these two equations. We can eliminate variable y, receive one equation x4 − x = 0 and solve it. We obtain real values x1 = 0, x2 = 1 and write
the domain Ω as a normal domain with respect √
to the x−axis. It is bounded by
2
lines x = 0, x = 1 and curves y = x and y = x (see Fig. 4.) We can express
Fig. 4. Domain Ω in Example 1.5.
inequations for Ω in the form:
0 ≤ x ≤ 1,
√
x2 ≤ y ≤
x
6
and by Fubini’s Theorem
RR
f (x, y) dxdy =
R1
√
dx
0
Ω
Rx
f (x, y) dy.
x2
We can use a similar procedure and express the integral as an integral over
normal domain with respect to the y−axis with inequations
0 ≤ y ≤ 1,
√
y.
y2 ≤ x ≤
The double integral then takes form
RR
f (x, y) dxdy =
R1
√
dy
0
Ω
Ry
f (x, y) dx.
y2
Example 1.6.
RR
Determine integration limits for
f (x, y) dxdy over the domain Ω, which is a
Ω
triangle bounded by equations y = 1, y = x + 4, y = 6 − x.
Solution: First, we express the domain as normal with respect to the x−axis. If
we bound the domain by −3 ≤ x ≤ 5, the upper limit of inner integral can’t be
written as one curve and we need to divide the domain Ω into two subdomains
Ω1 , Ω2 by line x = 1 (see Fig. 5.) with inequations
Ω1 : −3 ≤ x ≤ 1,
Ω2 :
1 ≤ y ≤ x + 4,
1 ≤ x ≤ 5,
1 ≤ y ≤ 6 − x.
Fig. 5. Domain Ω in Example 1.6.
Using Fubini’s Theorem we can express
Z1
ZZ
f (x, y) dxdy =
Ω
−3
Zx+4
Z5
Z6−x
dx
f (x, y) dy + dx
f (x, y) dy.
1
1
7
1
However, it is much better to express the domain as normal with respect to the
y−axis. There is no reason to split the domain which is now bounded by inequations
1 ≤ y ≤ 5,
y − 4 ≤ x ≤ 6 − y,
where we have expressed a variable x from boundary equations. The integral is
written in the form
Z5
ZZ
f (x, y) dxdy =
Z6−y
dy
f (x, y) dx.
1
Ω
y−4
Example RR
1.7.
√
Compute
xy dxdy over the domain Ω bounded by y = x2 , y = x, x ≥ 2.
Ω
√
Solution: By solving the system of equations y = x2 , y = x we receive intersections of both curves in x = 0, x = 4 according to Fig. 6. It is better to express the
Fig. 6. Domain Ω in Example 1.7.
domain as normal with respect to x−axis with boundaries
2 ≤ x ≤ 4,
√
x
≤y≤
x
2
and compute the integral:
√
√
RR
R4
Rx
R4 h y2 i x
R4 x2
xy dxdy = dx xy dy = x 2
dx =
−
2
Ω
2
x/2
x/2
2
8
2
x3
8
dx =
h
x3
6
x4
32
i4
=
−
= 11
.
6
2
The second approach requires splitting the domain into two subdomains. It is a
good exercise to compute the example this way.
Example 1.8.
RR
Compute
(2x − y 2 ) dxdy over the domain Ω bounded by y = 1 − x, y =
Ω
x + 1, y = 3.
Solution: According to Fig.7. we compute the integral over the normal domain
Fig. 7. Domain Ω in Example 1.8.
with respect to y−axis. Limits of the integral:
1 ≤ y ≤ 3,
1 − y ≤ x ≤ y − 1.
Computation:
y−1
RR
R
R3
R3
y−1
(2x − y 2 ) dxdy = dy (2x − y 2 ) dx = [x2 − xy 2 ]1−y dy =
1
Ω
1−y
1
R3
[(y − 1)2 − (y − 1)y 2 − (1 − y)2 + (1 − y)y 2 ] dy =
i3
h1 4
y
2 3
= − 2 + 3y
= − 68
.
3
=
R3
(−2y 3 + 2y 2 ) dy =
1
1
Exercises
Compute the following integrals over their domains Ω.
RR
1.
(5x2 − 2xy) dxdy, where Ω is triangle ABC, where A = [0, 0], B =
Ω
[2, 0], C = [0, 1].
[3]
9
2.
RR
x2 dxdy, where Ω is bounded by curves y =
Ω
3.
RR
16
,
x
y = x, x = 8.
6xy dxdy, where Ω is bounded by curves y = 0, x = 2, y = x2 .
[576]
[32]
Ω
4.
RR
5.
RR
xy dxdy, where Ω is given by x2 + 4y 2 ≤ 4, x ≥ 0, y ≥ 0.
[ 21 ]
(3x − 2y) dxdy, where Ω is given by x2 + y 2 ≤ 4.
[0]
Ω
Ω
1.3
Double Integrals in Polar Coordinates
At this moment we are able to compute the double integral over an arbitrary domain. In this section we want to look at some domains that are easier to describe
in terms of polar coordinates. We might have a domain that is a disc, ring or part
of a disc or ring. Let us consider a double integral of an arbitrary function over
the disc with the center in the origin of coordinates and the radius r = 2 (same
domain that is used in last Exercise No.5). Using Cartesian coordinates we obtain
limits of the integral
−2 ≤ x ≤ 2,
√
√
− 4 − x2 ≤ y ≤
4 − x2 .
and by Fubini’s Theorem the integral can be written in the form
√
Z4−x2
Z2
ZZ
f (x, y) dxdy =
dx
−2
Ω
f (x, y) dy.
√
− 4−x2
In these cases using Cartesian coordinates can be tedious. However, we are able
to replace Cartesian coordinates x, y by polar coordinates ρ, ϕ, where ρ denotes
a distance between the point [x, y] and the origin of coordinates and is called a
radius, and ϕ denotes the positively oriented angle between positive part of the
x−axis and the radius vector and is called angular coordinate or azimuth (see
Fig. 8). Hence, we obtain transformation equations
x = ρ cos ϕ,
y = ρ sin ϕ.
(5)
Transformation to polar coordinates is a special case of mapping region Ω onto Ω∗
that is an image of Ω in polar coordinates in our case. For example a disc with the
center in the origin of coordinates and the radius r = 2, Ω : {[x, y] : x2 +y 2 ≤ 4},
is mapped onto Ω∗ : {[ρ, ϕ] : ρ ∈ (0, 2i, ϕ ∈ h0, 2π)}.
10
Fig. 8. Polar Coordinates ρ, ϕ.
Theorem 1.5.
• Let equations x = u(r, s), y = v(r, s) map the region Ω bijectively to the
region Ω∗ .
• Let function f (x, y) be continuous and bounded on Ω and functions x =
u(r, s), y = v(r, s) have continuous partial derivatives on Ω̂ that fulfills
Ω∗ ⊂ Ω̂.
∂u ∂u ∂r ∂s 6= 0 in Ω∗ .
• Let J(u, v) = ∂v
∂v
∂r ∂s Then
ZZ
ZZ
f (x, y)dxdy =
(6)
Ω∗
Ω
Determinant J(u, v) = f (u(r, s), v(r, s))|J(u, v)|drds
∂u
∂r
∂u
∂s
∂v
∂r
∂v
∂s
is called Jacobian or Jacobi determinant.
We will use this theorem for transformation of the double integral to polar coordinates as well as the triple integral to cylindrical and spherical coordinates.
According to Theorem 1.5. we replace square element dxdy by |J| dρ dϕ,
where the Jacobian of the transformation to polar coordinates satisfies
∂x ∂x ∂ρ ∂ϕ cos ϕ −ρ sin ϕ =
= ρ.
J(ρ, ϕ) = ∂y ∂y
sin ϕ ρ cos ϕ ∂ρ
∂ϕ
11
Transformation of the double integral to polar coordinates can then be written in
the form
ZZ
ZZ
f (ρ cos ϕ, ρ sin ϕ)ρ dρ dϕ.
(7)
f (x, y)dxdy =
Ω
Ω∗
Example RR
1.9.
Compute
y dxdy over the domain Ω : x2 + y 2 ≤ 9, y ≥ 0 using transformation
Ω
to polar coordinates.
Solution: The domain Ω is an upper half of the disc with the center in the origin
of coordinates and radius r = 3. We use transformation to polar coordinates and
the domain Ω∗ = {[ρ, ϕ] : ρ ∈ (0, 3i, ϕ ∈ h0, πi}. Using (7) we obtain
h 3 i3
RR
RR
R3
Rπ
y dxdy =
ρ sin ϕ ρ dρ dϕ = ρ2 dρ sin ϕ dϕ = ρ3 · [− cos ϕ]π0 = 18.
Ω∗
Ω
0
0
0
Example RR
1.10.
Compute
x dxdy over the domain Ω : 4 ≤ x2 + y 2 ≤ 9, y ≥ x, x ≥ 0.
Ω
Solution: We can see real advantage of the transformation on this domain (see
Fig. 9). While using Cartesian coordinates would be complicated, domain Ω∗ =
{[ρ, ϕ] : ρ ∈ h2, 3i, ϕ ∈ h π4 , π2 i} for polar coordinates is rectangular.
Fig. 9. Domain Ω in Example 1.10.
h 3 i3
π/2
RR
R3
R
π/2
x dxdy =
ρ cos ϕ ρ dρ dϕ = ρ2 dρ cos ϕ dϕ = ρ3 · [sin ϕ]π/4 =
2
2
Ω
Ω∗
π/4
√ = 19
1 − 22 .
2
RR
12
Example 1.11.
Calculate limits of the integral transformed to polar coordinates for the domain
Ω : x2 + y 2 ≤ 2ax.
Solution: First, we find the center and radius of the disc.
x2 + y 2 ≤ 2ax ⇒ x2 − 2ax + a2 + y 2 ≤ a2 ⇒ (x − a)2 + y 2 ≤ a2 .
