5.1
5.1.1
Double Integrals
Volume of an enclosed region
Consider the diagram in Figure 1. It shows a curve in two variables z =
f (x, y) that lies above some region R on the xy-plane. How can we calculate
the volume enclosed between the surface and the region R? Recall, that to
z=fHx,yL
-5
0
5
5
0
-5
5
0
R
-5
Figure 1: Volume between z = f (x, y) and the region R.
calculate the area under a curve of one variable we integrate. We should
do the same here. We break up the region R into infinitesimal areas ∆Ak
around the point (x∗k , yk∗ ), and integrate over these areas. This is represented
in Figure 2. We therefore define the double integral in this region to be
given by
ZZ
n
X
f (x, y) dA = lim
(1)
f (x∗k , yk∗ ) ∆Ak ,
n→∞
R
k=1
which is expressed through Riemann sums as was the case for single integrals.
From these, we can calculate the volume between the region R and the surface
z = f (x, y) ≥ 0 to be
ZZ
V =
f (x, y) dA .
R
1
(2)
z=fHx,yL
-5
0
5
5
0
-5
5
0
-5
Hx* k , y* k L
Figure 2: Volume between z = f (x, y) and the region R with an infinitesimal
area sum indicated.
If the condition f (x, y) ≥ 0 does not hold, i.e. f (x, y) does not lie above R
for all values considered, we find the difference in volumes above and below
the xy-plane and we call this the net signed volume. We note that in this
definition f (x, y) appears to be a “height”.
5.1.2
Double integrals over rectangular regions
In order to progress, we must understand what it means to integrate over
a region R. If R is a rectangle with x-coordinates in the range a to b and
y-coordinates in the range c to d, then these will be the respective limits
for the integrations. The area element dA = dxdy for a rectangular region,
and we must understand that we can perform partial definite integration
by treating y as a constant when integrating over x and vice-versa in the
integrals
Z
Z
b
d
f (x, y) dx and
a
f (x, y) dy .
c
2
(3)
To understand this, let us look at partial integrals of x3 y.
1
Z 1
Z 1
y
x4 y 3
3
= ,
x y dx = y
x dx =
4 x=0 4
0
0
Z 1
Z 1
3 2 1
xy x3
3
3
y dy =
x y dy = x
=
.
2 y=0
2
0
0
(4)
Furthermore, a partial definite integral with respect to x can subsequently
be integrated with respect to y, or vice-versa. We call this process iterated
integration and is one of the means by which we perform double integrals,
called iterated integrals. Thus, we have
Z dZ b
Z d Z b
f (x, y) dx dy =
f (x, y) dx dy ,
c
a
c
a
(5)
Z bZ d
Z b Z d
f (x, y) dy dx =
f (x, y) dy dx .
a
c
a
Example: Calculate
Z 2Z 3
(x2 y − 2y) dx dy
1
c
Z
3
Z
and
0
0
2
(x2 y − 2y) dy dx .
1
Solution: Taking the first of these, we have
Z
2
Z
3
Z
2
2
(x y − 2y) dx dy =
1
0
1
Z
x3 y
− 2xy
3
3
dy
x=0
2
[9y − 6y − (0 − 0)] dy
=
1
Z
=
2
3y dy
2
3 2
= y
2
y=1
3
9
= (4 − 1) = .
2
2
1
3
The other gives
3
Z
2
Z
Z
2
3
(x y − 2y) dy dx =
0
0
1
x2 y 2
− y2
2
3
2
dx
y=1
2
x
2
=
2x − 4 −
−1
dx
2
0
Z 3
3 2
=
x − 3 dx
2
0
3
3
x
=
− 3x
2
x=0
27
27 − 18
9
=
− 9 − (0 − 0) =
= .
2
2
2
Z
Notice that both orders of integration give the same result! This is not a
coincidence. It is a consequence of the following theorem:
Fubini’s Theorem:
Let R be the rectangle defined by
a ≤ x ≤ b,
c ≤ y ≤ d.
