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1 Double Integral
1.1 Double Integral over Rectangular Domain
As the definite integral of a positive function of one variable represents the areaof the region between the graph and the x-asis, the double integral of a positivefunction of two variables represents the volume of the region between the surfacedefined 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 ∈ 〈a, b〉, y ∈ 〈c, d〉 onthe xy−plane according to Fig 1.We devide interval 〈a, b〉, (resp. 〈c, d〉) 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〈xi−1, xi〉, i = 1, 2, . . . ,m, (resp. 〈yj−1, yj〉, j = 1, 2, . . . , n). We denote sizesof each component ∆xi = xi − xi−1, ∆yj = yj − yj−1. This way is the wholerectangular 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 wecan evaluate the volume of a prism with basis Dij and and height z = f(ξi, ηj).
The sum of the volumesm∑i=1
n∑j=1
f(ξi, ηj) ·∆xi ·∆yj represents the volume of the
body consisted of such prisms over all rectangles Dij if f(x, y) ≥ 0 on D.
Definition 1.1.
If there exists limm∑i=1
n∑j=1
f(ξi, ηj) · ∆xi · ∆yj for m → ∞, n → ∞, ∆xi →
1
0, ∆yj → 0 for all i = 1, 2, . . . ,m, j = 1, 2, . . . , n, we call it double integral of
function f(x, y) over the rectangular domain D and denote it∫∫D
f(x, y) dxdy.
Theorem 1.1. (Dirichlet’s)Let D = [x, y] ∈ R2 : x ∈ 〈a, b〉, y ∈ 〈c, d〉. If function f(x, y) is continuouson rectangle D, then
∫∫D
f(x, y) dxdy =
b∫a
d∫c
f(x, y) dy
dx =
d∫c
b∫a
f(x, y) dx
dy. (1)
In fact there are two ways of computing the double integral. If the inner differ-ential is dy then the limits on the inner integral must have y limits of integrationand outer integral must have x limits of integration. We compute the integrald∫c
f(x, y) dy by holding x constant and integrating with respect to y as if this
were a single integral (similar approach is used for partial derivatives of functionof more than one variable). This will result as a function of a single variable xwhich we can integrate once again. We use similar approach for the second wayof computing of the double integral.We usually write
b∫a
d∫c
f(x, y) dy
dx =
b∫a
dx
d∫c
f(x, y) dy
andd∫c
b∫a
f(x, y) dx
dy =
d∫c
dy
b∫a
f(x, y) dx.
Theorem 1.2. (Properties of the double integral over a rectangular domain)
1.∫∫D
cf(x, y) dxdy = c
∫∫D
f(x, y) dxdy,
2.∫∫D
(f(x, y) + g(x, y)) dxdy =
∫∫D
f(x, y) dxdy +
∫∫D
g(x, y) dxdy,
2
3.∫∫D
f(x, y) dxdy =
∫∫D1
f(x, y) dxdy +
∫∫D2
f(x, y) dxdy,
where f(x, y), g(x, y) are continuous functions on D, c ∈ R and D1, D2 are rect-angles that fullfil D = D1 ∪D2.
Example 1.1.
Compute∫∫D
6xy2 dxdy over the domain D = [x, y] : 2 ≤ x ≤ 4, 1 ≤ y ≤ 2.
Solution: We will show both ways of the computinga) by integrating the inner integral with respect to variable y∫∫
D
6xy2 dxdy =
4∫2
dx
2∫1
6xy2 dy =
4∫2
[2xy3
]21dx =
4∫2
(16x− 2x) dx =
=
4∫2
14x dx =[7x2]4
2= 112− 28 = 84.
b) by integrating the inner integral with respect to variable x∫∫D
6xy2 dxdy =
2∫1
dy
4∫2
6xy2 dx =
2∫1
[3x2y2
]42dy =
2∫1
(48y2 − 12y2) dy =
=
2∫1
36y2 dy =[12y3
]21
= 96− 12 = 84.
If the integrand f(x, y) can be writen as a multiplication of two functions of onevariable f(x, y) = f1(x) · f2(y), it holds:∫∫
D
f(x, y) dxdy =
∫ b
a
f1(x) dx ·∫ d
c
f2(y) dy. (2)
Example 1.2.Compute the integral from Example 1.1. by using equation (2).
Solution:∫∫D
6xy2 dxdy = 6
4∫2
x dx ·2∫
1
y2 dy =[x2]4
2·[y3]2
1= 12 · 7 = 84.
3
If the decomposition is not possible, we can always use Dirichlet’s Theorem andrelation (1).
Example 1.3.
Compute∫∫D
(2x − 4y) dxdy over the domain D = [x, y] : 1 ≤ x ≤ 3, −1 ≤
y ≤ 1.
Solution:∫∫D
(2x− 4y) dxdy =
3∫1
dx
1∫−1
(2x− 4y) dy =
3∫1
[2xy − 2y2
]1−1dx =
=
3∫1
4x dx =[2x2]3
1= 16.
Although generally the order of integration doesn’t matter, in some cases the in-tegral can be easily solved by using one way of integration while it can be rathercomplicated using the other way. Everything depends on the integrand f(x, y)itself and the limits of integration.
Example 1.4.
Compute∫∫D
xy dxdy over the domain D = [x, y] : 0 ≤ x ≤ 1, 1 ≤ y ≤ 2.
Solution: We will use both ways of the computing to show the situation.a)∫∫D
xy dxdy =
2∫1
dy
1∫0
xy dx =
2∫1
[xy+1
y + 1
]1
0
dy =
2∫1
(1
y + 1− 0
y + 1
)dy =
=
2∫1
dy
y + 1= [ln |y + 1|]21 = ln 3− ln 2 = ln
3
2.
b)∫∫D
xy dxdy =
1∫0
dx
2∫1
xy dy =
1∫0
[xy
lnx
]2
1
dx =
1∫0
(x2
lnx− x
lnx
)dx = . . .
4
The further computation of the integral is tough.
ExercisesCompute the following integrals over their domains D.
1.∫∫D
(x2 + y2) dxdy, D = −2 ≤ x ≤ 0, −1 ≤ y ≤ 2. [14]
2.∫∫D
x√x2 + y dxdy, D = 0 ≤ x ≤ 1, 0 ≤ y ≤ 3. [
2
15(31− 9
√3)]
3.∫∫D
sin(2x+ y) dxdy, D = 0 ≤ x ≤ π,π
4≤ y ≤ π. [0]
4.∫∫D
1
(x+ y + 1)2dxdy, D = 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. [ln
4
3]
1.2 Double Integral over General Domain
There is no reason to limit our problem to rectangular regions. The integral do-main can be of a general shape. We extend the Riemann’s definition of the doubleintegral over rectangular domain to a closed connected bounded domain Ω with-out any problem. The domain is connected if we can connect every two pointsfrom it by curve that lies within the domain. We can always find a rectangle Dthat fullfils Ω ⊆ D (see Fig. 2.) and we can define function f ∗(x, y) by
f ∗(x, y) =
f(x, y) ∀ [x, y] ∈ Ω,
0 ∀ [x, y] ∈ D \ Ω.
Then it holds∫∫Ω
f(x, y) dxdy =
∫∫D
f ∗(x, y) dxdy.
The properties of the double integral over a general domain must correspond toTheorem 1.2.
Theorem 1.3. (Properties of the double integral over a general domain)
1.∫∫Ω
cf(x, y) dxdy = c
∫∫Ω
f(x, y) dxdy,
2.∫∫Ω
(f(x, y) + g(x, y)) dxdy =
∫∫Ω
f(x, y) dxdy +
∫∫Ω
g(x, y) dxdy,
5
Fig. 2. Extension to closed connected domain Ω.
3.∫∫Ω
f(x, y) dxdy =
∫∫Ω1
f(x, y) dxdy +
∫∫Ω2
f(x, y) dxdy,
where f(x, y), g(x, y) are continuous functions on Ω, c ∈ R and Ω1,Ω2 are do-mains 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 ∈ 〈a, b〉.
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 ∈ 〈c, d〉.
Theorem 1.4. (Fubini’s)
1. If the function f(x, y) is continuous on a normal domain with respect to thex−axis, then it holds
∫∫Ω
f(x, y) dxdy =
b∫a
dx
g2(x)∫g1(x)
f(x, y) dy. (3)
2. If the function f(x, y) is continuous on a normal domain with respect to the
6
Fig. 3. a) Normal domain with respect to the x−axis, b) Normal domain withrespect to the y−axis
y−axis, then it holds
∫∫Ω
f(x, y) dxdy =
d∫c
dy
h2(y)∫h1(y)
f(x, y) dx. (4)
Example 1.5.
Determine integration limits for∫∫Ω
f(x, y) dxdy over the domain Ω, which is
bounded by curves y = x2 and x = y2.
Solution: We need to find intersections of curves y = x2 and x = y2 by solving thesystem of these two equations. We can eliminate variable y, receive one equationx4 − x = 0 and solve it. We obtain real values x1 = 0, x2 = 1 and write thedomain Ω as a normal domain with respect to the x−axis. It is bounded by linesx = 0, x = 1 and curves y = x2 and y =
√x (see Fig. 4.) We can express
inequations for Ω in the form:
0 ≤ x ≤ 1,
x2 ≤ y ≤√x
and by Fubini’s Theorem∫∫Ω
f(x, y) dxdy =
1∫0
dx
√x∫
x2
f(x, y) dy.
We can use a similar procedure and express the integral as an integral overnormal domain with respect to the y−axis with inequations
0 ≤ y ≤ 1,
y2 ≤ x ≤ √y.
7
Fig. 4. Domain Ω in Example 1.5.
The double integral then takes form∫∫Ω
f(x, y) dxdy =
1∫0
dy
√y∫
y2
f(x, y) dx.
Example 1.6.
Determine integration limits for∫∫Ω
f(x, y) dxdy over the domain Ω, which is a
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. Ifwe bound the domain by −3 ≤ x ≤ 5, the upper limit of inner integral can’t bewritten as one curve and we need to divide the domain Ω into two subdomainsΩ1,Ω2 by line x = 1 (see Fig. 5.) with inequations
Ω1 : −3 ≤ x ≤ 1, Ω2 : 1 ≤ x ≤ 5,
1 ≤ y ≤ x+ 4, 1 ≤ y ≤ 6− x.Using Fubini’s Theorem we can express
∫∫Ω
f(x, y) dxdy =
1∫−3
dx
x+4∫1
f(x, y) dy +
5∫1
dx
6−x∫1
f(x, y) dy.