We have found that center S = [a, 0] and radius r = a (see Fig 10). The azimuth
Fig. 10. Domain Ω in Example 1.11.
must fulfill − π2 ≤ ϕ ≤ π2 . We can see that the upper limit of coordinate ρ depends
on the azimuth ϕ. We obtain the value of the limit by puting transformation equations (5) to boundary equations of Ω :
x2 + y 2 = 2ax ⇒ ρ2 cos2 ϕ + ρ2 sin2 ϕ = 2aρ cos ϕ ⇒ ρ2 = 2aρ cos ϕ ⇒
⇒ ρ(ρ − 2a cos ϕ) = 0 ⇒ ρ1 = 0, ρ2 = 2a cos ϕ.
We have just obtained the limits of the integral. However, it is neccessary to realise the dependancy of coordinate ρ on coordinate ϕ. We can’t no more calculate
integrals over such domains as an rectangular ones and we need to use Fubini’s
Theorem. The integral of an arbritrary function can be written as
Zπ/2
ZZ
f (x, y) dxdy =
Ω
Example 1.12.
RR q
4−
Compute
Ω
2aZcos ϕ
dϕ
0
−π/2
x2
9
−
y2
4
f (ρ cos ϕ, ρ sin ϕ) ρ dρ.
dxdy over the domain Ω : 4x2 + 9y 2 ≤ 36 using
transformation to polar coordinates.
13
2
2
Solution: The boundary of the domain can be written in the form x9 + y4 =
1. Therefore, the domain is ellipse with center in the origin of coordinates and
semi-axis a = 3, b = 2 (see Fig. 11). In such case we use generalized polar
Fig. 11. Domain Ω in Example 1.12.
coordinates in the form
x = aρ cos ϕ,
y = bρ sin ϕ.
For Jacobian of the transformation we obtain
∂x ∂x ∂ρ ∂ϕ a cos ϕ −aρ sin ϕ
=
J(ρ, ϕ) = ∂y
∂y ∂ρ ∂ϕ b sin ϕ bρ cos ϕ
(8)
= abρ.
Using generalized polar coordinates we obtained transformated domain Ω∗ =
{[ρ, ϕ]
: ρ ∈ (0, 1i, ϕ ∈ h0, 2π)}
qand we can solve the integral:
RR q
RR
2
2
ϕ)2
ϕ)2
y
4 − x9 − 4 dxdy =
4 − (3ρ cos
− (2ρ sin
6ρ dρ dϕ =
9
4
Ω
Ω∗
R2π
√
R1 p
4 − ρ2 ρ dρ dϕ = 6 dϕ
4 − ρ2 ρ dρ = 6 · 2π · 13 (8 − 3 3) =
0
0
Ω∗
√
= 4π(8 − 3 3).
Integral over coordinate ρ was calculated using substitution
4 − ρ2 = t, ρ dρ = − 21 dt.
=6
RR p
Exercises
Compute the following integrals over their domains Ω.
RR
(1 − 2x − 3y) dxdy, Ω : x2 + y 2 ≤ 2.
1.
[2π]
Ω
2.
RR p
1 − x2 − y 2 dxdy, Ω : x2 + y 2 ≤ 1, x ≥ 0, y ≥ 0.
Ω
14
[ π6 ]
3.
RR
sin
p
x2 + y 2 dxdy, Ω : π 2 ≤ x2 + y 2 ≤ 4π 2 .
[−6π 2 ]
Ω
4.
RR
5.
RR
xy dxdy, Ω : x2 + y 2 ≤ 4y, y ≥ x ≥ 0.
[ 28
]
3
(2x + y) dxdy, Ω : 4x2 + y 2 ≤ 16, y ≥ x, y ≥ 0.
[ 32
]
3
Ω
Ω
1.4
1.4.1
Practical Applications of the Double Integral
Volume of a Body
According to section 1.1 we know that
RR
f (x, y) dxdy of a positive function
D
f (x, y) > 0 over a rectangular domain D has a meaning of the volume of the
prism with a rectangular base D bounded from above by
RR the function f (x, y). If
we replace rectangle D by general domain Ω, we obtain
f (x, y) dxdy (see Fig.
Ω
2) and we calculate volume of the cylindrical body with basis Ω and bounded from
above by the function f (x, y). Therefore, we are able to define the volume of the
cylindrical body with basis Ω and bounded by an arbitrary function f (x, y) as
ZZ
V =
|f (x, y)| dxdy.
(9)
Ω
Example 1.13.
Calculate the volume of the body bounded by surfaces 2x + 3y = 12, 2z =
y 2 , x = 0, y = 0, z = 0.
Solution: The basis of the body lies in the plane z = 0. Planes 2x + 3y = 12,
x = 0, y = 0 are perpendicular to the basis, thus they define the triangular domain
2
Ω (see Fig. 12). Surface z = y2 ≥ 0 for all [x, y] ∈ Ω, therefore it bounds the
body from above. We write the domain as a normal with respect to the x−axis
with inequations for Ω in the form:
0 ≤ x ≤ 6,
2
0 ≤ y ≤ 4 − x.
3
Using (9) we compute the volume of the body
4− 32 x
6
3
RR y2
R
R 2
R6 h y3 i4− 23 x
R6
1
1
1
2
V =
dxdy
=
dx
y
dy
=
dx
=
4
−
x
dx =
2
2
2
3
6
3
0
0
0
0
0
Ω
h 2 4 i6
(4− 3 x)
= 61 · (− 32 )
= 16.
4
0
15
Fig. 12. Domain Ω of the body in Example 1.13.
Exercises
Compute the volumes of the bodies bounded by surfaces:
√
1. z = 0, z = xy, y = 0, y = x, x + y = 2.
[ 256
]
21
2. z = 0, 2y = x2 , z = 4 − y 2 .
[ π2 ]
3. z = 0, z = 1 − x2 − y 2 .
1.4.2
[ 38 ]
Area of a Region
Let us consider a cylindrical body with a domain Ω bounded from above by a
constant function f (x, y) = 1. The volume of such a body as well as the area of
a region Ω can be expressed as
ZZ
A=
dxdy.
(10)
Ω
Example 1.14.
Calculate the area of a region Ω bounded by curves y = x2 , y = 4 − x2 .
2
2
Solution: We
√ need to find intersections of both parabolas y = x , y = 4 − x that
are x = ± 2. We write the domain as a normal with respect to the x−axis with
inequations in the form (see Fig. 13):
√
√
− 2 ≤x≤
2,
2
x ≤ y ≤ 4 − x2 .
Using (10) we compute the area of our region using symmetry of the domain with
respect to the y axis
16
Fig. 13. Domain Ω of the area in Example 1.14.
√
A=
RR
Ω
dxdy =
R2
√
− 2
dx
4−x
R 2
dy = 2
x2
√
R2
2
(4 − 2x ) dx = 2 4x −
0
√2
2 3
x 0
3
=
√
16 2
.
3
Exercises
Compute the areas of the regions bounded by surfaces:
1. y = x, y = 5x, x = 1.
[2]
2. y = 2x , y = 2−2x , y = 4.
[12 −
√
[ 83 π − 2 3]
3. x2 + y 2 = 4, x2 + y 2 = 4y.
1.4.3
9
]
ln 4
Surface Area
We are able to compute the area of the surface z = f (x, y) where [x, y] is a point
in the region Ω. Function z = f (x, y) must have continuous partial derivatives on
Ω. In this case surface area is given by
s
2 2
ZZ
∂z
∂z
S=
1+
+
dxdy.
(11)
∂x
∂y
Ω
Example 1.15.
Calculate the area of a surface x + y + z = 4 bounded by planes x = 0, y = 0,
x = 2, y = 2.
Solution: We express the surface in the form z = 4 − x − y and hence both partial
∂z
∂z
derivatives ∂x
= −1, ∂y
= −1. According to Fig. 14 the domain Ω is a rectangle
given by inequations 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 and we calculate the surface area
17
Fig. 14. Domain Ω of the area in Example 1.15.
r
S=
RR
1+
Ω
∂z 2
∂x
+
2
∂z
∂y
RR p
1 + (−1)2 + (−1)2 dxdy =
dxdy =
Ω
√
√ R2 R2
√
= 3 dx dy = 3[x]20 [y]20 = 4 3.
0
0
Exercises
Compute the areas of the surfaces given by:
√
1. z = 2xy bounded by planes x = 1, y = 1, x = 2, y = 4.
[12 −
2. y 2 + z 2 = 9 bounded by planes x = 0, x = 2, y = −3, y = 3.
√
2]
[6π]
√
[ 32 π(5 5 − 1)]
3. z = xy bounded by cylinder x2 + y 2 = 4.
1.4.4
16
3
Center of Mass
Let σ(x, y) be a surface density defined for each [x, y] ∈ Ω. The mass of a domain
Ω is defined by
ZZ
m=
σ(x, y) dxdy.
Ω
Static moment of the domain Ω with respect to the x−axis (resp. y-axis) is given
by
Sx =
RR
y σ(x, y) dxdy,
resp.
Sy =
Ω
RR
x σ(x, y) dxdy.
Ω
The coordinations of the center of mass C = [ξ, η] can be expressed as
ξ=
Sy
,
m
η=
18
Sx
.
m
(12)
Example 1.16.
Calculate the coordinates of the center of mass of homogeneous disc bounded by
equation (x − 1)2 + y 2 = 1.
Solution: The disc is homogeneous with constant surface density σ. All integrals
for the calculation of the mass of the disc and its static moments will be transformed to the polar coordinates with the same limits
−
π
π
≤ϕ≤
,
2
2
0 ≤ ρ ≤ 2 cos ϕ.
We know that the center of mass of the homogeneous disc must be in the center
of the disc. We will prove it by computing the mass of the disc
π/2
2 cos
RR
RR
R
R ϕ
σ(x, y) dxdy = σ
ρ dρ dϕ = σ
dϕ
ρ dρ =
m=
=σ
Ω
π/2
R
h 2 i2 cos ϕ
−π/2
0
ρ
2
Ω∗
π/2
R
dϕ = σ
0
−π/2
2 cos2 ϕ dϕ = 2σ
−π/2
π/2
R
−π/2
1
(1
2
+ cos 2ϕ) dϕ =
π/2
= σ (ϕ + 21 sin 2ϕ) −π/2 = σπ
and both static moments
π/2
2 cos
RR
RR
R
R ϕ 2
Sx =
y σ(x, y) dxdy = σ
ρ sin ϕ ρ dρ dϕ = σ
sin ϕ dϕ
ρ dρ =
=σ
Ω
π/2
R
Ω∗
sin ϕ
−π/2
h 3 i2 cos ϕ
ρ
3
0
dϕ = 83 σ
0
−π/2
π/2
R
sin ϕ cos3 ϕ dϕ = 0.