(6)
If f (x, y) is continuous on this rectangle then
Z
ZZ
d
Z
c
a
a
d
f (x, y) dy dx .
f (x, y) dx dy =
f (x, y) dA =
R
Z bZ
b
(7)
c
In other words, over rectangular regions, we can exchange the order of integration without changing the result.
Example: Find the volume enclosed between the surface z = 4 − x − y and
the region R = [0, 1] × [0, 2].
4
Solution:
Z
1
Z
2
f (x, y) dy dx
V =
0
Z
0
1
2
Z
(4 − x − y) dy dx
2
Z 1
y 2 4y − xy −
=
dx
2 y=0
0
Z 1
=
[8 − 2x − 2] dx
0
Z 1
=
[6 − 2x] dx
0
1
= 6x − x2 x=0 = 6 − 1 = 5 .
=
0
0
(8)
You might like to verify that the opposite order of integration gives the same
result.
5.1.2.1
Properties of double integrals
Below are listed the properties of double integrals
1.
ZZ
ZZ
c f (x, y) dA = c
R
f (x, y) dA ,
where c is constant.
(9)
R
2.
ZZ
f (x, y) ± g(x, y) dA =
R
ZZ
ZZ
f (x, y) dA ±
R
g(x, y) dA .
(10)
R
3. If R1 and R2 are regions such that R = R1 + R2 (see Figure 3), then
ZZ
ZZ
ZZ
f (x, y) dA =
f (x, y) dA +
f (x, y) dA .
(11)
R
R1
R2
5
��
��
Figure 3: Sum of two regions.
Example: Find the values of
ZZ
x dA ,
ZZ
and
x dA ,
R1
R2
for R1 = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2} and R2 = {(x, y) : 1 ≤ x ≤ 2, 0 ≤
y ≤ 2} and check if the sum agrees with
ZZ
x dA
R
where R = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 2}.
Solution: The region R1 gives
ZZ
Z 1Z 2
Z
x dA =
x dy dx =
0
R1
Z
0
1
0
1
Z
(2x − 0) dx =
=
0
0
6
1
2
xy 0 dx
1
2x dx = x2 0 = 1 ,
and the region R2 gives
Z
ZZ
x dA =
2
Z
2
x dy
1
R2
Z
0
2
=
1
2
2x dx = x2 1 = 4 − 1 = 3 .
The sum of the integrals over these regions is therefore 4. Note that R =
R1 + R2 , and so we expect that we should get this result for the calculation
over R.
ZZ
Z 2Z 2
x dA =
x dy
0
R
Z
=
0
0
2
2
2x dx = x2 0 = 4 − 0 = 4 ,
as expected.
5.1.3
Double integrals over non-rectangular regions
In general, we can consider integrating over a region of arbitrary shape and
complexity. In this course we will only consider two types of region:
• A type I region is bounded to the left and right by vertical lines x = a
and x = b, and is bounded below and above by curves y = g1 (x) and
y = g2 (x).
• A type II region is bounded below and above by horizontal lines y = c
and y = d, and is bounded to the left and right by curves x = h1 (y)
and x = h2 (y).
In other words we have the following double integrals
Z bZ
ZZ
Type I region:
g2 (x)
f (x, y) dA =
R
ZZ
Type II region:
f (x, y) dy dx ,
a
Z
g1 (x)
d
Z
f (x, y) dA =
f (x, y) dx dy .
c
R
7
(12)
h2 (y)
h1 (y)
An important thing to notice here is that the variable with limits that depend
on the other variable is integrated over first. So, in a type I region, we always
integrate over y first because the limits depend on x. If we swapped the order
the x integration would miss this contribution and we would get the wrong
result. Some regions are both type I and type II regions. The region bounded
by the circle x2 + y 2 = a2 is a good example. We can integrate in either order
provided we treat the limits for the integrals correctly. This means that to
integrate over y first we must treat it as a type I region and to integrate over
x we treat it as a type II region. The following example should make this
clearer.