However, it is much better to express the domain as normal with respect to they−axis. There is no reason to split the domain which is now bounded by inequa-tions
1 ≤ y ≤ 5,
y − 4 ≤ x ≤ 6− y,
8
Fig. 5. Domain Ω in Example 1.6.
where we have expressed a variable x from boundary equations. The integral iswritten in the form ∫∫
Ω
f(x, y) dxdy =
5∫1
dy
6−y∫y−4
f(x, y) dx.
Example 1.7.
Compute∫∫Ω
xy dxdy over the domain Ω bounded by y =x
2, y =
√x, x ≥ 2.
Solution: By solving the system of equations y =x
2, y =
√x we receive inter-
sections of both curves in x = 0, x = 4 according to Fig. 6. It is better to expressthe domain as normal with respect to x−axis with boundaries
2 ≤ x ≤ 4,x
2≤ y ≤
√x
and compute the integral
∫∫Ω
xy dxdy =
4∫2
dx
√x∫
x/2
xy dy =
4∫2
x
[y2
2
]√xx/2
dx =
4∫2
(x2
2− x3
8
)dx =
=
[x3
6− x4
32
]4
2
=11
6.
9
Fig. 6. Domain Ω in Example 1.7.
The second approach requires splitting the domain into two subdomains. It is agood exercise to compute the example this way.
Example 1.8.
Compute∫∫Ω
(2x − y2) dxdy over the domain Ω bounded by y = 1 − x, y =
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.
10
Computation:
∫∫Ω
(2x− y2) dxdy =
3∫1
dy
y−1∫1−y
(2x− y2) dx =
3∫1
[x2 − xy2
]y−1
1−y dy =
=
3∫1
[(y − 1)2 − (y − 1)y2 − (1− y)2 + (1− y)y2] dy =
3∫1
(−2y3 + 2y2) dy =
=
[−y
4
2+
2
3y3
]3
1
= −68
3.
ExercisesCompute the following integrals over their domains Ω.
1.∫∫Ω
(5x2 − 2xy) dxdy, where Ω is triangle ABC, where A = [0, 0], B =
[2, 0], C = [0, 1]. [3]
2.∫∫Ω
x2 dxdy, where Ω is bounded by curves y =16
x, y = x, x = 8. [576]
3.∫∫Ω
6xy dxdy, where Ω is bounded by curves y = 0, x = 2, y = x2. [32]
4.∫∫Ω
xy dxdy, where Ω is given by x2 + 4y2 ≤ 4, x ≥ 0, y ≥ 0. [12]
5.∫∫Ω
(3x− 2y) dxdy, where Ω is given by x2 + y2 ≤ 4. [0]
1.3 Double Integrals in Polar Coordinates
At this moment we are able to compute the double integral over an arbitrary do-main. In this section we want to look at some domains that are easier to describein terms of polar coordinates. We might have a domain that is a disc, ring or partof a disc or ring. Let us consider a double integral of an arbitrary function overthe disc with the center in the origin of coordinates and the radius r = 2 (same
11
domain that is used in last Exercise No.5). Using Cartesian coordinates we obtainlimits of the integral
−2 ≤ x ≤ 2,
−√
4− x2 ≤ y ≤√
4− x2.
and by Fubini’s Theorem the integral can be written in the form
∫∫Ω
f(x, y) dxdy =
2∫−2
dx
√4−x2∫
−√
4−x2
f(x, y) dy.
In these cases using Cartesian coordinates can be tedious. However, we are ableto replace Cartesian coordinates x, y by polar coordinates ρ, ϕ, where ρ denotesa distance between the point [x, y] and the origin of coordinates and is called aradius, and ϕ denotes the positively oriented angle between positive part of thex−axis and the radius vector and is called angular coordinate or azimuth (seeFig. 8). Hence, we obtain transformation equations
Fig. 8. Polar Coordinates ρ, ϕ.
x = ρ cosϕ, y = ρ sinϕ. (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 thecenter in the origin of coordinates and the radius r = 2,Ω : [x, y] : x2 +y2 ≤ 4,is mapped onto Ω∗ : [ρ, ϕ] : ρ ∈ (0, 2〉, ϕ ∈ 〈0, 2π).
12
Theorem 1.5.
• Let equations x = u(r, s), y = v(r, s) map the region Ω bijectively to theregion 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 Ω that fulfillsΩ∗ ⊂ Ω.
• Let J(u, v) =
∣∣∣∣∣∣∂u∂r
∂u∂s
∂v∂r
∂v∂s
∣∣∣∣∣∣ 6= 0 in Ω∗.
Then ∫∫Ω
f(x, y)dxdy =
∫∫Ω∗
f(u(r, s), v(r, s))|J(u, v)|drds (6)
Determinant J(u, v) =
∣∣∣∣∣∣∂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 coor-dinates 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
J(ρ, ϕ) =
∣∣∣∣∣∣∂x∂ρ
∂x∂ϕ
∂y∂ρ
∂y∂ϕ
∣∣∣∣∣∣ =
∣∣∣∣∣∣ cosϕ −ρ sinϕ
sinϕ ρ cosϕ
∣∣∣∣∣∣ = ρ.
Transformation of the double integral to polar coordinates can then be written inthe form ∫∫
Ω
f(x, y) dxdy =
∫∫Ω∗
f(ρ cosϕ, ρ sinϕ)ρ dρ dϕ. (7)
Example 1.9.
Compute∫∫Ω
y dxdy over the domain Ω : x2 + y2 ≤ 9, y ≥ 0 using transforma-
tion to polar coordinates.
Solution: The domain Ω is an upper half of the disc with the center in the origin
13
of coordinates and radius r = 3. We use transformation to polar coordinates andthe domain Ω∗ = [ρ, ϕ] : ρ ∈ (0, 3〉, ϕ ∈ 〈0, π〉. Using (7) we obtain
∫∫Ω
y dxdy =
∫∫Ω∗
ρ sinϕρ dρ dϕ =
3∫0
ρ2dρ
π∫0
sinϕdϕ =
[ρ3
3
]3
0
· [− cosϕ]π0 =
= 18.
Example 1.10.
Compute∫∫Ω
x dxdy over the domain Ω : 4 ≤ x2 + y2 ≤ 9, y ≥ x, x ≥ 0.
Solution: We can see real advantage of the transformation on this domain (seeFig. 9). While using Cartesian coordinates would be complicated, domain Ω∗ =
[ρ, ϕ] : ρ ∈ 〈2, 3〉, ϕ ∈ 〈π4,π
2〉 for polar coordinates is rectangular.
Fig. 9. Domain Ω in Example 1.10.
∫∫Ω
x dxdy =
∫∫Ω∗
ρ cosϕρ dρ dϕ =
3∫2
ρ2dρ
π/2∫π/4
cosϕdϕ =
[ρ3
3
]3
2
· [sinϕ]π/2π/4 =
=19
2
(1−√
2
2
).
14
Example 1.11.Calculate limits of the integral transformed to polar coordinates for the domainΩ : x2 + y2 ≤ 2ax.
Solution: First, we find the center and radius of the disc.
x2 + y2 ≤ 2ax
x2 − 2ax+ a2 + y2 ≤ a2
(x− a)2 + y2 ≤ 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 ≤ π
2. We can see that the upper limit of coordinate ρ de-
pends on the azimuth ϕ. We obtain the value of the limit by puting transformationequations (5) to boundary equations of Ω
x2 + y2 = 2ax
ρ2 cos2 ϕ+ ρ2 sin2 ϕ = 2aρ cosϕ
ρ2 = 2aρ cosϕ
ρ(ρ− 2a cosϕ) = 0
ρ1 = 0, ρ2 = 2a cosϕ.
We have just obtained the limits of the integral. However, it is necessary to re-alise the dependancy of coordinate ρ on coordinate ϕ. We can’t no more calculateintegrals over such domains as an rectangular ones and we need to use Fubini’s
15
Theorem. The integral of an arbritrary function can be written as∫∫Ω
f(x, y) dxdy =
π/2∫−π/2
dϕ
2a cosϕ∫0
f(ρ cosϕ, ρ sinϕ) ρ dρ.
Example 1.12.
Compute∫∫Ω
√4− x2
9− y2
4dxdy over the domain Ω : 4x2 + 9y2 ≤ 36 using
transformation to polar coordinates.
Solution: The boundary of the domain can be written in the formx2
9+y2
4=
1. Therefore, the domain is ellipse with center in the origin of coordinates andsemi-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 cosϕ
y = bρ sinϕ. (8)
For Jacobian of the transformation we obtain
J(ρ, ϕ) =
∣∣∣∣∣∣∂x∂ρ
∂x∂ϕ
∂y∂ρ
∂y∂ϕ
∣∣∣∣∣∣ =
∣∣∣∣∣∣ a cosϕ −aρ sinϕ
b sinϕ bρ cosϕ
∣∣∣∣∣∣ = abρ.
Using generalized polar coordinates we obtained transformated domain Ω∗ =[ρ, ϕ] : ρ ∈ (0, 1〉, ϕ ∈ 〈0, 2π) and we can solve the integral now.∫∫
Ω
√4− x2
9− y2
4dxdy =
∫∫Ω∗
√4− (3ρ cosϕ)2
9− (2ρ sinϕ)2
46ρ dρ dϕ =
16
= 6
∫∫Ω∗
√4− ρ2ρ dρ dϕ = 6
2π∫0
dϕ
1∫0
√4− ρ2ρ dρ = 6 · 2π · 1
3(8− 3
√3) =
= 4π(8− 3√
3).
Integral over coordinate ρ was calculated using substitution
4− ρ2 = t
−2ρ dρ = dt
ExercisesCompute the following integrals over their domains Ω.