−π/2
The last integral is an integral of an odd function on a symetric interval, therefore
must be equal to 0.
π/2
2 cos
RR
RR
R
R ϕ 2
Sy =
x σ(x, y) dxdy = σ
ρ cos ϕ ρ dρ dϕ = σ
cos ϕ dϕ
ρ dρ =
=σ
Ω
π/2
R
Ω∗
cos ϕ
−π/2
π/2
R
= 23 σ
−π/2
h 3 i2 cos ϕ
ρ
3
0
dϕ = 38 σ
0
−π/2
π/2
R
−π/2
cos4 ϕ dϕ = 83 σ
(1+2 cos 2ϕ+cos2 2ϕ) dϕ = 32 σ
π/2
R
−π/2
π/2
R
−π/2
[ 12 (1+cos 2ϕ)]2 dϕ =
(1+2 cos 2ϕ+ 12 + 12 cos 4ϕ) dϕ =
π/2
= σ 2 ϕ + sin 2ϕ + sin 4ϕ −π/2 = 23 σ · 32 π = σπ.
0
Now, we can use (12) to show that ξ = σπ
= 1, η = σπ
= 0 to prove our hypothσπ
esis. C = [1, 0].
2
3
3
1
8
Exercises
Compute the coordinates of the center of mass of the homogeneous regions bounded
by curves:
19
1. y = x, y = x2
[C = [ 12 , 25 ]]
2. y = x2 , x = 4, y = 0
[C = [3, 24
]]
5
2
2.1
Triple Integral
Triple Integral over a Rectangular Hexahedron
Now, that we know how to integrate over a two-dimensional region, we need to
move on to the integration over a three-dimensional regions. While we used a double integral to integrate over a two-dimensional regions, we use triple integral to
integrate over three-dimensional regions. The definition of the three-dimensional
integral over a rectangular hexahedron is similar to Definition 1.1. we used for
double integral over a rectangle.
Let u = f (x, y, z) is a function of three variables that is continuous and
bounded on the rectangular hexahedron G = {[x, y, x] ∈ R3 : x ∈ ha, bi, y ∈
hc, di, z ∈ he, hi}. We devide intervals ha, bi, hc, di, he, hi by three sequences of
points a = x0 < x1 < x2 < . . . < xm = b, c = y0 < y1 < y2 < . . . <
yn = d and e = z0 < z1 < z2 < . . . < zp = h to intervals hxi−1 , xi i, i =
1, 2, . . . , m, hyj−1 , yj i, j = 1, 2, . . . , n, hzk−1 , zk i, k = 1, 2, . . . , p. We denote
∆xi = xi − xi−1 , ∆yj = yj − yj−1 , ∆zk = zk − zk−1 . The planes that lead
through points xi (resp. yj or zk ) parallel to coordinate planes devide the hexahedron G to m · n · p small hexahedrons Gijk (see. Fig. 15) with volume of each
∆Gijk = ∆xi · ∆yj · ∆zk . We choose an arbitrary point [ξi , ηj , ζk ] in each hex-
Fig. 15. Triple integral on a rectangular hexahedron
ahedron Gijk and we create products f (ξi , ηj , ζk ) · ∆Gijk = f (ξi , ηj , ζk ) · ∆xi ·
∆yj · ∆zk that for positive function f (x, y, z) ≥ 0 has a physical meaning of the
mass of the hexahedron Gijk with density f (ξi , ηj , ζk ). The sum of these products
20
p
m P
n P
P
f (ξi , ηj , ζk ) · ∆xi · ∆yj · ∆zk represents the mass of the body consisted
i=1 j=1 k=1
of such hexahedrons.
Definition 2.1.
p
m P
n P
P
f (ξi , ηj , ζk )·∆xi ·∆yj ·∆zk for m → ∞, n → ∞, p →
If there exists lim
i=1 j=1 k=1
∞, ∆xi → 0, ∆yj → 0, ∆zk → 0 for all i = 1, 2, . . . , m, j = 1, 2, . . . , n, z =
1, 2, . . . , k, we call it a triple integral
of function f (x, y, z) over the rectangular
RRR
hexahedron G and denote it by
f (x, y, z)dxdydz.
G
The triple integral over a hexahedron G of a positive function f (x, y, z) ≥ 0 has
a meaning of the mass of a hexahedron G with density f (x, y, z).
Theorem 2.1. (Dirichlet’s)
Let G = {[x, y, z] ∈ R3 : x ∈ ha, bi, y ∈ hc, di, z ∈ he, hi}. If a function
f (x, y, z) is continuous on the hexahedron G, then
 
 
ZZZ
Zb Zd Zh
(13)
f (x, y, z)dxdydz =   f (x, y, z)dz  dy  dx.
a
G
c
e
The Theorem is similar to two-dimensional Dirichlet’s Theorem 1.1. We can
rewrite (13) by using a different order of integration in five more ways. The triple
integral is then converted to three one-dimensional integrals. Similarly to the double integral, we can write
 
 
Zb Zd Zh
Zb
Zd Zh
  f (x, y, z)dz  dy  dx = dx dy f (x, y, z)dz.
(14)
a
c
e
a
c
e
Theorem 2.2. (Properties of the triple integral over a rectangular hexahedron)
RRR
RRR
1.
cf (x, y, z)dxdydz = c
f (x, y, z)dxdydz,
G
2.
RRR
3.
RRR
G
(f (x, y, z)+g(x, y, z))dxdydz =
G
G
RRR
RRR
f (x, y, z)dxdydz+
g(x, y, z)dxdydz,
G
f (x, y, z)dxdydz =
RRR
f (x, y, z)dxdydz +
G1
G
RRR
G2
21
f (x, y, z)dxdydz,
where f (x, y, z), g(x, y, z) are continuous functions on G, c ∈ R and G1 , G2 are
hexahedrons that fullfil G = G1 ∪ G2 .
If the integrand f (x, y, z) can be writen as a product of three functions of one
variable f (x, y, z) = f1 (x) · f2 (y) · f3 (z), it holds:
ZZZ
Z
b
Z
f1 (x)dx ·
f (x, y, z)dxdydz =
a
G
d
Z
h
f2 (y)dy ·
c
f3 (z)dz.
(15)
e
Example RRR
2.1.
Compute
xy 2 zdxdydz over the rectangular hexahedron G = {[x, y, z] : x ∈
G
h0, 2i, y ∈ h1, 3i, z ∈ h1, 2i}.
RRR
Solution:
= 2 · (9 −
G
1
)
3
xy 2 zdxdydz =
· (2 −
1
)
2
=2·
R2
x dx ·
0
26 3
·2
3
R3
y 2 dy ·
1
R2
1
z dz =
h 2 i2 h 3 i3 h 2 i2
x
· y3 · z2 =
2
0
1
1
= 26.
If it is not possible to decompose the integrand using (15), we can always use
Dirichlet’s Theorem and formula (13).
Example RRR
2.2.
Compute
(x + y) dxdydz over the rectangular hexahedron G = {[x, y, z] :
G
x ∈ h0, 1i, y ∈ h0, 2i, z ∈ h0, 3i}.
RRR
Solution:
(x + y) dxdydz =
R2
0
0
dx
0
G
R1
R1
R1 h
= 3 dx (x + y) dy = 3 xy +
=6·
3
2
0
R2
R3
R1 R2
dy (x + y) dz = dx (x + y) [z]30 dy =
0
0
i
2 2
y
2
0
R1
0
dx = 3 (2x + 2) dx = 6
0
h
0
x2
2
+x
i1
=
0
= 9.
Exercises
Compute the following integrals over their domains G.
RRR 1 2 3
1.
( x + y + z ) dxdydz, G = {x ∈ h1, 2i, y ∈ h1, 2i, z ∈ h1, 2i}. [6 ln 2]
G
2.
RRR
3.
RRR
G
1
1−x−y
[10 ln 54 ]
dxdydz, G = {x ∈ h0, 1i, y ∈ h2, 5i, z ∈ h2, 4i}.
√
xy 2 z dxdydz, G = {x ∈ h−2, 1i, y ∈ h1, 3i, z ∈ h2, 4i}.
√
[ 52
( 2 − 4)]
3
G
22
2.2
Triple Integral over a General Domain
Similarly like in two-dimensional case, we are able to generalize our problem
of solving triple integrals over any three-dimensional region Ω that is bounded
by a closed surface. We consider only such surfaces that don’t intersect themselves and lines parallel with z−axis leading through an arbitrary inner point
of the surface intersect with the surface in two points. Such domain will be
called normal domain with respect to the coordinate plane x, y (see Fig. 16).
We create an orthogonal projection Ω1 of the domain Ω into coordinate plane
Fig. 16. Normal domain with respect to coordinate plane x, y
x, y. A variable z must fulfill f1 (x, y) ≤ z ≤ f2 (x, y). The domain Ω1 is either normal with respect to x−axis or y−axis and we describe it using approach
from section 1.2 by inequalities x1 ≤ x ≤ x2 , g1 (x) ≤ y ≤ g2 (x) (resp.
y1 ≤ y ≤ y2 , h1 (y) ≤ x ≤ h2 (y)). If the function f (x, y, z) is continuous on
Ω we use a method similar to Fubini’s Theorem 1.4 and express
gZ2 (x)
Zx2
ZZZ
f (x, y, z) dxdydz =
dx
x1
Ω
f2Z(x,y)
dy
g1 (x)
f (x, y, z) dz
(16)
f1 (x,y)
or
f (x, y, z) dxdydz =
Ω
hZ2 (y)
Zy2
ZZZ
dy
y1
h1 (y)
f2Z(x,y)
dx
f (x, y, z) dz
(17)
f1 (x,y)
We start with integration over variable z with bounds that are functions of two
variables x, y and after that we calculate a double integral over a regular domain
Ω1 . We are able to use analogical approach and create an orthogonal projection
of the domain Ω into coordinate planes either x, z or y, z. That way we can use
23
six different orders of integration for an arbitrary domain. The triple integral over
a general closed domain has analogical properties as the triple integral over a
rectangular hexahedron.