Example: Compute the volume of the solid bounded above by z = xy 2 and
below by R, show in Figure 4. This region is between y = x3 above and
y = −x2 below, for values 0 ≤ x ≤ 1.
Solution: The volume is given by
y
1.0
0.5
0.2
0.4
0.6
0.8
1.0
x
-0.5
-1.0
Figure 4: Region R bounded above by y = x3 , below by y = −x2 for values
0 ≤ x ≤ 1.
8
ZZ
V =
xy 2 dA =
1
Z
"Z
1
Z
=
0
1
Z
xy 2 dy dx
−x2
0
R
#
x3
=
xy 3
3
hx
x 3
dx
y=−x2
i
(x9 + x6 ) dx
3
10
1
=
x + x7 dx
3 0
1
1 x11 x8
=
+
3 11
8 x=0
1 1
1
19
19
=
+
=
.
=
3 11 8
3 · 88
264
0
Z
1
Let’s now look at a slightly harder example where you need to work out what
R is.
Example: Find the volume of the tetrahedron bounded by the coordinate
planes and the plane z = 6 − 2x − 3y. The solid and it’s projection to the
xy-plane is sketched in Figure 5.
Solution: Firstly, we see that we can treat this using either a type I or type
II region. Let’s consider the projection to the xy-plane as a type I region.
Then, the region extends from 0 ≤ x ≤ 3 and g1 (x) ≤ y ≤ g2 (x). We find
this by setting z = 0, which gives 0 = 6 − 2x − 3y. To find the interception
with the x-axis, we set y = 0 and find x = 3. Therefore the limits are
0 ≤ x ≤ 3 since we consider the positive x direction only. Similarly, we
consider the positive y-direction, so the lower bound for the y integration
must be g1 (x) = 0. However, the upper bound is given by 0 = 6 − 2x − 3y
and can be rewritten as y = −2x/3+2. Therefore, the integral for the volume
9
6
y=-2x3+2
y
2.0
4
z
1.5
2
1.0
0
0
1
0
0.5
2
1
y
3
2
x
34
0.5
1.0
1.5
2.0
2.5
3.0
x
Figure 5: The tetrahedron z = 6 − 2x − 3y for x, y, z ≥ 0 and the projection
to z = 0.
is given by
ZZ
Z
3
Z
g2 (x)
(6 − 2x − 3y) dA =
V =
(6 − 2x − 3y) dy dx
0
R
Z
g1 (x)
3
Z
−2x/3+2
(6 − 2x − 3y) dy dx
=
0
Z
0
3
=
0
3
3
6y − 2xy − y 2
2
−2x/3+2
dx
y=0
3
2
=
6(−2x/3 + 2) − 2x(−2x/3 + 2) − (−2x/3 + 2) dx
2
0
Z 3
2
2x
dx
=
6 − 4x +
3
0
3
2x3
2
= 6x − 2x +
= 18 − 18 + 6 = 6 .
9 x=0
Z
10
5.1.4
Area as a double integral
We can also calculate area by means of a double integral. If we are given
some surface z = f (x, y), we can set z = 1. and then, we find that
ZZ
1 · dA = area of R .
(13)
V =A·h=A·1=A=
R
Thus
ZZ
Area of R =
dA .
(14)
R
Example: Calculate the area between the curves y = x3 and y =
0 ≤ x ≤ 1, shown in Figure 6.
Solution: The area is given by
y
1.0
0.5
0.2
0.4
0.6
0.8
1.0
x
-0.5
-1.0
Figure 6: The curves y = x3 and y =
ZZ
A=
Z
1
√
Z
dA =
√
x for 0 ≤ x ≤ 1.
x
dy dx
x3
0
R
Z
1
=
Z0 1
=
√x
y y=x3 dx
√
( x − x3 ) dx
0
2x3/2 x4
=
−
3
4
11
1
=
x=0
2 1
5
− =
.
3 4
12
√
x for
5.1.5
Parametric surfaces
We have already encountered the idea of a parametric curve. We have a
parameter that determines where on the curve a point is. It looks like
x = x(t) ,
y = y(t) ,
z = z(t) .