1.∫∫Ω
(1− 2x− 3y) dxdy, Ω : x2 + y2 ≤ 2. [2π]
2.∫∫Ω
√1− x2 − y2 dxdy, Ω : x2 + y2 ≤ 1, x ≥ 0, y ≥ 0. [
π
6]
3.∫∫Ω
sin√x2 + y2 dxdy, Ω : π2 ≤ x2 + y2 ≤ 4π2. [−6π2]
4.∫∫Ω
xy dxdy, Ω : x2 + y2 ≤ 4y, y ≥ x ≥ 0. [28
3]
5.∫∫Ω
(2x+ y) dxdy, Ω : 4x2 + y2 ≤ 16, y ≤ 0, x ≤ 0. [−32]
1.4 Practical Applications of the Double Integral1.4.1 Volume of a Body
According to section 1.1, we know that∫∫D
f(x, y) dxdy of a positive function
f(x, y) > 0 over a rectangular domain D has a meaning of the volume of theprism with a rectangular base D bounded from above by the function f(x, y). If
we replace rectangle D by general domain Ω, we obtain∫∫Ω
f(x, y) dxdy (see
Fig. 2) and we calculate volume of the cylindrical body with basis Ω and boundedfrom above by the function f(x, y). Therefore, we are able to define the volume of
17
the cylindrical body with basis Ω and bounded by an arbitrary function f(x, y)as
V =
∫∫Ω
|f(x, y)| dxdy. (9)
Example 1.13.Calculate the volume of the body bounded by surfaces 2x + 3y = 12, 2z =y2, 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
Ω (see Fig. 12). Surface z =y2
2≥ 0 for all [x, y] ∈ Ω, therefore it bounds the
Fig. 12. Domain Ω of the body in Example 1.13.
body from above. We write the domain as a normal with respect to the x−axiswith inequations for Ω in the form:
0 ≤ x ≤ 6,
0 ≤ y ≤ 4− 2
3x.
Using (9) we compute the volume of the body
V =
∫∫Ω
y2
2dxdy =
1
2
6∫0
dx
4− 23x∫
0
y2dy =1
2
6∫0
[y3
3
]4− 23x
0
dx =
=1
6
6∫0
(4− 2
3x
)3
dx =1
6·(−3
2
)[(4− 2
3x)4
4
]6
0
= 16.
18
ExercisesCompute the volumes of the bodies bounded by surfaces:
1. z = 0, z = xy, y = 0, y =√x, x+ y = 2. [
3
8]
2. z = 0, 2y = x2, z = 4− y2. [256
21]
3. z = 0, z = 1− x2 − y2. [π
2]
1.4.2 Area of a Region
Let us consider a cylindrical body with a domain Ω bounded from above by aconstant function f(x, y) = 1. The volume of such a body as well as the area ofa region Ω can be expressed as
A =
∫∫Ω
dxdy. (10)
Example 1.14.Calculate the area of a region Ω bounded by curves y = x2, y = 4− x2.
Solution: We need to find intersections of both parabolas y = x2, y = 4− x2 thatare x = ±
√2. We write the domain as a normal with respect to the x−axis with
Fig. 13. Domain Ω of the area in Example 1.14.
inequations in the form (see Fig. 13):
−√
2 ≤ x ≤√
2,
x2 ≤ y ≤ 4− x2.
19
Using (10) we compute the area of our region using symmetry of the domain withrespect to the y axis
A =
∫∫Ω
dxdy =
√2∫
−√
2
dx
4−x2∫x2
dy = 2
√2∫
0
(4−2x2) dx = 2
[4x− 2
3x3
]√2
0
=16√
2
3.
ExercisesCompute 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− 9
ln 4]
3. x2 + y2 = 4, x2 + y2 = 4y. [8
3π − 2
√3]
1.4.3 Surface Area
We are able to compute the area of the surface z = f(x, y) where [x, y] is a pointin the region Ω. Function z = f(x, y) must have continuous partial derivatives onΩ. In this case surface area is given by
S =
∫∫Ω
√1 +
(∂z
∂x
)2
+
(∂z
∂y
)2
dxdy. (11)
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 derivatives∂z
∂x= −1,
∂z
∂y= −1. According to Fig. 14 the domain Ω is
a rectangle given by inequations 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 and we calculate thesurface area
S =
∫∫Ω
√1 +
(∂z
∂x
)2
+
(∂z
∂y
)2
dxdy =
∫∫Ω
√1 + (−1)2 + (−1)2dxdy =
=√
3
2∫0
dx
2∫0
dy =√
3[x]20[y]20 = 4√
3.
20
Fig. 14. Domain Ω of the area in Example 1.15.
ExercisesCompute the areas of the surfaces given by:
1. z =√
2xy bounded by planes x = 1, y = 1, x = 2, y = 4. [12− 16
3
√2]
2. y2 + z2 = 9 bounded by planes x = 0, x = 2, y = −3, y = 3. [6π]
3. z = xy bounded by cylinder x2 + y2 = 4. [2
3π(5√
5− 1)]
1.4.4 Center of Mass
Let σ(x, y) be a surface density defined for each [x, y] ∈ Ω. The mass of a domainΩ is defined by
m =
∫∫Ω
σ(x, y) dxdy.
Static moment of the domain Ω with respect to the x−axis (resp. y-axis) is givenby
Sx =∫∫Ω
y σ(x, y) dxdy, resp. Sy =∫∫Ω
x σ(x, y) dxdy.
The coordinations of the center of mass C = [ξ, η] can be expressed as
ξ =Sym, η =
Sxm. (12)
21
Example 1.16.Calculate the coordinates of the center of mass of homogeneous disc bounded byequation (x− 1)2 + y2 = 1.
Solution: The disc is homogeneous with constant surface density σ. All integralsfor the calculation of the mass of the disc and its static moments will be trans-formed to the polar coordinates with the same limits
−π2≤ ϕ ≤ π
2,
0 ≤ ρ ≤ 2 cosϕ.
We know that the center of mass of the homogeneous disc must be in the centerof the disc. We will prove it by computing the mass of the disc
m =
∫∫Ω
σ(x, y) dxdy = σ
∫∫Ω∗
ρ dρ dϕ = σ
π/2∫−π/2
dϕ
2 cosϕ∫0
ρ dρ =
= σ
π/2∫−π/2
[ρ2
2
]2 cosϕ
0
dϕ = σ
π/2∫−π/2
2 cos2 ϕdϕ = 2σ
π/2∫−π/2
1
2(1 + cos 2ϕ) dϕ =
= σ
[ϕ+
1
2sin 2ϕ
]π/2−π/2
= σπ
and both static moments.
Sx =
∫∫Ω
y σ(x, y) dxdy = σ
∫∫Ω∗
ρ sinϕρ dρ dϕ = σ
π/2∫−π/2
sinϕdϕ
2 cosϕ∫0
ρ2 dρ =
= σ
π/2∫−π/2
sinϕ
[ρ3
3
]2 cosϕ
0
dϕ =8
3σ
π/2∫−π/2
sinϕ cos3 ϕdϕ = 0.
The last integral is an integral of an odd function on a symetric interval, thereforemust be equal to 0.
Sy =
∫∫Ω
x σ(x, y) dxdy = σ
∫∫Ω∗
ρ cosϕρ dρ dϕ = σ
π/2∫−π/2
cosϕdϕ
2 cosϕ∫0
ρ2 dρ =
22
= σ
π/2∫−π/2
cosϕ
[ρ3
3
]2 cosϕ
0
dϕ =8
3σ
π/2∫−π/2
cos4 ϕdϕ =
=8
3σ
π/2∫−π/2
[1
2(1 + cos 2ϕ)
]2
dϕ =2
3σ
π/2∫−π/2
(1 + 2 cos 2ϕ+ cos2 2ϕ) dϕ =
=2
3σ
π/2∫−π/2
(1 + 2 cos 2ϕ+1
2+
1
2cos 4ϕ) dϕ =
=2
3σ
[3
2ϕ+ sin 2ϕ+
1
8sin 4ϕ
]π/2−π/2
=2
3σ · 3
2π = σπ.
Now, we can use (12) to show that ξ =σπ
σπ= 1, η =
0
σπ= 0 to prove our
hypothesis. C = [1, 0].
ExercisesCompute the coordinates of the center of mass of the homogeneous regions boundedby curves:
1. y = x, y = x2 [C = [1
2,2
5]]
2. y = x2, x = 4, y = 0 [C = [3,24
5]]
2 Triple Integral
2.1 Triple Integral over a Rectangular Hexahedron
Now, that we know how to integrate over a two-dimensional region, we need tomove on to the integration over a three-dimensional regions. While we used a dou-ble integral to integrate over a two-dimensional regions, we use triple integral tointegrate over three-dimensional regions. The definition of the three-dimensionalintegral over a rectangular hexahedron is similar to Definition 1.1. we used fordouble integral over a rectangle.
Let u = f(x, y, z) is a function of three variables that is continuous andbounded on the rectangular hexahedron G = [x, y, x] ∈ R3 : x ∈ 〈a, b〉, y ∈〈c, d〉, z ∈ 〈e, h〉. We devide intervals 〈a, b〉, 〈c, d〉, 〈e, h〉 by three sequences ofpoints a = x0 < x1 < x2 < . . . < xm = b, c = y0 < y1 < y2 < . . . <
23
yn = d and e = z0 < z1 < z2 < . . . < zp = h to intervals 〈xi−1, xi〉, i =1, 2, . . . ,m, 〈yj−1, yj〉, j = 1, 2, . . . , n, 〈zk−1, zk〉, k = 1, 2, . . . , p. We denote∆xi = xi − xi−1, ∆yj = yj − yj−1, ∆zk = zk − zk−1. The planes that leadthrough points xi (resp. yj or zk) parallel to coordinate planes devide the hexahe-dron 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 themass of the hexahedron Gijk with density f(ξi, ηj, ζk). The sum of these productsm∑i=1
n∑j=1
p∑k=1
f(ξi, ηj, ζk) ·∆xi ·∆yj ·∆zk represents the mass of the body consisted
of such hexahedrons.
Definition 2.1.
If there exists limm∑i=1
n∑j=1
p∑k=1
f(ξi, ηj, ζk) · ∆xi · ∆yj · ∆zk for m → ∞, n →
∞, p → ∞, ∆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 hexahedron G and denote it by∫∫∫G
f(x, y, z) dxdydz.