Theorem 2.3. (Properties of the triple integral over a general domain)
RRR
RRR
1.
cf (x, y, z)dxdydz = c
f (x, y, z)dxdydz,
Ω
2.
RRR
3.
RRR
Ω
(f (x, y, z)+g(x, y, z))dxdydz =
Ω
Ω
RRR
RRR
f (x, y, z)dxdydz+
g(x, y, z)dxdydz,
Ω
f (x, y, z)dxdydz =
RRR
f (x, y, z)dxdydz +
Ω1
Ω
RRR
f (x, y, z)dxdydz,
Ω2
where f (x, y, z), g(x, y, z) are continuous functions on G, c ∈ R and Ω1 , Ω2 are
domains that fullfil Ω = Ω1 ∪ Ω2 .
Example 2.3.
RRR
Determine integration limits for
f (x, y, z) dxdydz over the domain Ω that is
Ω
bounded by surfaces z = 21 (x2 + y 2 ), z = 4 − 21 (x2 + y 2 ).
Solution: Both surfaces are paraboloids and it is obvious that 21 (x2 + y 2 ) ≤ z ≤
4 − 12 (x2 + y 2 ). We can see the projection of the domain Ω to coordinate plane
y, z on Fig. 17. We obtain an equation of the intersection of both surfaces from
Fig. 17. Projection of the Domain Ω to coordinate plane y, z in Example 2.3.
the equation 12 (x2 + y 2 ) = 4 − 12 (x2 + y 2 ) ⇒ x2 + y 2 = 4. Therefore, the
orthogonal projection Ω1 of the domain Ω to coordinate plane x, y is a ring with
the center in the origin of coordinates and radius r = 2. That domain can be
treated as√a normal domain
√ with respect to the x−axis with inequations −2 ≤
2
x ≤ 2, − 4 − x ≤ y ≤ 4 − x2 as well
pas a normal domain
p with respect to the
y−axis with inequations −2 ≤ y ≤ 2, − 4 − y 2 ≤ x ≤ 4 − y 2 .
24
Example 2.4.
RRR
Compute
x dxdydz over the domain Ω bounded by surfaces x = 0, y =
Ω
0, z = 0, 2x + 2y + z − 6 = 0.
Solution: The domain Ω must fulfill 0 ≤ z ≤ 6 − 2x − 2y (see Fig. 18). The
Fig. 18. Domain Ω in Example 2.4.
orthogonal projection Ω1 of the domain Ω to coordinate plane x, y is the triangle
in plane z = 0 bounded by lines x = 0, y = 0, y = 3 − x. The last equation is
obtained from 2x + 2y + z − 6 = 0 by letting z = 0. We describe Ω1 as normal
with respect to x−axis by inequations 0 ≤ x ≤ 3, 0 ≤ y ≤ 3−x and we calculate
the integral using (16)
6−2x−2y
3−x
3−x
RRR
R
R3
R
R3
R
dy =
x dz = x dx
[z]6−2x−2y
x dxdydz = dx
dy
0
0
Ω
=
R3
x dx
0
R3
0
0
3−x
R
R3
0
0
(6 − 2x − 2y) dy =
0
x [6y − 2xy −
0
3−x
y 2 ]0
dx =
R3
x[6(3 − x) − 2x(3 − x) − (3 − x)2 ] dx = (x3 − 6x2 + 9x) dx =
0
h0 4
i3
= x4 − 2x3 + 29 x2 = 27
.
4
=
0
Exercises
Compute the following integrals over their domains Ω.
RRR 3 2
1.
x y z dxdydz, Ω = {y ≥ 0, y ≤ x, x ≤ 1, z ≥ 0, z ≤ xy}.
1
[ 110
]
Ω
2.
RRR
3.
RRR
Ω
1
1+x+y
dxdydz, Ω = {x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1}. [ 23 − 2 ln 2]
y cos(x+z) dxdydz, Ω = {y ≤
Ω
25
√
2
x, y ≥ 0, z ≥ 0, x+z ≤ π2 }. [ π16 − 12 ]
4.
RRR
dxdydz, Ω = {x + y ≤ 1, y ≥ 0, y ≤ 2x, z ≥ 0, z ≤ 1 − x2 }.
Ω
5.
RRR
41
[ 162
]
dxdydz, Ω = {z ≥ 3x2 + 3y 2 , z ≤ 1 − x2 − y 2 }.
Ω
2.3
[ π8 ]
Transformation of the Triple Integral
Similarly to the double integral, using Cartesian coordinates for some domains
can be rather complicated. Especially in case of cylinders, cones or spheres. We
formulate theorem analogical to Theorem 1.5. that describes general transformation of the Triple Integral.
Theorem 2.4.
• Let equations x = u(r, s, t), y = v(r, s, t), z = w(r, s, t) map the region Ω
bijectively to the region Ω∗ .
• Let function f (x, y, z) be continuous and bounded on Ω and functions x =
u(r, s, t), y = v(r, s, t), z = w(r, s, t) have continuous partial derivatives
on Ω̂ that fulfills Ω∗ ⊂ Ω̂.
∂u ∂u ∂u ∂r ∂s ∂t ∂v
∂v 6= 0 in Ω∗ .
• Let J(u, v, w) = ∂v
∂r ∂s ∂t ∂w ∂w ∂w ∂r ∂s ∂t Then
ZZZ
ZZZ
f (x, y, z) dxdydz =
Ω∗
Ω
Determinant J(u, v, w) = terminant.
2.3.1
f (u(r, s, t), v(r, s, t), w(r, s, t))|J|drdsdt
∂u
∂r
∂u
∂s
∂u
∂t
∂v
∂r
∂v
∂s
∂v
∂t
∂w
∂r
∂w
∂s
∂w
∂t
(18)
is again called Jacobian or Jacobi de
Transformation to Cylindrical Coordinates
Trasformation to cylindrical coordinates is suitable for integration domains such
as cylinders, cones, their parts and it is used in cases when orthogonal projection
26
Ω1 of the domain Ω to coordinate plane x, y is a disc or a part of a disc. We replace
Cartesian coordinates x, y, z by cylindrical coordinates ρ, ϕ, z (see Fig. 19). The
Fig. 19. Transformation to cylindrical coordinates
meaning of coordinates ρ, ϕ is the same as we have already used for polar coordinates and the third coordinate z doesn’t change. Therefore, the transformation to
cylindrical coordinates is given by transformation equations:
x = ρ cos ϕ,
y = ρ sin ϕ,
z = z.
(19)
According to Theorem 2.4. we replace volume element dxdydz by |J|dρdϕdz,
where the Jacobian of the trasformation to cylindrical coordinates satisfies
∂x ∂x ∂x ∂ρ ∂ϕ ∂z cos ϕ −ρ sin ϕ 0 ∂y
∂y = J(ρ, ϕ, z) = ∂y
sin ϕ ρ cos ϕ 0 = ρ.
∂ρ ∂ϕ ∂z ∂z ∂z ∂z ∂ρ ∂ϕ ∂z 0
0
1 The transformation of the triple integral to cylindrical coordinates can then be
written in the form
ZZZ
ZZZ
f (x, y, z)dxdydz =
f (ρ cos ϕ, ρ sin ϕ, z)ρ dρ dϕ dz.
(20)
Ω∗
Ω
Example RRR
2.5.
Compute
dxdydz over the domain Ω bounded by surfaces x2 + y 2 = 1, z =
Ω
0, z = 1.
Solution: The domain Ω is the cylinder symetrical with respect to the z−axis, with
radius of the base ρ = 1 and height z = 1 (see Fig. 20). We need to determine the
27
Fig. 20. Domain Ω in Example 2.5.
bounds of the transformated domain Ω∗ . It is obvious that 0 ≤ z ≤ 1. Inequations
for coordinates ρ, ϕ are the same as for transformation to polar coordinates, i.e.
0 ≤ ρ ≤ 1, 0 ≤ ϕ ≤ 2π. Therefore
h 2 i1
RRR
RRR
R2π R1
R1
2π
dxdydz =
ρ dρ dϕ dz = dϕ ρ dρ z dz = [ϕ]0 · ρ2 · [z]10 = π.
Ω∗
Ω
0
0
0
0
Example RRR
2.6.
Compute
dxdydz over the domain Ω bounded by surfaces z ≥ 3x2 +3y 2 , z ≤
Ω
1 − x2 − y 2 .
Solution: The domain is similar to the domain in Example 2.3. Both surfaces are
paraboloids and the orthogonal projection Ω1 of the domain Ω to coordinate plane
x, y is a ring, whose equation we obtain from the intersection of both paraboloids
3x2 + 3y 2 = 1 − x2 − y 2 ⇒ x2 + y 2 = 14 . Therefore, inequations for coordinates
ρ, ϕ must fulfill 0 ≤ ρ ≤ 12 , 0 ≤ ϕ ≤ 2π. We obtain limits of the variable z from
equations of both paraboloids by substituting of variables:
z = 3x2 + 3y 2 = 3ρ2 cos2 ϕ + 3ρ2 sin2 ϕ = 3ρ2 .
z = 1 − x2 − y 2 = 1 − ρ2 cos2 ϕ − ρ2 sin2 ϕ = 1 − ρ2 .
Hence, 3ρ2 ≤ z ≤ 1 − ρ2 and we compute the integral by using (20)
1/2
1/2
1−ρ
RRR
RRR
R2π
R
R 2
R
2
dxdydz =
ρ dρ dϕ dz = dϕ ρ dρ
dz = 2π ρ [z]1−ρ
3ρ2 dρ =
Ω∗
Ω
= 2π
1/2
R
0
ρ(1 − 4ρ2 ) dρ = 2π
0
0
1/2
R
(ρ − 4ρ3 ) dρ = 2π
0
0
3ρ2
h
ρ2
2
Exercises
Compute the following integrals over their domains Ω.