(15)
It is also possible to have a parametric surface, which is a surface with
two parameters, u and v that determines where the point on the surface is.
It looks like
x = x(u, v) , y = y(u, v) , z = z(u, v) .
(16)
Because it is a surface rather than a curve, one parameter would not be
sufficient to uniquely determine the position. As a simple example, take a
straight line in three dimensions. It is a (very simple) curve, and obviously
one parameter, say its length, suffices to describe where we are, regardless
of which direction it is oriented in. Of course, we must have a reference
point, but once we have that it is easy to know where we are based on the
parameter length. If we now consider a square in three dimensions, it again
can be oriented in any direction. We need a reference point again and then
two parameters, say length and width describe where a point is. If we only
had the length say, then all we would know is that the point is somewhere of a
series of parallel lines. Some important and common parametric coordinates
are:
• Rectangular coordinates (also called Cartesian coordinates):
x = u,
y = v.
(17)
y = ρ sin θ .
(18)
• Polar coordinates:
x = ρ cos θ ,
• Cylindrical coordinates:
x = ρ cos θ ,
y = ρ sin θ ,
z.
(19)
z = ρ cos φ .
(20)
• Spherical coordinates:
x = ρ sin φ cos θ ,
y = ρ sin φ sin θ ,
The ranges of the parameters are ρ > 0, 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π.
These are just the usual coordinates we met before. The only difference
is that we can now also think of them as a way to parameterise a surface.
12
2.0
1.5
0.0
0.5
1.0
1.0
0.5
1.0
v
0.0
0.5
0.0
1.0
2
t
0.5
0
0.0
0.0
0.5
-2
u
1.0
Figure 7: A parametric line and a parametric square in three dimensions.
Also note that in polar coordinates, a double integral is
#
ZZ
Z β "Z ρ2 (θ)
f (x, y) dA =
f (ρ, θ) ρ dρ dθ ,
α
R
(21)
ρ1 (θ)
where an extra ρ appears, which will be explained later when we discuss
Jacobians, and the limits are determined by the allowed values of ρ and θ.
Let’s look at some examples.
Example: Change z = x2 + y 2 − 9 to rectangular and polar coordinates.
Solution: In rectangular coordinates, it is simply x = u2 + v 2 − 9. In polar
coordinates, it is
z = ρ2 cos2 θ + ρ2 sin2 θ − 9 = ρ2 − 9 .
(22)
Example: Change x2 + y 2 + z 2 = 9 to spherical polar coordinates.
Solution: The coordinates for this surface are for ρ = 3 and so
x = 3 sin φ cos θ ,
y = 3 sin φ sin θ ,
where θ and φ are the parameters.
13
z = 3 cos φ ,
(23)
5.1.5.1
Surfaces of revolution
If we take a curve y = f (x), we can rotate it about the x-axis to trace out
a surface in three dimensions. The parameter of revolution will be v and we
set
x = u , y = f (u) cos v , z = f (u) sin v .
(24)
The concept is shown in Figure 9 and examples are
(a) :
f (u) = u ,
(25)
which gives a double-napped cone, and
(b) :
f (u) = u ,
v = u,
(26)
which is an interesting shape, both of which are shown in Figure 8.
y
2
1
0
HaL
-1
-2
2
HbL
1
z
x
0
0
-1
-2
5
-2
-4
0
-5
-2
z
-1
0
0
y
x
-5
1
2
5
Figure 8: Surfaces of revolution.
5.1.5.2
Vector-valued functions of two variables
Since we have extended curves of one parameter to surfaces of two parameters, we can likewise extend a vector-valued function of one variable to a
vector-valued function of two variables:
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k .
14
(27)
This is the vector form of a parametric surface of two variables. It has partial
derivatives
∂r
∂x
∂y
∂z
=
i+
j+
k,
∂u
∂u
∂u
∂u
5.1.6
∂r
∂x
∂y
∂z
=
i+
j+
k.