The triple integral over a hexahedron G of a positive function f(x, y, z) ≥ 0 hasa 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 ∈ 〈a, b〉, y ∈ 〈c, d〉, z ∈ 〈e, h〉. If a function
24
f(x, y, z) is continuous on the hexahedron G, then
∫∫∫G
f(x, y, z) dxdydz =
b∫a
d∫c
h∫e
f(x, y, z) dz
dy
dx. (13)
The Theorem is similar to two-dimensional Dirichlet’s Theorem 1.1. We canrewrite (13) by using a different order of integration in five more ways. The tripleintegral is then converted to three one-dimensional integrals. Similarly to the dou-ble integral, we can write
b∫a
d∫c
h∫e
f(x, y, z)dz
dy
dx =
b∫a
dx
d∫c
dy
h∫e
f(x, y, z)dz. (14)
Theorem 2.2. (Properties of the triple integral over a rectangular hexahe-dron)
1.∫∫∫G
cf(x, y, z) dxdydz = c
∫∫∫G
f(x, y, z) dxdydz,
2.∫∫∫G
(f(x, y, z) + g(x, y, z)) dxdydz =
=
∫∫∫G
f(x, y, z) dxdydz +
∫∫∫G
g(x, y, z) dxdydz,
3.∫∫∫G
f(x, y, z) dxdydz =
=
∫∫∫G1
f(x, y, z) dxdydz +
∫∫∫G2
f(x, y, z) dxdydz,
where f(x, y, z), g(x, y, z) are continuous functions on G, c ∈ R and G1, G2 arehexahedrons that fullfil G = G1 ∪G2.
If the integrand f(x, y, z) can be writen as a product of three functions of onevariable f(x, y, z) = f1(x) · f2(y) · f3(z), it holds:∫∫∫
G
f(x, y, z)dxdydz =
∫ b
a
f1(x)dx ·∫ d
c
f2(y)dy ·∫ h
e
f3(z)dz. (15)
25
Example 2.1.
Compute∫∫∫G
xy2zdxdydz over the rectangular hexahedron G = [x, y, z] :
x ∈ 〈0, 2〉, y ∈ 〈1, 3〉, z ∈ 〈1, 2〉.
Solution:∫∫∫G
xy2zdxdydz =
2∫0
x dx ·3∫
1
y2 dy ·2∫
1
z dz =
[x2
2
]2
0
·[y3
3
]3
1
·[z2
2
]2
1
=
= 2 ·(
9− 1
3
)·(
2− 1
2
)= 2 · 26
3· 3
2= 26.
If it is not possible to decompose the integrand using (15), we can always useDirichlet’s Theorem and formula (13).
Example 2.2.
Compute∫∫∫G
(x + y) dxdydz over the rectangular hexahedron G = [x, y, z] :
x ∈ 〈0, 1〉, y ∈ 〈0, 2〉, z ∈ 〈0, 3〉.
Solution:∫∫∫G
(x+ y) dxdydz =
1∫0
dx
2∫0
dy
3∫0
(x+ y) dz =
1∫0
dx
2∫0
(x+ y) [z]30 dy =
= 3
1∫0
dx
2∫0
(x+y) dy = 3
1∫0
[xy +
y2
2
]2
0
dx = 3
1∫0
(2x+2) dx = 6
[x2
2+ x
]1
0
=
= 6 · 3
2= 9.
ExercisesCompute the following integrals over their domains G.
1.∫∫∫G
(1
x+
2
y+
3
z) dxdydz, G = x ∈ 〈1, 2〉, y ∈ 〈1, 2〉, z ∈ 〈1, 2〉.
[6 ln 2]
26
2.∫∫∫G
1
1− x− ydxdydz, G = x ∈ 〈0, 1〉, y ∈ 〈2, 5〉, z ∈ 〈2, 4〉. [10 ln
4
5]
3.∫∫∫G
xy2√z dxdydz, G = x ∈ 〈−2, 1〉, y ∈ 〈1, 3〉, z ∈ 〈2, 4〉.
[52
3(√
2− 4)]
2.2 Triple Integral over a General Domain
Similarly like in two-dimensional case, we are able to generalize our problemof solving triple integrals over any three-dimensional region Ω that is boundedby a closed surface. We consider only such surfaces that don’t intersect them-selves and lines parallel with z−axis leading through an arbitrary inner pointof the surface intersect with the surface in two points. Such domain will becalled 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 ei-ther normal with respect to x−axis or y−axis and we describe it using approachfrom 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 onΩ we use a method similar to Fubini’s Theorem 1.4 and express
∫∫∫Ω
f(x, y, z) dxdydz =
x2∫x1
dx
g2(x)∫g1(x)
dy
f2(x,y)∫f1(x,y)
f(x, y, z) dz (16)
27
or ∫∫∫Ω
f(x, y, z) dxdydz =
y2∫y1
dy
h2(y)∫h1(y)
dx
f2(x,y)∫f1(x,y)
f(x, y, z) dz (17)
We start with integration over variable z with bounds that are functions of twovariables x, y and after that we calculate a double integral over a regular domainΩ1. We are able to use analogical approach and create an orthogonal projectionof the domain Ω into coordinate planes either x, z or y, z. That way we can usesix different orders of integration for an arbitrary domain. The triple integral overa general closed domain has analogical properties as the triple integral over arectangular hexahedron.
Theorem 2.3. (Properties of the triple integral over a general domain)
1.∫∫∫
Ω
cf(x, y, z) dxdydz = c
∫∫∫Ω
f(x, y, z) dxdydz,
2.∫∫∫
Ω
(f(x, y, z) + g(x, y, z)) dxdydz =∫∫∫Ω
f(x, y, z) dxdydz +
∫∫∫Ω
g(x, y, z) dxdydz,
3.∫∫∫
Ω
f(x, y, z) dxdydz =
=
∫∫∫Ω1
f(x, y, z) dxdydz +
∫∫∫Ω2
f(x, y, z) dxdydz,
where f(x, y, z), g(x, y, z) are continuous functions on G, c ∈ R and Ω1,Ω2 aredomains that fullfil Ω = Ω1 ∪ Ω2.
Example 2.3.
Determine integration limits for∫∫∫
Ω
f(x, y, z) dxdydz over the domain Ω that
is bounded by surfaces z =1
2(x2 + y2), z = 4− 1
2(x2 + y2).
Solution: Both surfaces are paraboloids and it is obvious that1
2(x2 + y2) ≤ z ≤
4 − 1
2(x2 + y2). 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
28
Fig. 17. Projection of the Domain Ω to coordinate plane y, z in Example 2.3.
the equation1
2(x2 + y2) = 4− 1
2(x2 + y2) which leads to x2 + y2 = 4. Therefore,
the orthogonal projection Ω1 of the domain Ω to coordinate plane x, y is a ringwith the center in the origin of coordinates and radius r = 2. That domain canbe treated as a normal domain with respect to the x−axis with inequations −2 ≤x ≤ 2, −
√4− x2 ≤ y ≤
√4− x2 as well as a normal domain with respect to the
y−axis with inequations −2 ≤ y ≤ 2, −√
4− y2 ≤ x ≤√
4− y2.
Example 2.4.
Compute∫∫∫
Ω
x dxdydz over the domain Ω bounded by surfaces x = 0, y =
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
29
in plane z = 0 bounded by lines x = 0, y = 0, y = 3 − x. The last equation isobtained from 2x + 2y + z − 6 = 0 by letting z = 0. We describe Ω1 as normalwith respect to x−axis by inequations 0 ≤ x ≤ 3, 0 ≤ y ≤ 3−x and we calculatethe integral using (16).
∫∫∫Ω
x dxdydz =
3∫0
dx
3−x∫0
dy
6−2x−2y∫0
x dz =
3∫0
x dx
3−x∫0
[z]6−2x−2y0 dy =
=
3∫0
x dx
3−x∫0
(6− 2x− 2y) dy =
3∫0
x[6y − 2xy − y2
]3−x0
dx =
=
3∫0
x[6(3− x)− 2x(3− x)− (3− x)2
]dx =
3∫0
(x3 − 6x2 + 9x) dx =
=
[x4
4− 2x3 +
9
2x2
]3
0
=27
4.
ExercisesCompute the following integrals over their domains Ω.
1.∫∫∫
Ω
x3y2z dxdydz, Ω = y ≥ 0, y ≤ x, x ≤ 1, z ≥ 0, z ≤ xy. [1
110]
2.∫∫∫
Ω
1
1 + x+ ydxdydz, Ω = x ≥ 0, y ≥ 0, z ≥ 0, x+ y + z ≤ 1.
[3
2− 2 ln 2]
3.∫∫∫
Ω
y cos(x+ z) dxdydz, Ω = y ≤√x, y ≥ 0, z ≥ 0, x+ z ≤ π
2.
[π2
16− 1
2]
4.∫∫∫
Ω
dxdydz, Ω = x+ y ≤ 1, y ≥ 0, y ≤ 2x, z ≥ 0, z ≤ 1− x2. [41
162]
5.∫∫∫
Ω
dxdydz, Ω = z ≥ 3x2 + 3y2, z ≤ 1− x2 − y2. [π
8]
30
2.3 Transformation of the Triple Integral
Similarly to the double integral, using Cartesian coordinates for some domainscan be rather complicated. Especially in case of cylinders, cones or spheres. Weformulate theorem analogical to Theorem 1.5. that describes general transforma-tion 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 Ωbijectively 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 derivativeson Ω that fulfills Ω∗ ⊂ Ω.
• Let J(u, v, w) =
∣∣∣∣∣∣∣∣∣∂u∂r
∂u∂s
∂u∂t
∂v∂r
∂v∂s
∂v∂t
∂w∂r
∂w∂s
∂w∂t
∣∣∣∣∣∣∣∣∣ 6= 0 in Ω∗.
Then∫∫∫Ω
f(x, y, z) dxdydz =
∫∫∫Ω∗
f(u(r, s, t), v(r, s, t), w(r, s, t))|J |drdsdt
(18)
Determinant J(u, v, w) =
∣∣∣∣∣∣∣∣∣∂u∂r
∂u∂s
∂u∂t
∂v∂r
∂v∂s
∂v∂t
∂w∂r
∂w∂s
∂w∂t
∣∣∣∣∣∣∣∣∣ is again called Jacobian or Jacobi de-
terminant.