28
− ρ4
i1/2
0
= 2π ·
1
16
= π8 .
1.
RRR
z dxdydz, Ω = {x2 + y 2 ≤ 4, 0 ≤ z ≤ 2, y ≥ 0}.
[4π]
dxdydz, Ω = {x2 + y 2 ≤ 1, x ≥ 0, 0 ≤ z ≤ 6}.
[3π]
RRR p
z x2 + y 2 dxdydz, Ω = {x2 + y 2 ≤ 2x, 0 ≤ z ≤ 1}.
[ 98 ]
Ω
2.
RRR
Ω
3.
Ω
2.3.2
Transformation to Spherical Coordinates
Transformation to spherical coordinates is suitable for integrals where integration
domains are spheres, ellipsoids or their parts or in cases when orthogonal projection Ω1 of the domain Ω to coordinate plane x, y is a disc or a part of a disc.
We replace Cartesian coordinates x, y, z by spherical coordinates ρ, ϕ, ϑ (see Fig.
21). The coordinate ρ denotes a distance between the point [x, y, z] and the origin
Fig. 21. Transformation to spherical coordinates
of the coordinates, ϕ denotes positively oriented angle in coordinate plane x, y
between positive part of the x−axis and the projection ρ1 of the radius vector ρ
to coordinate plane x, y and ϑ denotes positively oriented angle between positive
part of the z−axis and the radius vector ρ. We obtain transformation equations
from Fig. 21
x = ρ cos ϕ sin ϑ,
y = ρ sin ϕ sin ϑ,
z = ρ cos ϑ.
Jacobian of the transformation to spherical coordinates satisfies
∂x ∂x ∂x ∂ρ ∂ϕ ∂ϑ cos ϕ sin ϑ −ρ sin ϕ sin ϑ ρ cos ϕ cos ϑ
∂y
∂y = J = ∂y
sin ϕ sin ϑ ρ cos ϕ sin ϑ ρ sin ϕ cos ϑ
∂ρ ∂ϕ ∂ϑ ∂z ∂z ∂z ∂ρ ∂ϕ ∂ϑ cos ϑ
0
−ρ sin ϑ
29
(21)
= −ρ2 sin ϑ.
Transformation of the triple integral to spherical coordinates can be written in the
form
ZZZ
ZZZ
f (ρ cos ϕ sin ϑ, ρ sin ϕ sin ϑ, ρ cos ϑ)ρ2 sin ϑ dρ dϕ dϑ.
f (x, y, z)dxdydz =
Ω∗
Ω
(22)
The sphere with the center in the origin of coordinates and radius a transformed
to spherical coordinates is mapped to domain Ω∗ given by inequations 0 ≤ ρ ≤
a, 0 ≤ ϕ ≤ 2π, 0 ≤ ϑ ≤ π.
Example 2.7.
RRR 2
Compute
(x + y 2 + z 2 ) dxdydz over the domain Ω bounded by surfaces
Ω
x2 + y 2 + z 2 ≥ 1, x2 + y 2 + z 2 ≤ 4.
Solution: The domain Ω is bounded by two spherical surfaces with the center in
the origin of coordinates and radiuses a = 1, b = 2. We use transformation to
spherical coordinates with inequations 1 ≤ ρ ≤ 2, 0 ≤ ϕ ≤ 2π, 0 ≤ ϑ ≤ π and
calculate
using (22)
RRR 2 the 2integral
2
(x + y + z ) dxdydz =
ΩRRR
=
(ρ2 cos2 ϕ sin2 ϑ + ρ2 sin2 ϕ sin2 ϑ + ρ2 cos2 ϑ)ρ2 sin ϑ dρ dϕ dϑ =
Ω∗
=
=
R2
1
31
5
dρ
R2π
0
h 5 i2
Rπ
R2
R2π Rπ
π
dϕ ρ4 sin ϑ dϑ = ρ4 dρ dϕ sin ϑ dϑ = ρ5 ·[ϕ]2π
0 ·[− cos ϑ]0 =
0
1
· 2π · 2 =
0
1
0
124
π.
5
Example RRR
2.8.
Compute
z dxdydz over the domain Ω bounded by surfaces x2 + y 2 + z 2 ≤
Ω
4, x ≥ 0, y ≥ 0, z ≥ 0.
Solution: The domain Ω is one eighth of the sphere in the first octant with the
center in the origin of coordinates and radius a = 2. Therefore, 0 ≤ ρ ≤ 2, 0 ≤
ϕRRR
≤ π2 , 0 ≤ ϑ ≤ π2RRR
and
z dxdydz =
ρ cos ϑ ρ2 sin ϑ dρ dϕ dϑ =
Ω∗
Ω
π/2
R
R2
π/2
R
3
R2
3
π/2
R
= dρ dϕ ρ sin ϑ cos ϑ dϑ = ρ dρ dϕ
0
0
0
h 40i2 0
π/2
π/2
ρ
· [ϕ]0 · − 14 cos 2ϑ 0 = 4 · π2 · 21 = π.
4
π/2
R
0
0
Exercises
Compute the following integrals over their domains Ω.
30
1
2
sin 2ϑ dϑ =
1.
RRR
Ω
2.
π
z(x2 + y 2 ) dxdydz, Ω = {x2 + y 2 + z 2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0}. [ 48
]
RRR p
x2 + y 2 + z 2 dxdydz, Ω = {x2 +y 2 +z 2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0}.
Ω
3.
RRR
[ π8 ]
(x2 + y 2 ) dxdydz, Ω = {4 ≤ x2 + y 2 + z 2 ≤ 9, z ≥ 0}.
Ω
2.4
π]
[ 844
15
Practical Applications of the Triple Integral
2.4.1
Volume of a Body
As we have already mentioned in section 2.1., the physical meaning of the triple
integral of the function f (x, y, z) ≥ 0 over the domain Ω is the mass of the domain
Ω with the density σ = f (x, y, z). If σ = 1, the triple integral has a meaning of
volume of the body Ω
ZZZ
V =
dxdydz
(23)
Ω
Example 2.9.
Compute the volume of the body bounded by cylindrical surfaces z = 5 − y 2 , z =
y 2 + 3 and planes x = 0, x = 2.
Solution: While limits of variables x, z are obvious, we need to calculate intersections of cylindrical surfaces to determine the limits of variable y.
5 − y 2 = y 2 + 3 ⇒ 2y 2 = 2 ⇒ y = ±1.
Therefore, inequations for the domain Ω are in the form:
Ω : 0 ≤ x ≤ 2, −1 ≤ y ≤ 1, y 2 + 3 ≤ z ≤ 5 − y 2 .
We calculate volume of the body using (23):
5−y
RRR
R2
R1
R 2
R1
R1
2
V =
dxdydz = dx dy
dz = [x]20 [z]5−y
dy
=
2
(2 − 2y 2 ) dy =
2
y +3
=4 y−
−1
0
Ω
h
i
3 1
y
3
−1
=4·
4
3
=
−1
y 2 +3
−1
16
.
3
Exercises
Compute the volumes of bodies bounded by their surfaces.
1. z = x2 + y 2 , z = y.
π
[ 32
]
2. x − y + z = 6, x + y = 2, x = y, y = 0, z = 0.
[ 16
]
3
√
[ 3215 2 ]
3. y = x2 , z = 0, y + z = 2.
31
2.4.2
Mass of a Body
Mass of the domain Ω is given by
ZZZ
m=
σ(x, y, z) dxdydz,
(24)
Ω
where σ(x, y, z) denotes volume density in each point of the domain Ω.
Example 2.10.
Compute the mass of the body bounded by spherical surfaces x2 + y 2 + z 2 =
1, x2 + y 2 + z 2 = 4 with the density defined by σ = x2 + y 2 + z 2 .
Solution: The problem was solved in Example 2.7.
Exercises
Compute the mass of bodies.
1. x2 + y 2 + z 2 ≤ 4, σ =
p
x2 + y 2 + z 2 .
2. x2 + y 2 + z 2 ≤ 1, σ =
2
.
x2 +y 2 +z 2
[8π]
3. x2 = 2y, y + z = 1, 2y + z = 2, σ = y.
3
[16π]
√
[ 8352 ]
Line Integral
3.1
The Curve and Its Orientation
To study the line integral we must be first able to express analytically its domain
(the curve in two- or three-dimensional space) and implement orientation of the
curve.
Definition 3.1.
Let x = x(t), y = y(t), z = z(t) be continuous functions for t ∈ ha, bi.
1. The curve k with parametrical equations x = x(t), y = y(t), z = z(t), t ∈
ha, bi is called positively oriented with respect to the parameter t, if and
only if its points are ordered so that for arbitrary values t1 , t2 ∈ ha, bi, t1 <
t2 , the point M1 = [x(t1 ), y(t1 ), z(t1 )] lies before the point
M2 = [x(t2 ), y(t2 ), z(t2 )], i.e.
∀t1 , t2 ∈ ha, bi : t1 < t2 ⇔ M1 ≺ M2 .
Reversely,
∀t1 , t2 ∈ ha, bi : t1 < t2 ⇔ M2 ≺ M1 ,
the curve is called negatively oriented with respect to the parameter t.
32
2. If the curve k is positively oriented with respect to the parameter t ∈ ha, bi,
then the point A = [x(a), y(a), z(a)] is called the starting point of the
curve and the point B = [x(b), y(b), z(b)] is called the end point of the
curve.
3. The curve is called closed, if A = B.
4. The curve is called smooth in ha, bi, if there exists nonzero derivative of at
least one of parametrical equations x0 = x0 (t), y 0 = y 0 (t), z 0 = z 0 (t), ∀t ∈
ha, bi.
5. The curve is called simple in ha, bi, if it doesn’t intersect itselfs, i.e.
∀t1 , t2 ∈ ha, bi : t1 6= t2 ⇒ M1 6= M2 .
The symbol ≺ means "precedes" or "lies before".
Example 3.1.
Write parametrizations of
1. the line segment between points A = [a1 , a2 , a3 ], B = [b1 , b2 , b3 ],
2. the circle x2 + y 2 = r2 , r > 0,
3. the circle (x − m)2 + (y − n)2 = r2 , r > 0,
4. the ellipse
x2
a2
+
y2
b2
= 1, a, b > 0.