∂v
∂v
∂v
∂v
(28)
Tangent planes to parametric surfaces
Let σ be a parametric surface in three dimensions. A plane is tangent to σ
at P0 if a line through P0 lies in the plane if and only if it is a tangent line
at P0 of a curve on σ.
drdu‰drdv
drdu
15
10
drdv
5
2
0
0
-2
0
-2
2
Figure 9: Revolution around the x-axis producing a surface of revolution.
Let r(u, v) be a curve on the parametric surface σ and P0 = (a, b, c) a point
on σ with a = x(u0 , v0 ), b = y(u0 , v0 ) and c = z(u0 , v0 ). Then:
If
If
∂r
∂u
∂r
∂v
6= 0, it is tangent to the constant v-curve.
6= 0, it is tangent to the constant u-curve.
∂r
∂r
From this, we see that if ∂u
× ∂v
6= 0, at P0 then it is orthogonal to both
tangent vectors and it therefore normal to the tangent plane at P0 . We define
the principal unit normal vector to the surface r(u, v) at (u0 , v0 ) to be
∂r
×
∂u
n = ∂r
×
∂u
∂r
∂v ∂r ∂v
,
(29)
∂r
∂r
× ∂v
6= 0. Figure 10 shows the tangent vectors, tangent plane,
provided ∂u
and normal vector.
15
drdu‰drdv
drdu
15
10
drdv
5
2
0
0
-2
0
-2
2
Figure 10: Tangent plane and normal vector.
Example: Find the equation of the tangent plane for the surface
r(u, v) = ui + 2v 2 j + (u2 + v)k ,
(30)
at the point (2, 2, 3).
Solution: First we find the normal vector.
∂r
∂r
= i + 2uk ,
= 4vj + k ,
∂u
∂v
∂r ∂r
⇒
×
= −8uvi − j + 4vk .
∂u ∂v
At the point (2, 2, 3), we see that we should have u = 2, 2v 2 = 2 and
u2 + v = 3. Solving these gives u = 2 and v = −1. Therefore, we have the
normal vector
n = 16i − j − 4k ,
(31)
from which we can write the tangent plane using equation (??) as
16(x − 2) − (y − 2) − 4(z − 3) = 0
⇒ 16x − y − 4z = 18 .
(32)
First, note that we can use any normal vector to get the tangent plane.
If we used the unit normal for example, we would just divide the entire
equation by a constant, which leaves it unchanged. Also, it is now more
useful to think of the equation for the tangent plane given by equation (??)
16
in a slightly different way. Denoting the normal vector at P0 = (a, b, c) by
n = (nx , ny , nz ), the equation of the tangent plane at P0 becomes
nx (x − a) + ny (y − b) + nz (z − c) = 0 .
5.1.7
(33)
Surface area
In addition to the volume of a solid, we also might want to know its surface
area. Let r(u, v) be a smooth parametric surface on a region of the plane
∂r
∂r
× ∂v
6= 0 on R. Due to this condition, a tangent plane exist for
R, with ∂u
every point (u, v) on R. Consider the surface in Figure 11. If we consider
4
2
0
-2
-4
30
20
10
0
-4
-2
0
2
4
Figure 11: The projection of a parametric surface.
the area of the surface defined on a sub-region contained in one of the boxes
defined by the gridlines in the region R, we find the area of a box of size
∆uk × ∆vk at a corner point (uk , vk ) is ∆Ak = ∆uk ∆vk . Then, considering
the (approximately flat) parallelogram on the surface for this region, it will
have sides
∂r
∆uk ,
(34)
r(uk + ∆uk , vk ) − r(uk , vk ) ≈
∂u
and
∂r
r(uk , vk + ∆vk ) − r(uk , vk ) ≈
∆vk ,
(35)
∂v
17
for ∆uk and ∆vk small. Then, the area of this parallelogram is
∂r
∂r
∆vk ∆Sk = ∆uk ×
∂u
∂v
∂r ∂r ∆uk ∆vk
= ×
∂u ∂v ∂r ∂r ∆Ak .