2.3.1 Transformation to Cylindrical Coordinates
Trasformation to cylindrical coordinates is suitable for integration domains suchas cylinders, cones, their parts and it is used in cases when orthogonal projectionΩ1 of the domain Ω to coordinate plane x, y is a disc or a part of a disc. We replaceCartesian coordinates x, y, z by cylindrical coordinates ρ, ϕ, z (see Fig. 19). Themeaning of coordinates ρ, ϕ is the same as we have already used for polar coordi-nates and the third coordinate z doesn’t change. Therefore, the transformation to
31
Fig. 19. Transformation to cylindrical coordinates
cylindrical coordinates is given by transformation equations
x = ρ cosϕ
y = ρ sinϕ (19)z = z
According to Theorem 2.4. we replace volume element dxdydz by |J |dρdϕdz,where the Jacobian of the trasformation to cylindrical coordinates satisfies
J(ρ, ϕ, z) =
∣∣∣∣∣∣∣∣∣∂x∂ρ
∂x∂ϕ
∂x∂z
∂y∂ρ
∂y∂ϕ
∂y∂z
∂z∂ρ
∂z∂ϕ
∂z∂z
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣cosϕ −ρ sinϕ 0
sinϕ ρ cosϕ 0
0 0 1
∣∣∣∣∣∣∣∣∣ = ρ.
The transformation of the triple integral to cylindrical coordinates can then bewritten in the form∫∫∫
Ω
f(x, y, z)dxdydz =
∫∫∫Ω∗
f(ρ cosϕ, ρ sinϕ, z)ρ dρ dϕ dz. (20)
Example 2.5.
Compute∫∫∫
Ω
dxdydz over the domain Ω bounded by surfaces x2 +y2 = 1, z =
0, z = 1.
Solution: The domain Ω is the cylinder symetrical with respect to the z−axis, withradius of the base ρ = 1 and height z = 1 (see Fig. 20). We need to determine thebounds of the transformated domain Ω∗. It is obvious that 0 ≤ z ≤ 1. Inequations
32
Fig. 20. Domain Ω in Example 2.5.
for coordinates ρ, ϕ are the same as for transformation to polar coordinates, i.e.0 ≤ ρ ≤ 1, 0 ≤ ϕ ≤ 2π. Therefore
∫∫∫Ω
dxdydz =
∫∫∫Ω∗
ρ dρ dϕ dz =
2π∫0
dϕ
1∫0
ρ dρ
1∫0
z dz =
= [ϕ]2π0 ·[ρ2
2
]1
0
· [z]10 = π.
Example 2.6.
Compute∫∫∫
Ω
dxdydz over the domain Ω bounded by surfaces z ≥ 3x2 +
3y2, z ≤ 1− x2 − y2.
Solution: The domain is similar to the domain in Example 2.3. Both surfaces areparaboloids and the orthogonal projection Ω1 of the domain Ω to coordinate planex, y is a ring. We obtain its equation from the intersection of both paraboloids
3x2 + 3y2 = 1− x2 − y2
x2 + y2 =1
4.
Therefore, inequations for coordinates ρ, ϕ must fulfill 0 ≤ ρ ≤ 1
2, 0 ≤ ϕ ≤ 2π.
We obtain limits of the variable z from equations of both paraboloids by substi-tuting of variables
z = 3x2 + 3y2 = 3ρ2 cos2 ϕ+ 3ρ2 sin2 ϕ = 3ρ2,
33
z = 1− x2 − y2 = 1− ρ2 cos2 ϕ− ρ2 sin2 ϕ = 1− ρ2.
Hence, 3ρ2 ≤ z ≤ 1− ρ2 and we compute the integral by using (20)
∫∫∫Ω
dxdydz =
∫∫∫Ω∗
ρ dρ dϕ dz =
2π∫0
dϕ
1/2∫0
ρ dρ
1−ρ2∫3ρ2
dz =
= 2π
1/2∫0
ρ [z]1−ρ2
3ρ2 dρ = 2π
1/2∫0
ρ(1− 4ρ2) dρ = 2π
1/2∫0
(ρ− 4ρ3) dρ =
= 2π
[ρ2
2− ρ4
]1/2
0
= 2π · 1
16=π
8.
ExercisesCompute the following integrals over their domains Ω.
1.∫∫∫
Ω
z dxdydz, Ω = x2 + y2 ≤ 4, 0 ≤ z ≤ 2, y ≥ 0. [4π]
2.∫∫∫
Ω
dxdydz, Ω = x2 + y2 ≤ 1, x ≥ 0, 0 ≤ z ≤ 6. [3π]
3.∫∫∫
Ω
z√x2 + y2 dxdydz, Ω = x2 + y2 ≤ 2x, 0 ≤ z ≤ 1. [
16
9]
4.∫∫∫
Ω
z dxdydzΩ = z ≥√x2 + y2, z ≤ 1. [
π
4]
2.3.2 Transformation to Spherical Coordinates
Transformation to spherical coordinates is suitable for integrals where integrationdomains are spheres, ellipsoids or their parts or in cases when orthogonal pro-jection 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 originof the coordinates, ϕ denotes positively oriented angle in coordinate plane x, ybetween positive part of the x−axis and the projection ρ1 of the radius vector ρto coordinate plane x, y and ϑ denotes positively oriented angle between positive
34
Fig. 21. Transformation to spherical coordinates
part of the z−axis and the radius vector ρ. We obtain transformation equationsfrom Fig. 21
x = ρ cosϕ sinϑ
y = ρ sinϕ sinϑ (21)z = ρ cosϑ
Jacobian of the transformation to spherical coordinates satisfies
J =
∣∣∣∣∣∣∣∣∣∂x∂ρ
∂x∂ϕ
∂x∂ϑ
∂y∂ρ
∂y∂ϕ
∂y∂ϑ
∂z∂ρ
∂z∂ϕ
∂z∂ϑ
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣cosϕ sinϑ −ρ sinϕ sinϑ ρ cosϕ cosϑ
sinϕ sinϑ ρ cosϕ sinϑ ρ sinϕ cosϑ
cosϑ 0 −ρ sinϑ
∣∣∣∣∣∣∣∣∣ = −ρ2 sinϑ.
Transformation of the triple integral to spherical coordinates can be written in theform ∫∫∫
Ω
f(x, y, z)dxdydz =
=
∫∫∫Ω∗
f(ρ cosϕ sinϑ, ρ sinϕ sinϑ, ρ cosϑ) ρ2 sinϑ dρ dϕ dϑ. (22)
The sphere with the center in the origin of coordinates and radius a transformedto spherical coordinates is mapped to domain Ω∗ given by inequations 0 ≤ ρ ≤a, 0 ≤ ϕ ≤ 2π, 0 ≤ ϑ ≤ π.
Example 2.7.
Compute∫∫∫
Ω
(x2 + y2 + z2) dxdydz over the domain Ω bounded by surfaces
35
x2 + y2 + z2 ≥ 1, x2 + y2 + z2 ≤ 4.
Solution: The domain Ω is bounded by two spherical surfaces with the center inthe origin of coordinates and radiuses a = 1, b = 2. We use transformation tospherical coordinates with inequations 1 ≤ ρ ≤ 2, 0 ≤ ϕ ≤ 2π, 0 ≤ ϑ ≤ π andcalculate the integral by using (22)∫∫∫
Ω
(x2 + y2 + z2) dxdydz =
=
∫∫∫Ω∗
(ρ2 cos2 ϕ sin2 ϑ+ ρ2 sin2 ϕ sin2 ϑ+ ρ2 cos2 ϑ)ρ2 sinϑ dρ dϕ dϑ =
=
2∫1
dρ
2π∫0
dϕ
π∫0
ρ4 sinϑ dϑ =
2∫1
ρ4dρ
2π∫0
dϕ
π∫0
sinϑ dϑ =
=
[ρ5
5
]2
1
· [ϕ]2π0 · [− cosϑ]π0 =31
5· 2π · 2 =
124
5π.
Example 2.8.
Compute∫∫∫
Ω
z dxdydz over the domain Ω bounded by surfaces x2 + y2 + z2 ≤
4, x ≥ 0, y ≥ 0, z ≥ 0.
Solution: The domain Ω is one eighth of the sphere in the first octant with thecenter in the origin of coordinates and radius a = 2. Therefore, 0 ≤ ρ ≤ 2, 0 ≤ϕ ≤ π
2, 0 ≤ ϑ ≤ π
2and we can calculate the integral∫∫∫
Ω
z dxdydz =
∫∫∫Ω∗
ρ cosϑ ρ2 sinϑ dρ dϕ dϑ =
=
2∫0
dρ
π/2∫0
dϕ
π/2∫0
ρ3 sinϑ cosϑ dϑ =
2∫0
ρ3 dρ
π/2∫0
dϕ
π/2∫0
1
2sin 2ϑ dϑ =
[ρ4
4
]2
0
· [ϕ]π/20 ·[−1
4cos 2ϑ
]π/20
= 4 · π2· 1
2= π.
ExercisesCompute the following integrals over their domains Ω.
36
1.∫∫∫
Ω
z(x2 + y2) dxdydz, Ω = x2 + y2 + z2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0.
[π
48]
2.∫∫∫
Ω
√x2 + y2 + z2 dxdydz, Ω = x2 + y2 + z2 ≤ 1, x ≥ 0, y ≥ 0, z ≥
0. [π
8]
3.∫∫∫
Ω
(x2 + y2) dxdydz, Ω = 4 ≤ x2 + y2 + z2 ≤ 9, z ≥ 0. [844
15π]
2.4 Practical Applications of the Triple Integral2.4.1 Volume of a Body
As we have already mentioned in section 2.1., the physical meaning of the tripleintegral of the function f(x, y, z) ≥ 0 over the domain Ω is the mass of the domainΩ with the density σ = f(x, y, z). If σ = 1, the triple integral has a meaning ofvolume of the body Ω
V =
∫∫∫Ω
dxdydz (23)
Example 2.9.Compute the volume of the body bounded by cylindrical surfaces z = 5−y2, z =y2 + 3 and planes x = 0, x = 2.