Solution:
1. We should know from the analytical geometry that the parametrical equations of the line segment between points A = [a1 , a2 , a3 ], B = [b1 , b2 , b3 ]
are in the form:
x = a1 + u1 · t, y = a2 + u2 · t, z = a3 + u3 · t, t ∈ h0, 1i,
~ = (u1 , u2 , u3 ) is the vector parallel to the line segment AB.
where u = AB
2. Parametrical equations of the circle with the center in the origin of coordinates and radius r are in the form:
x = r cos t, y = r sin t, t ∈ h0, 2π).
3. Analogically for the circle with the center in the point C = [m, n] and radius
r, the parametrical equations are in the form:
x = m + r cos t, y = n + r sin t, t ∈ h0, 2π).
4. Parametrical equations of the ellipse with the center in the origin of coordinates and semi-axis a, b are in the form:
x = a cos t, y = b sin t, t ∈ h0, 2π).
33
3.2
Line Integral of a Scalar Field
Analogically to the other types of integral, we need to divide our domain - the
curve k - into small elements. Let us consider the simple smooth curve k with
parametrization x = x(t), y = y(t), z = z(t), t ∈ ha, bi positively oriented
with respect to the parameter t. We divide interval ha, bi by sequence of points
a = t0 < t1 < · · · < tn = b into n partial curves k1 , k2 , . . . , kn (see Fig. 22). For
Fig. 22. Dividing of the curve k - the domain of the line integral
each i = 1, . . . , n, we denote by ∆si the length of each element ki and we choose
an arbitrary point Mi = [x(ti ), y(ti ), z(ti )] in each element ki . Moreover, we are
able to construct positively oriented unitary tangential vector τi (Mi ) at each point
Mi that we will use in section 3.3. to define the line integral of a vector field. The
curve k lies within the domain Ω and we consider a bounded continuous scalar
function u(X) = u(x, y, z) defined for each X ∈ Ω. Now we can create the sum
n
n
P
P
of products
u(Mi ) · ∆si =
u(x(ti ), y(ti ), z(ti )) · ∆si and define the line
i=1
i=1
integral of a scalar field.
Definition 3.2.
n
P
If there exists lim u(x(ti ), y(ti ), z(ti )) · ∆si for n → ∞ and ∆si → 0 we call
i=1
it
R the line integral of a scalar field u(x, y, z) along the curve k and denote it
u(x, y, z) ds.
k
The line integral of the scalar field doesn’t depend on the orientation of the
curve, because the lengths of all components ∆si are always positive. The element of the curve ds in the three-dimensional space is the body diagonal of the
rectangular
dx, dy, dz. Therefore we obtain
p hexahedron with sidesp
ds = (dx)2 + (dy)2 + (dz)2 = (ẋ(t)dt)2 + (ẏ(t)dt)2 + (ż(t)dt)2 =
34
p
= (ẋ(t))2 + (ẏ(t))2 + (ż(t))2 dt
and the line integral of the scalar field can then be written in the form
Zb
Z
u(x, y, z) ds =
p
u(x(t), y(t), z(t)) (ẋ(t))2 + (ẏ(t))2 + (ż(t))2 dt.
(25)
a
k
That way we transform the line integral of the scalar field into the one-dimensional
definite integral.
Theorem 3.1. (Properties of the line integral of a scalar field)
R
R
1. cu(X)ds = c u(X)ds,
k
2.
R
k
(u(X) + v(X))ds =
k
3.
R
R
u(X)ds +
k
u(X)ds =
k
R
u(X)ds +
k1
R
v(X)ds,
k
R
u(X)ds,
k2
where c ∈ R, k1 , k2 are curves such that curve k fulfills k = k1 ∪ k2 and
u(X), v(X) are bounded continuous scalar functions for all X ∈ Ω that contains the curve k.
Example 3.2.
R
Calculate the line integral (x + z) ds along the line segment between points
k
A = [1, 2, 3], B = [3, 2, 1].
Solution: We create the parametrization of the line segment:
x = 1 + 2t, y = 2, z = 3 − 2t, t ∈ h0, 1i
and its derivatives: ẋ = 2, ẏ =
p0, ż = −2.
√
√
We express the element ds = (ẋ(t))2 + (ẏ(t))2 + (ż(t))2 dt = 8 dt = 2 2 dt.
We use (25) to calculate the line integral:
√
√ R1
√
R
R1
(x + z) ds = (1 + 2t + 3 − 2t) · 2 2 dt = 8 2 dt = 8 2.
0
0
k
If we consider p
just the two-dimensional problem, i.e. the curve is in x, y plane,
the element ds = (ẋ(t))2 + (ẏ(t))2 dt and the line integral of the scalar field is
then in the form
Zb
Z
u(x, y) ds =
k
p
u(x(t), y(t)) (ẋ(t))2 + (ẏ(t))2 dt.
a
35
(26)
Example 3.3.
R
Calculate the line integral 1 ds, where k is the first arc of cycloid x = a(t −
k
sin t), y = a(1 − cos t), a > 0, t ∈ h0, 2πi.
Solution: We express derivatives ẋ = a(1 − cos t), ẏ = a sin t and the element of
the curve
p
p
ds =p (ẋ(t))2 + (ẏ(t))2 dt = a2 (1 − cos t)2 + a2 sin2 t dt =
√ √
√
− 2 cos t + cos2 t + sin2 t dt = a 2 − 2 cos t dt = 2a 1 − cos t dt =
= a 1q
√
= 2a 2 sin2 2t dt = 2a sin 2t dt.
Now, we calculate the integral using (26)
2π
R
R2π
1 ds = 2a sin 2t dt = 2a −2 cos 2t 0 = −4a · (−1 − 1) = 8a.
0
k
Suppose the curve in x, y plane can be expressed explicitly by y = y(x), x ∈
ha, bi.p
The element of thep
curve is then given by p
2
2
ds = (dx) + (dy) = (dx)2 + (y 0 (x)dx)2 = 1 + (y 0 (x))2 dx
and the line integral is then in the form
Zb
Z
u(x, y) ds =
p
u(x, y(x)) 1 + (y 0 (x))2 dx.
(27)
a
k
Example 3.4.
R
Calculate the line integral y ds, where k is a part of the function y = x3 between
k
points A = [0, 0], B = [1, 1].
Solution: A derivative of the function:
y 0 = 3x2 . √
p
An element of the curve: ds = 1 + (3x2 )2 dx = 1 + 9x4 dx.
Calculation of the integral by using (27):
h√ i10
√
R
R10 √
R1 3 √
1
1
2
1
4
y ds = x 1 + 9x dx = 36
t dt = 36 · 3
t3
= 54
(10 10 − 1).
0
k
1
4
1
The integral was calculated by using substitution 1 + 9x = t, 36x3 dx = dt.
Exercises
Compute the line integrals of scalar fields along given curves.
R 2
1. x2z+y2 ds, k is one thread of the spiral x = cos t, y = sin t, z = t
√
[ 83 2π 3 ]
k
2.
R
√
x ds,
k is a line segment between points A = [0, 0], B = [1, 2]
k
36
[
5
]
2
3.
R
x2 ds,
k is an upper half of the circle x2 + y 2 = a2 , a > 0
x2 ds,
k : y = ln x, x ∈ h1, 3i
k
4.
R
k
3.3
[ 12 πa3 ]
√
√
[ 31 (10 10 − 2 2)]
Line Integral of a Vector Field
In the previous section we looked at line integral of a scalar field u(x, y, z). In
this section we are going to evaluate line integral of vector fields. We consider
simple smooth curve k with the same parametrization and division to small elements as we have already discribed in section 3.2. Furthermore, we consider
bounded continuous vector field F(X) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k
defined for each X ∈ Ω. We create scalar products of the vector field and the
positively oriented tangential vector of each element of the curve F(Mi ) · ∆si ,
n
P
where ∆si = ∆si τi . Now we can create sum of such products
F(Mi ) · ∆si =
n
P
i=1
F(x(ti ), y(ti ), z(ti )) · ∆si τi and define the line integral of a vector field.
i=1
Definition 3.2. n
P
If there exists
F(x(ti ), y(ti ), z(ti )) · ∆si τi for n → ∞ and ∆si → 0, we call
i=1
itR the line integral of a vector field F(x, y, z) along the curve k+ and denote it
F(x, y, z) ds, where k+ denotes positively oriented curve k. While k− would
k+
denote negatively oriented curve k.
The line integral of the vector field depends on the orientation of the curve k,
because coordinates of the unitary tagential vectors τi depend on the orientation
of the curve. To derive the form of the line integral of the vector field, we need to
express the scalar product
F(x, y, z)·ds = (P (x, y, z), Q(x, y, z), R(x, y, z))·(dx, dy, dz) = P (x, y, z)dx+
Q(x, y, z)dy + R(x, y, z)dz
and the differentials dx = ẋ(t)dt, dy = ẏ(t)dt, dz = ż(t)dt by using parametrization of the curve. The line integral of the vector field then can be written in the
form
Zb
Z
F(x, y, z) ds = ε [P (x(t), y(t), z(t))ẋ(t) + Q(x(t), y(t), z(t))ẏ(t)+
k
a
+R(x(t), y(t), z(t))ż(t)] dt,
(28)
where ε = 1 in case of positively oriented curve k with respect to the parameter t,
while ε = −1 in case of negatively oriented curve k. This way we transform the
37
line integral of the vector field into the one-dimensional definite integral, similarly
to the case of the line integral of a scalar field.
Theorem 3.2. (Properties of the line integral of a vector field)
R
R
1. cF(X)ds = c F(X)ds,
k
2.
R
k
(F(X) + G(X))ds =
k
3.
R
R
F(X)ds +
k
F(X)ds =
k
4.
R
R
F(X)ds +
k1
F(X)ds = −
k+
R
G(X)ds,
k
R
F(X)ds,
k2
R
F(X)ds.
k−
where c ∈ R, k1 , k2 are curves such that curve k fulfills k = k1 ∪ k2 (considering
the same orientation of these curves) and F(X), G(X) are bounded continuous
vector functions for all X ∈ Ω that contains the curve k.