×
= ∂u ∂v (36)
This is the surface area of the parametric surface defined on the the box at
the point (uk , vk ). The total surface area for the region is the sum of all these
boxes, i.e
n X
∂r ∂r (37)
S=
∂u × ∂v ∆Ak ,
k=1
where n is the number of boxes in the region. If we take the limit, this gives
us the surface area of the parametric surface r(u, v) over the region
R
ZZ
Z Z ∂r ∂r S=
dS =
(38)
∂u × ∂v dA .
R
R
Example: Find the surface area of the sphere of radius 4 that lies in the
cylinder above the xy-plane with base x2 + y 2 = 12.
Solution: The diagram is in Figure 12. At first glance, this doesn’t look
much like what we need for equation (38) to work. We need to parameterise
the surface first. To do that we use spherical polar coordinates to give us the
parametric surface
r(θ, φ) = 4 sin φ cos θi + 4 sin φ sin θj + 4 cos φk .
We now need to work out the ranges of θ and φ. For θ, we see that we have
a cylinder that goes through an angle of 2π around the z-axis and therefore
we take 0 ≤ θ ≤ 2π. Finding the limits for φ requires more work. φ is the
rotation away from the z-axis, so the limits of φ will come from the limits
of z on the cylinder. We therefore substitute the values of x and y when the
18
-4
-2
0
2
4
4
2
0
-2
-4
-2
0
2
4
Figure 12: The intersection of the sphere x2 + y 2 + z 2 = 16 and the cylinder
with base x2 + y 2 = 12.
surface intersects the cylinder, i.e. when x2 + y 2 = 12 to find
x2 + y 2 + z 2 = 16
⇒12 + z 2 = 16
⇒z 2 = 4
⇒z = ±2 .
Then, because we are in spherical polar coordinates, we get
4 cos φ = ±2
1
⇒ cos φ = ±
2
π
⇒φ =
3
where we only use z = 2 because we are only interested in the area above
the xy-plane, and we therefore can’t have φ = 2π/3. The limits are then
0 ≤ φ ≤ π/3 since the angle runs from the x-axis (φ = 0) to the intersection
of the surfaces (φ = π/3). The derivatives of r are
rθ = −4 sin φ sin θi + 4 sin φ cos θj ,
rφ = 4 cos φ cos θi + 4 cos φ sin θj − 4 sin φk .
19
The cross product is
i
j
k
0
rθ × rφ = −4 sin φ sin θ 4 sin φ cos θ
4 cos φ cos θ 4 cos φ sin θ −4 sin φ
= −16 sin2 φ cos θi − 16 sin2 φ sin θj + (−16 sin φ cos φ sin2 θ
− 16 sin φ cos φ cos2 θ)k
= −16 sin2 φ cos θi − 16 sin2 φ sin θj − 16 sin φ cos φk .
The magnitude of this is
q
||rθ × rφ || = (−16 sin2 φ cos θ)2 + (−16 sin2 φ sin θ)2 + (−16 sin φ cos φ)2
q
= 16 sin4 φ cos2 θ + sin4 φ sin2 θ + sin2 φ cos2 φ
q
= 16 sin2 φ(sin2 φ cos2 θ + sin2 φ sin2 θ + cos2 φ)
q
= 16 sin2 φ(sin2 φ + cos2 φ)
q
= 16 sin2 φ
= 16| sin φ| .
20
For the region we are interested in, sin φ is always positive, so we can drop
the absolute value. The surface area is then given by
ZZ
16 sin φ dA
S=
R
Z
2π
"Z
#
π/3
=
16 sin φ dφ dθ
0
Z
0
2π
"Z
#
π/3
16 sin φ dφ dθ
=
0
Z
=
0
2π
[−16 cos φ]π/3
dθ
0
Z0 2π h
i
π
−16 cos − cos 0 dθ
3
Z0 2π 1
=
−16( − 1) dθ
2
0
Z 2π
8 dθ
=
=
0
= 8 θ|2π
0 = 16π .