Solution: While limits of variables x, z are obvious, we need to calculate intersec-tions of cylindrical surfaces to determine the limits of variable y.
5− y2 = y2 + 3
2y2 = 2
y = ±1
Therefore, inequations for the domain Ω are in the form
0 ≤ x ≤ 2
−1 ≤ y ≤ 1
y2 + 3 ≤ z ≤ 5− y2
37
We calculate volume of the body using (23)
V =
∫∫∫Ω
dxdydz =
2∫0
dx
1∫−1
dy
5−y2∫y2+3
dz = [x]20
1∫−1
[z]5−y2
y2+3dy =
= 2
1∫−1
(2− 2y2) dy = 4
[y − y3
3
]1
−1
= 4 · 4
3=
16
3.
ExercisesCompute the volumes of bodies bounded by their surfaces.
1. z = x2 + y2, z = y. [π
32]
2. x− y + z = 6, x+ y = 2, x = y, y = 0, z = 0. [16
3]
3. y = x2, z = 0, y + z = 2. [32√
2
15]
2.4.2 Mass of a Body
Mass of the domain Ω is given by
m =
∫∫∫Ω
σ(x, y, z) dxdydz, (24)
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 + y2 + z2 =1, x2 + y2 + z2 = 4 with the density defined by σ = x2 + y2 + z2.
Solution: The problem was solved in Example 2.7.
ExercisesCompute the mass of bodies.
1. x2 + y2 + z2 ≤ 4, σ =√x2 + y2 + z2. [16π]
2. x2 + y2 + z2 ≤ 1, σ =2
x2 + y2 + z2. [8π]
3. x2 = 2y, y + z = 1, 2y + z = 2, σ = y. [8√
2
35]
38
3 Line Integral
3.1 The Curve and Its OrientationTo 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 thecurve.
Definition 3.1.Let x = x(t), y = y(t), z = z(t) be continuous functions for t ∈ 〈a, b〉.
1. The curve k with parametrical equations x = x(t), y = y(t), z = z(t), t ∈〈a, b〉 is called positively oriented with respect to the parameter t, if andonly if its points are ordered so that for arbitrary values t1, t2 ∈ 〈a, b〉, t1 <t2, the point M1 = [x(t1), y(t1), z(t1)] lies before the pointM2 = [x(t2), y(t2), z(t2)], i.e.
∀t1, t2 ∈ 〈a, b〉 : t1 < t2 ⇔ M1 ≺M2.
Reversely,∀t1, t2 ∈ 〈a, b〉 : t1 < t2 ⇔ M2 ≺M1,
the curve is called negatively oriented with respect to the parameter t.
2. If the curve k is positively oriented with respect to the parameter t ∈ 〈a, b〉,then the point A = [x(a), y(a), z(a)] is called the starting point of thecurve and the point B = [x(b), y(b), z(b)] is called the end point of thecurve.
3. The curve is called closed, if A = B.
4. The curve is called smooth in 〈a, b〉, if there exists continuous derivativesof parametrical equations x′ = x′(t), y′ = y′(t), z′ = z′(t) and ∀t ∈ 〈a, b〉 :(x′(t), y′(t), z′(t)) 6= (0, 0, 0).
5. The curve is called piecewise smooth in 〈a, b〉, if it is smooth in 〈a, b〉 ex-cept for a finite number of points ti ∈ 〈a, b〉, i = 1, . . . , n.
6. The curve is called simple in 〈a, b〉, if it doesn’t intersect itselfs, i.e.
∀t1, t2 ∈ (a, b) : t1 6= t2 ⇒ M1 6= M2.
The symbol ≺ means "precedes" or "lies before".
Example 3.1.Write parametrizations of
39
1. the line segment between points A = [a1, a2, a3], B = [b1, b2, b3],
2. the circle x2 + y2 = r2, r > 0,
3. the circle (x−m)2 + (y − n)2 = r2, r > 0,
4. the ellipsex2
a2+y2
b2= 1, a, b > 0.
Solution:
1. We should know from the analytical geometry that the parametrical equa-tions 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, t ∈ 〈0, 1〉.z = a3 + u3 · t,
where u = ~AB = (u1, u2, u3) is the vector parallel to the line segment AB.
2. Parametrical equations of the circle with the center in the origin of coordi-nates and radius r are in the form
x = r cos t,
y = r sin t, t ∈ 〈0, 2π).
3. Analogically for the circle with the center in the pointC = [m,n] and radiusr, the parametrical equations are in the form
x = m+ r cos t,
y = n+ r sin t, t ∈ 〈0, 2π).
4. Parametrical equations of the ellipse with the center in the origin of coordi-nates and semi-axis a, b are in the form
x = a cos t,
y = b sin t, t ∈ 〈0, 2π).
40
Fig. 22. Dividing of the curve k - the domain of the line integral
3.2 Line Integral of a Scalar FieldAnalogically to the other types of integral, we need to divide our domain - thecurve k - into small elements. Let us consider the simple smooth curve k withparametrization x = x(t), y = y(t), z = z(t), t ∈ 〈a, b〉 positively orientedwith respect to the parameter t. We divide interval 〈a, b〉 by sequence of pointsa = t0 < t1 < · · · < tn = b into n partial curves k1, k2, . . . , kn (see Fig. 22). Foreach i = 1, . . . , n, we denote by ∆si the length of each element ki and we choosean arbitrary point Mi = [x(ti), y(ti), z(ti)] in each element ki. Moreover, we areable to construct positively oriented unitary tangential vector τi(Mi) at each pointMi that we will use in section 3.3. to define the line integral of a vector field. Thecurve k lies within the domain Ω and we consider a bounded continuous scalarfunction u(X) = u(x, y, z) defined for each X ∈ Ω. Now we can create the sumof products
n∑i=1
u(Mi) ·∆si =n∑i=1
u(x(ti), y(ti), z(ti)) ·∆si
and define the line integral of a scalar field.
Definition 3.2.
If there exists limn∑i=1
u(x(ti), y(ti), z(ti)) ·∆si for n→∞ and ∆si → 0 we call
it the line integral of a scalar field u(x, y, z) along the curve k and denote it∫k
u(x, y, z) ds.
The line integral of the scalar field doesn’t depend on the orientation of the
41
curve, because the lengths of all components ∆si are always positive. The ele-ment of the curve ds in the three-dimensional space is the body diagonal of therectangular hexahedron with sides dx, dy, dz. Therefore we obtain
ds =√
(dx)2 + (dy)2 + (dz)2 =√
(x(t)dt)2 + (y(t)dt)2 + (z(t)dt)2 =
=√
(x(t))2 + (y(t))2 + (z(t))2 dt
and the line integral of the scalar field can then be written in the form
∫k
u(x, y, z) ds =
b∫a
u(x(t), y(t), z(t))√
(x(t))2 + (y(t))2 + (z(t))2 dt. (25)
That way we transform the line integral of the scalar field into the one-dimensionaldefinite integral.
Theorem 3.1. (Properties of the line integral of a scalar field)
1.∫k
cu(X)ds = c
∫k
u(X)ds,
2.∫k
(u(X) + v(X))ds =
∫k
u(X)ds+
∫k
v(X)ds,
3.∫k
u(X)ds =
∫k1
u(X)ds+
∫k2
u(X)ds,
where c ∈ R, k1, k2 are curves such that curve k fulfills k = k1 ∪ k2 andu(X), v(X) are bounded continuous scalar functions for all X ∈ Ω that con-tains the curve k.
Example 3.2.
Calculate the line integral∫k
(x + z) ds along the line segment between points
A = [1, 2, 3], B = [3, 2, 1].
Solution: We create the parametrization of the line segment
x = 1 + 2t
y = 2 t ∈ 〈0, 1〉z = 3− 2t
42
and its derivatives: x = 2, y = 0, z = −2.We express the element ds
ds =√
(x(t))2 + (y(t))2 + (z(t))2 dt =√
8 dt = 2√
2 dt.
We use (25) to calculate the line integral
∫k
(x+ z) ds =
1∫0
(1 + 2t+ 3− 2t) · 2√
2 dt = 8√
2
1∫0
dt = 8√
2.
If we consider just the two-dimensional problem, i.e. the curve is in x, y plane,the element ds =
√(x(t))2 + (y(t))2 dt and the line integral of the scalar field is
then in the form∫k
u(x, y) ds =
b∫a
u(x(t), y(t))√
(x(t))2 + (y(t))2 dt. (26)
Example 3.3.
Calculate the line integral∫k
1 ds, where k is the first arc of cycloid
x = a(t− sin t), y = a(1− cos t), a > 0, t ∈ 〈0, 2π〉.
Solution: We express derivatives x = a(1− cos t), y = a sin t and the element ofthe curve
ds =√
(x(t))2 + (y(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 =
=√
2a
√2 sin2 t
2dt = 2a sin
t
2dt.
Now, we calculate the integral using (26)
∫k
1 ds =
2π∫0
2a sint
2dt = 2a
[−2 cos
t
2
]2π
0
= −4a · (−1− 1) = 8a.
43
Suppose the curve in x, y plane can be expressed explicitly by y = y(x), x ∈〈a, b〉. The element of the curve is then given by
ds =√
(dx)2 + (dy)2 =√
(dx)2 + (y′(x)dx)2 =√
1 + (y′(x))2 dx
and the line integral is then in the form
∫k
u(x, y) ds =
b∫a
u(x, y(x))√
1 + (y′(x))2 dx. (27)
Example 3.4.Calculate the line integral
∫k
y ds, where k is a part of the function y = x3 between
points A = [0, 0], B = [1, 1].
Solution: We calculate a derivative of the function
y′ = 3x2,
an element of the curve
ds =√
1 + (3x2)2 dx =√
1 + 9x4 dx
and finally the integral by using (27)
∫k
y ds =
1∫0
x3√
1 + 9x4 dx =1
36
10∫1
√t dt =
1
36· 23
[√t3]10
1=
1
54(10√
10−1),
where we used substitution 1 + 9x4 = t, 36x3 dx = dt.
ExercisesCompute the line integrals of scalar fields along given curves.