Example 3.5.
Calculate the line integral of the vector field F = (x, y, z) along the curve k, that
is one thread of the spiral x = 2 cos t, y = 2 sin t, z = 3t, t ∈ h0, 2πi.
Solution: Derivatives of parametrization: ẋ = −2 sin t, ẏ = 2 cos t, ż = 3.
Calculation
of theRintegral by using (28):
R
F(x, y, z)ds = x dx + y dy + z dz =
k
k
R2π
= [2 cos t · (−2 sin t) + 2 sin t · 2 cos t + 3t · 3] dt =
0
R2π
R2π
2π
= [−4 sin t cos t + 4 sin t cos t + 9t] dt = 9t dt = 29 [t2 ]0 = 18π 2 .
0
0
If we consider the two-dimensional problem, i.e. the curve is in x, y plane,
the vector field F = (P (x, y), Q(x, y)) and the tangential vector of the element
ds = (dx, dy) = (ẋ(t)dt, ẏ(t)dt). The line integral of the vector field is then in
the form
Zb
Z
F(x, y) ds = ε
k
[P (x(t), y(t))ẋ(t) + Q(x(t), y(t))ẏ(t)]dt
(29)
a
Suppose the curve in x, y plane is expressed explicitly by y = y(x), x ∈ ha, bi.
The tagential vector of the curve is given by ds = (dx, dy) = (dx, y 0 (x) dx) and
38
the line integral of a vector field is then in the form
Zb
Z
F(x, y) ds = ε
[P (x, y(x)) + Q(x, y(x))y 0 (x)] dx
(30)
a
k
Example 3.6.
R
Calculate the line integral (x + y) dx + (x − y) dy, where k : y = x1 , x ∈ h2, 3i.
k
Solution: The curve is given explicitly: y = x1 ⇒ y 0 = − x12 ⇒ dy = − x12 dx.
We calculate the integral by using (30)
R
R3 (x + y) dx + (x − y) dy =
(x + x1 ) + (x − x1 ) · (− x12 ) dx =
2
k
=
R3 x+
2
1
x
−
1
x
+
1
x3
dx =
R3
2
(x+ x13 ) dx
=
h
x2
2
−
1
2x2
i3
2
1
= 92 − 18
−2+ 18 =
185
.
72
Exercises
Compute the line integrals of vector fields along given curves.
R
~ : A = [1, 1, 1], B =
1. x dx−y dy+z dz, k is an oriented line segment AB
k
[4, 3, 2]
R
2. y dx + x dy,
[5]
k is one quarter of the circle x = a cos t, y = a sin t, a >
k
0, t ∈ h0, π2 i
R
3. (xy − 1) dx + x2 y dy,
[0]
k is an arc of the ellipse x = cos t, y = 2 sin t
k
from the starting point A = [1, 0] to the end point B = [0, 2].
[ 34 ]
R
4. xy dx + (y − x) dy, k is a part of the parabola y 2 = x from the starting
k
point A = [0, 0] to the end point B = [1, 1].
3.4
[ 17
]
30
Green’s Theorem
Green’s Theorem expresses the relation between a line integral of a vector field
along a plane (two-dimensional) closed curve and a double integral.
Definition 3.3.
Let Ω be a domain in a plane bounded by a simple closed curve k. The curve k is
orientated positively with respect to the domain Ω if traveling on the curve we
always have the domain Ω on the left side (see Fig. 23).
39
Fig. 23. Positively orientated curve k with respect to the domain Ω
Positive orientation of the curve means traveling in a counterclockwise direction, while negative orientation means traveling in a clockwise
direction. We
H
denote the line integral along a closed curve k by the symbol .
k
Theorem 3.3. (Green’s)
Let two-dimensional vector field F(x, y) = P (x, y)i + Q(x, y)j have continuous derivatives on the plane domain Ω, normal with respect to both axis x, y and
bounded by simple smooth closed curve k orientated positively with respect to the
domain Ω. Then
I
ZZ ∂Q(x, y) ∂P (x, y)
−
dxdy.
(31)
P (x, y)dx + Q(x, y)dy =
∂x
∂y
Ω
k
The Green’s Theorem transforms the line integral of a vector field along a
plane closed curve to a double integral over the domain Ω, that is bounded by the
curve k. It is especially useful in situations when k is a polygon and we have to
calculate as many line integrals as the lines the polygon consists of.
Example 3.7.
H
Calculate the integral (2xy − 5y) dx + (x2 + y) dy, where k is the circle with the
k
center in the origin of coordinates and radius r.
Solution: The disc x2 + y 2 ≤ r is two-dimensional domain that is normal with
respect to both coordinate axis. The curve k is simple and closed and we can
orientate it positively with respect to Ω. Also both functions P (x, y) = 2xy − 5y
and Q(x, y) = x2 + y fulfill assumptions of the Green’s Theorem.
We calculate derivatives Py0 = 2x − 5, Q0x = 2x and evaluate the integral by using
(31):
H
RR
RR
(2xy − 5y) dx + (x2 + y) dy = (2x − (2x − 5)) dxdy =
5 dxdy = 5πr2 .
k
Ω
Ω
40
Example 3.8.
H
Calculate the integral (x2 + y 2 ) dx + (x + y)2 dy, where k consists of the sides
k
of the triangle ABC: A = [1, 1], B = [1, 3], C = [3, 3].
Solution: All assumptions of the Green’s Theorem are fulfilled. The domain is
shown on the Fig. 24.
P (x, y) = x2 + y 2 , Q(x, y) = (x + y)2 ,
Fig. 24. Curve k and domain Ω in Example 3.8.
Py0 = 2y, Q0x = 2(x + y).
We calculate the double integral as a normal one with respect to the x−axis with
inequations for Ω in the form
1 ≤ x ≤ 3,
x ≤ y ≤ 3.
By using (31) we obtain:
H 2
RR
R3
R3
(x + y 2 ) dx + (x + y)2 dy = (2x + 2y − 2y) dxdy = 2 x dx dy =
1
Ω
k
R3
R3
= 2 x[y]3x dx = 2 (3x − x2 ) dx = 2
1
1
h
3 2
x
2
−
x3
3
i3
1
=
x
20
.
3
Exercises
Compute the line integrals of vector fields along given curves by using the Green’s
Theorem.
H
1. (x2 + y 2 ) dy, k are sides of the rectangle x = 0, y = 0, x = 2, y = 4.
k
[16]
41
2.
H
2y dx − (x + y) dy, k are sides of the triangle x = 0, y = 0, x + 2y = 4.
k
[−12]
3.
H
(x + y) dx − (x − y) dy, k is the ellipse 4x2 + 9y 2 = 36.
[−12π]
k
4.
H
(ex sin y − 16y) dx + (ex cos y + 16) dy, k is the circle x2 + y 2 = 2x. [16π]
k
3.5
Path Independence of the Line Integral
We can use another method of calculation of the line integral of a vector field in
situations when the value of the integral doesn’t depend on an integration curve
and depends only on its starting and end point.
Definition 3.4.
Let points A, B ∈ Ω. Let the vector function F(X) = P (X)i + Q(X)j + R(X)k
be continuous over the domain Ω. If the value of the line integral of the vector
field
Z
Z
F(X)ds = P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz
k
k
doesn’t depend on the integration curve k that lies within the domain Ω, we say
the integral is path independent between points A, B.
If this property is fulfilled for arbitrary points A, B ∈ Ω, we say integral is path
independent over Ω.
Theorem 3.4.
Let F(X) = P (X)i + Q(X)j + R(X)k have continuous partial derivatives in a
domain Ω. Let a curve k lie within Ω, A be the starting point of the curve k, while
B be its end point. Then:
R
R
1. The line integral F(X)ds = P (X)dx + Q(X)dy + R(X)dz is path
k
k
independent over Ω if and only if there exists some scalar function Φ(x, y, z)
over Ω such that F (x, y, z) = grad Φ(x, y, z), i.e.
P =
∂Φ
,
∂x
Q=
∂Φ
,
∂y
R=
∂Φ
.
∂z
The vector field F(x, y, z) is called a conservative field, function Φ(x, y, z)
is called a scalar potential.
42
2. The line integral
R
F(X)ds =
k
R
P (X)dx + Q(X)dy + R(X)dz is path
k
independent over Ω if and only if
i
∂
curl F = ∂x
P
j
∂
∂y
Q
k ∂ = o.
∂z R 3. In such case, the line integral of the vector field F(X) along a curve k from
the starting point A to the end point B is given by the difference of the
values of scalar potential in the end point B and starting point A :
Z
Z
F(X)ds = P (X)dx + Q(X)dy + R(X)dz = Φ(B) − Φ(A).
k
k
4. Hence, if the line integral is path independent and the curve k is closed
I
F(X)ds = 0.
k
If the problem is considered only in xy−plane, the path independent line integral of the vector field is given by
Z
Z
F(X)ds = P (x, y)dx + Q(x, y)dz = Φ(B) − Φ(A).
k
k
The test for determining if a vector field is conservative can be written in the form
∂Q
∂P
=
.
∂y
∂x
We will show the way of calculation of the scalar potential at following examples.
Example 3.9.
R
Prove that the line integral (3x2 − 2xy + y 2 ) dx − (x2 − 2xy + 3y 2 ) dy is path
k
independent. If it is, find the scalar potential.
Solution: Components of the vector field P = 3x2 −2xy+y 2 , Q = x2 −2xy+3y 2
and their partial derivatives ∂P
= ∂Q
= −2x + 2y are continuous. The test
∂y
∂x
condition is fulfilled. To find the potential we first integrate P (x, y) = 3x2 −
43
2xy + y 2 with
R respect to x.R
Φ(x, y) = P (x, y)dx = (3x2 − 2xy + y 2 ) dx = x3 − x2 y + xy 2 + C(y),
where C(y) is a function of variable y. We determine it by setting the partial
equal to Q(x, y).
derivative ∂Φ
∂y
∂Φ
2
= −x + 2xy + C 0 (y) = −x2 + 2xy − 3y 2 = Q(x, y).