5.1.7.1
Surface areas of surfaces of the form z = f (x, y)
If our surface is written in the form
r(x, y) = x(u, v)i + y(u, v)j + f (x(u, v), y(u, v))k ,
(39)
where z = f (x, y), then we can choose to parameterise using x = u and y = u
instead to give
r(u, v) = ui + vj + f (u, v)k .
(40)
The derivatives are then
∂z
k,
∂u
∂z
rv = j +
k,
∂v
ru = i +
21
(41)
from which we find
ru × rv = k −
∂z
∂z
i−
i.
∂u
∂v
(42)
We therefore have
s
||ru × rv || =
1+
∂z
∂u
2
+
∂z
∂v
2
.
(43)
Since x = u and y = v, this means we can rewrite the formula for surface
area as
s
2 2
ZZ
∂z
∂z
S=
1+
+
dA
(44)
∂x
∂y
R
Example: Find the surface area of the paraboloid z = x2 + y 2 below the
plane z = 1.
Solution: First, we note that z ≥ 0 because it is the sum of squares. Therefore we want the surface area between the xy-plane and the circle of radius
1 at z = 1. The surface area using equation (44) is
ZZ p
1 + 4x2 + 4y 2 dA .
S=
R
This is easier to calculate if we use polar coordinates:
x = ρ cos θ ,
y = ρ sin θ ,
which gives z = ρ2 and
p
p
1 + 4x2 + 4y 2 = 1 + 4ρ2 .
We can now easily calculate the area
Z 2π Z 1 p
2
S=
1 + 4ρ ρdρ dθ
0
0
Z 1
√
dt
1 + 4t
t = ρ2 ⇒ dt = 2ρdρ
= 2π
2
0
1
121
3/2 = 2π
(1 + 4t) 234
0
1 3/2
= 2π (5 − 1)
12
π √
= (5 5 − 1) = 5.33041 .
6
22
5.1.7.2
Lamina
A lamina is a region of space that has mass M and a variable density, given
by the density function δ(x, y). The mass is given by
ZZ
δ(x, y) dA .
(45)
M=
R
The centre of mass or centre of gravity of the lamina region is (x̄, ȳ)
where
ZZ
1
x̄ =
x δ(x, y) dA
M
R
ZZ
(46)
1
ȳ =
y δ(x, y) dA .
M
R
Similarly, we can define the centroid of the region R, which is the geometric
“centre” of R. It is given by (x̃, ỹ) where
RR
x dA
ZZ
1
R
x dA
x̃ = RR
=
A
dA
R
RRR
(47)
y dA
ZZ
1
=
ỹ = RRR
y dA .
A
dA
R
R
Note that for a homogeneous lamina, i.e. one with constant density, the
centre of gravity coincides with the centroid. Also note that in the textbook
the confusingly use (x̄, ȳ) for both the centre of gravity and centroid. The
notation (x̃, ỹ) for the centroid is used to avoid this confusion.
Example: Compute the mass and centre of gravity of the lamina inside the
unit circle, with density δ(x, y) = x2 + y 2 .
23
Solution: The mass is given by
ZZ
(x2 + y 2 ) dA
M=
R
Z
2π
1
Z
ρ2 · ρ dρ dθ
=
0
Z
= 2π
0
1
ρ3 dρ
0
ρ4 1
= 2π 0
4
π
= .
2
The x-component of the centre of gravity is given by
ZZ
1
x̄ =
x(x2 + y 2 ) dA
M
R
Z Z
2 2π 1
ρ cos θ · ρ2 · ρ dρ dθ
=
π 0
Z 2π Z0 1
2
=
ρ4 cos θ dρ dθ
π 0
Z 2π 0
2
1
=
cos θ dθ
π 0 5
2π
2
=
(− sin θ)0 = 0 .
5π
Similarly, ȳ = 0, and therefore the centre of gravity is at (0, 0). Note that
we could swap the order of integration because the limits were independent
of the other coordinates.
24