1.∫k
z2
x2 + y2ds, k is one thread of the spiral x = cos t, y = sin t, z = t
[8
3
√2π3]
2.∫k
x ds, k is a line segment between points A = [0, 0], B = [1, 2] [
√5
2]
44
3.∫k
x2 ds, k is an upper half of the circle x2 + y2 = a2, a > 0 [1
2πa3]
4.∫k
x2 ds, k : y = lnx, x ∈ 〈1, 3〉 [1
3(10√
10− 2√
2)]
3.3 Line Integral of a Vector FieldIn the previous section we looked at line integral of a scalar field u(x, y, z). In thissection we are going to evaluate line integral of vector fields. We consider simplesmooth curve k with the same parametrization and division to small elementsas we have already discribed in section 3.2. Furthermore, we consider boundedcontinuous vector field F(X) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k definedfor each X ∈ Ω. We create scalar products of the vector field and the positivelyoriented tangential vector of each element of the curve F(Mi) ·∆si, where ∆si =∆siτi. Now we can create sum of such products
n∑i=1
F(Mi) ·∆si =n∑i=1
F(x(ti), y(ti), z(ti)) ·∆siτi
and define the line integral of a vector field.
Definition 3.2.
If there existsn∑i=1
F(x(ti), y(ti), z(ti)) ·∆siτi for n → ∞ and ∆si → 0, we call
it the line integral of a vector field F(x, y, z) along the curve k+ and denote it∫k+
F(x, y, z) ds, where k+ denotes positively oriented curve k. While k− would
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 orientationof the curve. To derive the form of the line integral of the vector field, we need toexpress 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 = x(t) dt, dy = y(t) dt, dz = z(t) dt by using parametriza-tion of the curve. The line integral of the vector field then can be written in the
45
form∫k
F(x, y, z) ds = ε
b∫a
[P (x(t), y(t), z(t))x(t) +Q(x(t), y(t), z(t))y(t)+
+R(x(t), y(t), z(t))z(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 theline integral of the vector field into the one-dimensional definite integral, similarlyto the case of the line integral of a scalar field.
Theorem 3.2. (Properties of the line integral of a vector field)
1.∫k
cF(X)ds = c
∫k
F(X)ds,
2.∫k
(F(X) + G(X))ds =
∫k
F(X)ds +
∫k
G(X)ds,
3.∫k
F(X)ds =
∫k1
F(X)ds +
∫k2
F(X)ds,
4.∫k+
F(X)ds = −∫k−
F(X)ds.
where c ∈ R, k1, k2 are curves such that curve k fulfills k = k1 ∪ k2 (consideringthe same orientation of these curves) and F(X),G(X) are bounded continuousvector 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, thatis one thread of the spiral x = 2 cos t, y = 2 sin t, z = 3t, t ∈ 〈0, 2π〉.
Solution: We express derivatives of parametrization
x = −2 sin t
y = 2 cos t
z = 3
and we can calculate the integral by using (28)∫k
F(x, y, z)ds =
∫k
x dx+ y dy + z dz =
46
=
2π∫0
[2 cos t · (−2 sin t) + 2 sin t · 2 cos t+ 3t · 3] dt =
=
2π∫0
[−4 sin t cos t+ 4 sin t cos t+ 9t] dt =
2π∫0
9t dt =9
2
[t2]2π
0= 18π2.
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 elementds = (dx, dy) = (x(t)dt, y(t)dt). The line integral of the vector field is then inthe form ∫
k
F(x, y) ds = ε
b∫a
[P (x(t), y(t))x(t) +Q(x(t), y(t))y(t)]dt (29)
Suppose the curve in x, y plane is expressed explicitly by y = y(x), x ∈ 〈a, b〉.The tagential vector of the curve is given by ds = (dx, dy) = (dx, y′(x) dx) andthe line integral of a vector field is then in the form
∫k
F(x, y) ds = ε
b∫a
[P (x, y(x)) +Q(x, y(x))y′(x)] dx (30)
Example 3.6.
Calculate the line integral∫k
(x+y) dx+(x−y) dy, where k : y =1
x, x ∈ 〈2, 3〉.
Solution: The curve is given explicitly
y =1
x,
dy = − 1
x2dx.
We calculate the integral by using (30)
∫k
(x+ y) dx+ (x− y) dy =
3∫2
[(x+
1
x) + (x− 1
x) · (− 1
x2)
]dx =
47
=
3∫2
[x+
1
x− 1
x+
1
x3
]dx =
3∫2
(x+1
x3) dx =
[x2
2− 1
2x2
]3
2
=
=9
2− 1
18− 2 +
1
8=
185
72.
ExercisesCompute the line integrals of vector fields along given curves.
1.∫k
x dx−y dy+z dz, k is an oriented line segment ~AB : A = [1, 1, 1], B =
[4, 3, 2] [5]
2.∫k
y dx+ x dy, k is one quarter of the circle x = a cos t, y = a sin t, a >
0, t ∈ 〈0, π2〉 [0]
3.∫k
(xy − 1) dx + x2y dy, k is an arc of the ellipse x = cos t, y = 2 sin t
from the starting point A = [1, 0] to the end point B = [0, 2]. [4
3]
4.∫k
xy dx+ (y−x) dy, k is a part of the parabola y2 = x from the starting
point A = [0, 0] to the end point B = [1, 1]. [17
30]
3.4 Green’s TheoremGreen’s Theorem expresses the relation between a line integral of a vector fieldalong 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 isorientated positively with respect to the domain Ω if traveling on the curve wealways have the domain Ω on the left side (see Fig. 23).
Positive orientation of the curve means traveling in a counterclockwise di-rection, while negative orientation means traveling in a clockwise direction. We
denote the line integral along a closed curve k by the symbol∮k
.
48
Fig. 23. Positively orientated curve k with respect to the domain Ω
Theorem 3.3. (Green’s)Let two-dimensional vector field F(x, y) = P (x, y)i + Q(x, y)j have continuousderivatives on the plane domain Ω, which is bounded by simple piecewise smoothclosed curve k orientated positively with respect to the domain Ω. Then∮
k
P (x, y)dx+Q(x, y)dy =
∫∫Ω
[∂Q(x, y)
∂x− ∂P (x, y)
∂y
]dxdy. (31)
The Green’s Theorem transforms the line integral of a vector field along aplane closed curve to a double integral over the domain Ω, that is bounded by thecurve k. It is especially useful in situations when k is a polygon and we have tocalculate as many line integrals as the lines the polygon consists of.
Example 3.7.
Calculate the integral∮k
(2xy − 5y) dx + (x2 + y) dy, where k is the circle with
the center in the origin of coordinates and radius r.
Solution: The disc x2 + y2 ≤ r is two-dimensional domain that is normal withrespect to both coordinate axis. The curve k is simple and closed and we canorientate it positively with respect to Ω. Also both functions P (x, y) = 2xy − 5yand Q(x, y) = x2 + y fulfill assumptions of the Green’s Theorem. We calculatederivatives
P ′y = 2x− 5
Q′x = 2x
and evaluate the integral by using (31)∮k
(2xy−5y) dx+(x2 +y) dy =
∫∫Ω
(2x−(2x−5)) dxdy =
∫∫Ω
5 dxdy = 5πr2.
49
Fig. 24. Curve k and domain Ω in Example 3.8.
Example 3.8.
Calculate the integral∮k
(x2 + y2) dx+ (x+ y)2 dy, where k consists of the sides
of the triangle ABC: A = [1, 1], B = [1, 3], C = [3, 3].
Solution: All assumptions of the Green’s Theorem are fulfilled.
P = x2 + y2, Q = (x+ y)2,
P ′y = 2y, Q′x = 2(x+ y).
The domain is shown on the Fig. 24. We calculate the double integral as a normalone with respect to the x−axis with inequations for Ω in the form
1 ≤ x ≤ 3,
x ≤ y ≤ 3.
By using (31) we obtain∮k
(x2 + y2) dx+ (x+ y)2 dy =
∫∫Ω
(2x+ 2y − 2y) dxdy = 2
3∫1
x dx
3∫x
dy =
= 2
3∫1
x[y]3x dx = 2
3∫1
(3x− x2) dx = 2
[3
2x2 − x3
3
]3
1
=20
3.
ExercisesCompute the line integrals of vector fields along given curves by using the Green’sTheorem.
50
1.∮k
(x2 + y2) dy, k are sides of the rectangle x = 0, y = 0, x = 2, y = 4.
[16]
2.∮k
2y dx− (x+ y) dy, k are sides of the triangle x = 0, y = 0, x+ 2y = 4.
[−12]
3.∮k
(x+ y) dx− (x− y) dy, k is the ellipse 4x2 + 9y2 = 36. [−12π]
4.∮k
(ex sin y − 16y) dx+ (ex cos y + 16) dy, k is the circle x2 + y2 = 2x.
[16π]
3.5 Path Independence of the Line IntegralWe can use another method of calculation of the line integral of a vector field insituations when the value of the integral doesn’t depend on an integration curveand 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)kbe continuous over the domain Ω. If the value of the line integral of the vectorfield ∫
k
F(X) ds =
∫k
P (x, y, z) dx+Q(x, y, z) dy +R(x, y, z) dz
doesn’t depend on the integration curve k that lies within the domain Ω, we saythe integral is path independent between points A,B.If this property is fulfilled for arbitrary points A,B ∈ Ω, we say integral is pathindependent over Ω.
Theorem 3.4.Let F(X) = P (X)i + Q(X)j + R(X)k have continuous partial derivatives in adomain Ω. Let a curve k lie within Ω, A be the starting point of the curve k, whileB be its end point. Then:
1. The line integral∫k
F(X)ds =
∫k
P (X)dx + Q(X)dy + R(X)dz is path
independent over Ω if and only if there exists some scalar function Φ(x, y, z)
51
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.
2. The line integral∫k
F(X)ds =
∫k
P (X)dx + Q(X)dy + R(X)dz is path
independent over Ω if and only if
curlF =
∣∣∣∣∣∣∣∣∣i j k
∂∂x
∂∂y
∂∂z
P Q R
∣∣∣∣∣∣∣∣∣ = o.
3. In such case, the line integral of the vector field F(X) along a curve k fromthe starting point A to the end point B is given by the difference of thevalues of scalar potential in the end point B and starting point A :∫
k
F(X)ds =
∫k
P (X)dx+Q(X)dy +R(X)dz = Φ(B)− Φ(A).