∂y
R
Hence, C 0 (y) = −3y 2 . Then C(y) = − 3y 2 dy = −y 3 + C1 , where C1 is a
constant. The scalar potential is in the form
Φ(x, y) = x3 − x2 y + xy 2 − y 3 + C1 .
In general, the scalar potential over xy−plane can be written in the form
Zy
Zx
P (t, y) dt +
Φ(x, y) =
x0
Q(x0 , t) dt,
(32)
y0
where [x0 , y0 ] is an arbitrary point of a domain Ω. While in three-dimensional
space it can be written in the form
Zy
Zx
P (t, y, z) dt +
Φ(x, y, z) =
x0
Zz
Q(x0 , t, z) dt, +
y0
R(x0 , y0 , t) dt,
(33)
z0
where [x0 , y0 , z0 ] is an arbitrary point of a domain Ω.
We can calculate the scalar potential from previous example by using (32).
The origin of coordinates [0,0] was taken as the arbitrary point [x0 , y0 ]:
Ry 2
Rx 2
2
Φ(x, y) = (3t −2ty+y ) dt− (0 −2·0·t+3t2 ) dt = [t3 −t2 y+ty 2 ]x0 −[t3 ]y0 =
0
0
= x3 − x2 y + xy 2 − y 3 .
Example 3.10.
H
Calculate the integral x2 dx + y 2 dy along the closed curve k : x2 + y 2 = r2 .
k
Solution: Functions P = x2 and Q = y 2 fulfill assumptions of Theorem 3.4.
Partial derivatives ∂P
= ∂Q
= 0. Hence, the integral is path independant. The
∂y
∂x
curve is closed. Therefore, the integral must be equal to zero.
Example 3.11.
R
Calculate the integral (2x + yz) dx + (xz + z 2 ) dy + (xy + 2yz) dz from the
k
starting point A = [1, 1, 1] to the end point B = [1, 2, 3].
Solution: We determine functions P, Q, R and all needed partial derivatives:
P (x, y, z) = 2x + yz, ∂P (x,y,z)
= z, ∂P (x,y,z)
= y,
∂y
∂z
44
Q(x, y, z) = xz + z 2 , ∂Q(x,y,z)
= z, ∂Q(x,y,z)
= x + 2z,
∂x
∂z
∂R(x,y,z)
∂R(x,y,z)
R(x, y, z) = xy + 2yz,
= y,
= x + 2z.
∂x
∂y
∂Q
∂Q
∂P
∂R
∂P
∂R
= ∂y , ∂y = ∂z , ∂z = ∂x .
∂x
Hence, curl F = o and integral is path independent.
We calculate the scalar potential by using (33). The origin of coordinates [0, 0, 0]
is chosen as the arbitrary point for [x0 , y0 , z0 ].
Rz
Rx
Ry
Φ(x, y, z) = (2t + yz) dt + z 2 dt + 0 dt = x2 + xyz + yz 2 .
0
0
0
We
according to Theorem 3.4:
R calculate the integral
(2x+yz) dx+(xz+z 2 ) dy+(xy+2yz) dz = Φ(B)−Φ(A) = [x2 +xyz+yz 2 ]B
A =
k
= 12 + 1 · 2 · 3 + 2 · 32 − (12 + 1 · 1 · 1 + 1 · 12 ) = 22.
Exercises
Prove that following integrals are path independent. Then, calculate them if A is
the starting point and B is the end point.
R
dy
, A = [1, 2], B = [2, 3].
[4]
1. x dx+y
x2 +y 2
k
2.
R
(2y − 6xy 3 ) dx + (2x − 9x2 y 2 ) dy,
A = [1, 1], B = [4, 1].
[−39]
k
3.
R
yz dx + xz dy + xy dz,
A = [2, 2, 2], B = [2, 3, 4].
[16]
k
4.
R
k
3.6
3.6.1
dx+2 dy+3 dz
,
x+2y+3z
A = [0, 1, 0], B = [1, 0, 1].
[ln 2]
Practical Applications of the Line Integral
Area of a Cylindrical Region
Let function f (x, y) ≥ 0 be continuous on the domain Ω that contains the curve k.
We consider the cylindrical surface between the plane z = 0 and z = f (x, y)
above the curve k (see Fig. 25). We express the area of such a surface as
Z
A = f (x, y) ds.
(34)
k
It is the geometrical meaning of the line integral of a scalar field.
Example 3.12.
Calculate the area of the cylindrical surface x2 + y 2 = r2 bounded by z ≥ 0 and
z ≤ x.
45
Fig. 25. Area of a Cylindrical Region
Solution: The curve k is a part of the circle x2 + y 2 = r2 for x ≥ 0. Therefore,
the parametrization of the curve k is in the form x = r cos t, y = r sin t, where
t ∈ h− π2 , π2 i. We express derivatives of the parametric equations ẋ = −r sin t,
ẏ = rp
cos t and the element of the
√ curve2 is then given by
2
2
ds = (ẋ(t)) + (ẏ(t)) dt = r2 sin t + r2 cos2 t dt = r dt.
By using (34) we obtain
π/2
R
R
π/2
A = x ds =
r cos t rdt = r2 [sin t]−π/2 = 2r2 .
k
−π/2
Exercises
Calculate the area of cylindrical surfaces bounded by z = 0 and
1. x2 + y 2 = r2 , z =
xy
,
2r
x ≥ 0, y ≥ 0.
√
2. 9y 2 = 4(x − 1)3 , z = 2 − x.
[ 11
]
3
√
[ 16
(10
10 − 1)]
27
3. y = 38 x2 , z = x, x = 0, y = 6.
3.6.2
[ 41 r2 ]
Length of a Curve
Let k be simple, piecewise smooth curve. The length of the curve is numerically
equal to the area of the cylindrical surface above the curve k that is bounded by
planes z = 0 and z = 1. Hence, by letting f (x, y) = 1 from (34) we obtain
Z
L = 1 ds.
(35)
k
Example 3.13.
Calculate the length of one period of the cycloid x = a(t−sin t), y = a(1−cos t),
46
t ∈ h0, 2πi.
Solution: Similarly to the previous exercise we start with derivatives of the parametric equations of the cycloid: ẋ = a(1 − cos t), ẏ = a sin t.
Further,
of the curve:
p
p we continue with the element
ds =p (ẋ(t))2 + (ẏ(t))2 dt = a2 (1 − cos t)2 + a2 sin2 t dt =
√ √
√
= a 1 − 2 cos t + cos2 t + sin2 t dt = a 2 − 2 cos t dt = 2a 1 − cos t dt.
According to (35) we calculate the integral by using trigonometric identity
2 sin2 x = 1 − cos 2x :
√ R2π √
√ R2π q
R2π
R
L = 1 ds = 2a
1 − cos t dt = 2a
2 sin2 2t dt = 2a sin 2t dt =
0
k
0
0
= 2a[−2 cos 2t ]2π
0 = −4a · (−1 − 1) = 8a.
Exercises
Calculate the lengths of the curves
1. y = 21 ln x, z = 12 x2 , x ∈ h1, 2i
[ 12 (3 + ln 2)]
2. x = 2a cos t − a cos 2t, y = 2a sin t − a sin 2t, t ∈ h0, 2πi
3. y = 12 x2 , z = 16 x3 , x ∈ h0, 1i.
3.6.3
[16a]
[ 76 ]
Work in a Force Field
Suppose an object moving in a force field F = (P (x, y, z), Q(x, y, z), R(x, y, z))
along a curve k. Work done by the force F is then given by the line integral of a
vector field
Z
Z
W = F ds = P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz.
(36)
k
k
Example 3.14.
Calculate the work done by the force field F = (xy, x + y) on an object moving
along the line segment AB from the point A = [0, 0] to the point B = [1, 1].
Solution: We describe the line segment by function y = x, x ∈ h0, 1i and calculate the work according to (36)
R
R
R1
3
W = F ds = xydx + (x + y)dy = (x2 + 2x) dx = [ x3 + x2 ]10 = 13 + 1 = 43 .
k
k
0
Exercises
1. Calculate the work done by the force field F = (xy, x + y) on an object
moving from the point A = [0, 0] to the point B = [1, 1] if the curve k :
37
x = y2.
[ 30
]
47
2. Calculate the work done by the force field F = (x + y, 2x) on an object
moving around the closed curve k : x2 + y 2 = r2 in a positive direction.
[πr2 ]
3. Calculate the work done by the force field F = (x2 , y 2 , z 2 ) on an object
moving around the screw line k : x = cos t, y = sin t, z = t, t ∈ h0, π2 i.
3
[ π24 ]
3.6.4
Mass of a Curve
Let k be a simple, piecewise smooth curve and continuous function ρ(x, y, z) be
its linear density. The mass of a curve (e.g. mass of a wire) is given by the line
integral of a scalar field
Z
m=
ρ(x, y, z) ds.
(37)
k
Example 3.15.
Calculate the mass of one thread of the screw line k : x = cos t, y = sin t, z = t,
t ∈ h0, 2πi if its density is given by ρ = x2 + y 2 + z 2 .
Solution: Derivatives of the parametric equations:
ẋ = − sin t, ẏ = cos t, ż = 1.
Element
√
p of the curve:
√
ds = (ẋ(t))2 + (ẏ(t))2 + (ż(t))2 dt = sin2 t + cos2 t + 12 dt = 2 dt.
Now we can use (37) and express:
√ R2π
√ R2π
R
m = (x2 + y 2 + z 2 ) ds = 2 (cos2 t + sin2 t + t2 ) dt = 2 (t2 + 1) dt =
0
0
√ 8π3
√
√k 3
4 3
=
2(
+
2π)
=
2
2π(
π
+
1).
= 2[ t3 + t]2π
0
3
3
Exercises
1. Calculate the mass of one quarter of the circle x = a sin t, y = a cos t,
t ∈ h0, π2 i if the density in each point is equal to its y−coordinate.
[a2 ]
2. Calculate the mass of the parabola y = 12 x2 between the points A = [0, 0],
√
B = [2, 2]. The density ρ = xy .
[ 16 (5 5 − 1)]
3. Calculate the mass of the curve y = ln x, where x ∈ h1, 2i. The
√ density
√ at
1
each point is equal to the square of its x−coordinate.
[ 3 (5 5 − 2 2)]
48