4. Hence, if the line integral is path independent and the curve k is closed∮k
F(X)ds = 0.
If the problem is considered only in xy−plane, the path independent line inte-gral of the vector field is given by∫
k
F(X)ds =
∫k
P (x, y)dx+Q(x, y)dz = Φ(B)− Φ(A).
The test for determining if a vector field is conservative can be written in the form
∂P
∂y=∂Q
∂x.
We will show the way of calculation of the scalar potential at following examples.
52
Example 3.9.
Prove that the line integral∫k
(3x2 − 2xy + y2) dx− (x2 − 2xy + 3y2) dy is path
independent. If it is, find the scalar potential.
Solution: Components of the vector field
P = 3x2 − 2xy + y2
Q = x2 − 2xy + 3y2
and their partial derivatives
∂P
∂y=∂Q
∂x= −2x+ 2y
are continuous. The test condition is fulfilled. To find the potential we first inte-grate P (x, y) = 3x2 − 2xy + y2 with respect to x.
Φ(x, y) =
∫P (x, y)dx =
∫(3x2 − 2xy + y2) dx = x3 − x2y + xy2 + C(y),
where C(y) is a function of variable y. We determine it by setting the partial
derivative∂Φ
∂yequal to Q(x, y).
∂Φ
∂y= −x2 + 2xy + C ′(y) = −x2 + 2xy − 3y2 = Q(x, y).
Hence, C ′(y) = −3y2. Then
C(y) = −∫
3y2 dy = −y3 + C1,
where C1 is a constant. The scalar potential is in the form
Φ(x, y) = x3 − x2y + xy2 − y3 + C1.
In general, the scalar potential over xy−plane can be written in the form
Φ(x, y) =
x∫x0
P (t, y) dt+
y∫y0
Q(x0, t) dt, (32)
53
where [x0, y0] is an arbitrary point of a domain Ω. While in three-dimensionalspace it can be written in the form
Φ(x, y, z) =
x∫x0
P (t, y, z) dt+
y∫y0
Q(x0, t, z) dt,+
z∫z0
R(x0, y0, t) dt, (33)
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]
Φ(x, y) =
x∫0
(3t2−2ty+y2) dt−y∫
0
(02−2·0·t+3t2) dt = [t3−t2y+ty2]x0−[t3]y0 =
= x3 − x2y + xy2 − y3.
Example 3.10.
Calculate the integral∮k
x2 dx+ y2 dy along the closed curve k : x2 + y2 = r2.
Solution: Functions P = x2 and Q = y2 fulfill assumptions of Theorem 3.4.
Partial derivatives∂P
∂y=∂Q
∂x= 0. Hence, the integral is path independant. The
curve is closed. Therefore, the integral must be equal to zero.
Example 3.11.
Calculate the integral∫k
(2x + yz) dx + (xz + z2) dy + (xy + 2yz) dz from the
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)
∂y= z,
∂P (x, y, z)
∂z= y,
Q(x, y, z) = xz + z2,∂Q(x, y, z)
∂x= z,
∂Q(x, y, z)
∂z= x+ 2z,
R(x, y, z) = xy + 2yz,∂R(x, y, z)
∂x= y,
∂R(x, y, z)
∂y= x+ 2z.
We can see that∂Q
∂x=∂P
∂y,
∂R
∂y=∂Q
∂z,
∂P
∂z=∂R
∂x.
54
Hence, curlF = o and integral is path independent. We calculate the scalar po-tential by using (33). The origin of coordinates [0, 0, 0] is chosen as the arbitrarypoint for [x0, y0, z0].
Φ(x, y, z) =
x∫0
(2t+ yz) dt+
y∫0
z2 dt+
z∫0
0 dt = x2 + xyz + yz2.
We calculate the integral according to Theorem 3.4∫k
(2x+ yz) dx+ (xz + z2) dy + (xy + 2yz) dz = Φ(B)− Φ(A) =
= [x2 + xyz + yz2]BA = 12 + 1 · 2 · 3 + 2 · 32 − (12 + 1 · 1 · 1 + 1 · 12) = 22.
ExercisesProve that following integrals are path independent. Then, calculate them if A isthe starting point and B is the end point.
1.∫k
x dx+ y dy
x2 + y2, A = [1, 2], B = [2, 3]. [
1
2ln
13
5]
2.∫k
(2y − 6xy3) dx+ (2x− 9x2y2) dy, A = [1, 1], B = [4, 1]. [−39]
3.∫k
yz dx+ xz dy + xy dz, A = [2, 2, 2], B = [2, 3, 4]. [16]
4.∫k
dx+ 2 dy + 3 dz
x+ 2y + 3z, A = [0, 1, 0], B = [1, 0, 1]. [ln 2]
3.6 Practical Applications of the Line Integral3.6.1 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
A =
∫k
f(x, y) ds. (34)
It is the geometrical meaning of the line integral of a scalar field.
55
Fig. 25. Area of a Cylindrical Region
Example 3.12.Calculate the area of the cylindrical surface x2 + y2 = r2 bounded by z ≥ 0 andz ≤ x.
Solution: The curve k is a part of the circle x2 + y2 = 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 ∈ 〈−π2,π
2〉. We express derivatives of the parametric equations
x = −r sin t,
y = r cos t
and the element of the curve is then given by
ds =√
(x(t))2 + (y(t))2 dt =√r2 sin2 t+ r2 cos2 t dt = r dt.
By using (34) we obtain
A =
∫k
x ds =
π/2∫−π/2
r cos t rdt = r2[sin t]π/2−π/2 = 2r2.
ExercisesCalculate the area of cylindrical surfaces bounded by z = 0 and
1. x2 + y2 = r2, z =xy
2r, x ≥ 0, y ≥ 0. [
1
4r2]
56
2. 9y2 = 4(x− 1)3, z = 2−√x. [
11
3]
3. y =3
8x2, z = x, x = 0, y = 6. [
16
27(10√
10− 1)]
3.6.2 Length of a Curve
Let k be simple, piecewise smooth curve. The length of the curve is numericallyequal to the area of the cylindrical surface above the curve k that is bounded byplanes z = 0 and z = 1. Hence, by letting f(x, y) = 1 from (34) we obtain
L =
∫k
1 ds. (35)
Example 3.13.Calculate the length of one period of the cycloid x = a(t−sin t), y = a(1−cos t),t ∈ 〈0, 2π〉.
Solution: Similarly to the previous exercise we start with derivatives of the para-metric equations of the cycloid
x = a(1− cos t),
y = a sin t.
Further, we continue with the element of the curve
ds =√
(x(t))2 + (y(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 identity2 sin2 x = 1− cos 2x :
L =
∫k
1 ds =√
2a
2π∫0
√1− cos t dt =
√2a
2π∫0
√2 sin2 t
2dt = 2a
2π∫0
sint
2dt =
= 2a[−2 cost
2]2π0 = −4a · (−1− 1) = 8a.
ExercisesCalculate the lengths of the curves
1. first arc of a cycloid x = a(t− sin t), y = a(1− cos t), t ∈ 〈0, 2π〉, a > 0[8a]
57
2. cardioid x = 2a cos t− a cos 2t, y = 2a sin t− a sin 2t, t ∈ 〈0, 2π〉, a > 0[16a]
3. y =1
2lnx, z =
1
2x2, x ∈ 〈1, 2〉 [
1
2(3 + ln 2)]
4. y =1
2x2, z =
1
6x3, x ∈ 〈0, 1〉. [
7
6]
3.6.3 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 avector field
W =
∫k
F ds =
∫k
P (x, y, z)dx+Q(x, y, z)dy +R(x, y, z)dz. (36)
Example 3.14.Calculate the work done by the force field F = (xy, x + y) on an object movingalong 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 ∈ 〈0, 1〉 and calcu-late the work according to (36)
W =
∫k
F ds =
∫k
xydx+(x+y)dy =
1∫0
(x2+2x) dx = [x3
3+x2]10 =
1
3+1 =
4
3.
Exercises
1. Calculate the work done by the force field F = (xy, x + y) on an objectmoving from the point A = [0, 0] to the point B = [1, 1] if the curve k :
x = y2. [37
30]
2. Calculate the work done by the force field F = (x + y, 2x) on an objectmoving around the closed curve k : x2 + y2 = r2 in a positive direction.
[πr2]
3. Calculate the work done by the force field F = (x2, y2, z2) on an objectmoving around the screw line k : x = cos t, y = sin t, z = t, t ∈ 〈0, π
2〉.
[π3
24]
58
3.6.4 Mass of a Curve
Let k be a simple, piecewise smooth curve and continuous function ρ(x, y, z) beits linear density. The mass of a curve (e.g. mass of a wire) is given by the lineintegral of a scalar field
m =
∫k
ρ(x, y, z) ds. (37)
Example 3.15.Calculate the mass of one thread of the screw line k : x = cos t, y = sin t, z = t,t ∈ 〈0, 2π〉 if its density is given by ρ = x2 + y2 + z2.
Solution: Derivatives of the parametric equations are in the form
x = − sin t,
y = cos t,
z = 1.
We express element of the curve
ds =√
(x(t))2 + (y(t))2 + (z(t))2 dt =√
sin2 t+ cos2 t+ 12 dt =√
2 dt.
Now we can use (37) and calculate the mass of the given curve
m =
∫k
(x2 + y2 + z2) ds =√
2
2π∫0
(cos2 t+ sin2 t+ t2) dt =√
2
2π∫0
(t2 + 1) dt =
=√
2
[t3
3+ t
]2π
0
=√
2
(8π3
3+ 2π
)= 2√
2π
(4
3π3 + 1
).
Exercises
1. Calculate the mass of one quarter of the circle x = a sin t, y = a cos t,
t ∈ 〈0, π2〉 if the density in each point is equal to its y−coordinate. [a2]
2. Calculate the mass of the parabola y =1
2x2 between the points A = [0, 0],
B = [2, 2]. The density ρ =y
x. [
1
6(5√
5− 1)]
3. Calculate the mass of the curve y = lnx, where x ∈ 〈1, 2〉. The density at
each point is equal to the square of its x−coordinate. [1
3(5√
5− 2√
2)]
59
