ENGR 2422 Engineering Mathematics 2
Brief Notes on Chapter 6

Note that this section provides only a very brief introduction to the topic of multiple integration.

6.1 Double Integrals (Cartesian Coordinates)
6.2 Double Integrals (Plane Polar and Other Coordinates)
6.3 Triple Integrals

6.1 Double Integrals (Cartesian Coordinates)

In Cartesian coordinates on the xy-plane, the rectangular element of area is
ΔA = Δx Δy .

Summing all such elements of area along a vertical strip,
the area of the elementary strip is
(Sum[y=g(x) to h(x)] Dy) Dx
Summing all the strips across the region R,
the total area of the region is:
(Sum[x=a to b] (Sum[y=g(x) to h(x)] Dy) Dx)
In the limit as the elements Δx and Δy shrink to zero, this sum becomes
Integ[x=a to b] (Integ[y=g(x) to h(x)] 1 dy) dx
region bounded by y=h(x) above,
y=g(x) below,
x=a to the left and
x=b to the right

If the surface density σ within the region is a function of location, σ = f (x, y), then the mass of the region is
m = Integ[x=a to b] (Integ[y=g(x) to h(x)] f dy) dx

The inner integral must be evaluated first.

Re-iteration

One may reverse the order of integration by summing the elements of area ΔA horizontally first, then adding the strips across the region from bottom to top. This generates the double integral for the total area of the region
Integ[y=c to d] (Integ[x=p(y) to q(y)] 1 dx) dy
The mass becomes

region bounded by y=d above,
y=c below,
x=p(y) to the left and
x=q(y) to the right

Choose the orientation of elementary strips that generates the simpler double integration.
An example is Problem Set 10, Question 1:
vertical strips is preferable to horizontal strips .

Where a boundary can be represented only as the union of different functions, the double integral has to be split accordingly. An example is Problem Set 10, Question 7:

vertical strips; region must be
divided into three double integrals. Integral f dA =
Integral(sqrt(1/3) to 1)(1/x to 3x) f(x,y) dy dx
+ Integral(1 to sqrt(3))(1/x to 3/x) f(x,y) dy dx
+ Integral(sqrt(3) to 3)(x/3 to 3/x) f(x,y) dy dx

where no less than three double integrals are needed in order to integrate across the full region, regardless of the orientation of the strips (vertical, as illustrated, or horizontal). A change of variables can transform a region into an easier shape for integration (as is the case in Problem Set 10, Question 7).

A double integral may be factored completely into the product of two single integrals only if

all boundaries are parallel to coordinate axes; and
the integrand can be separated into factors, each of only one variable.

Example 1:

Find the mass of a rectangle, two of whose opposite corners are at (1, 2) and (5, 7), whose sides are parallel to the coordinate axes and whose surface density is σ = x²y .

mass = Integral(1 to 5)(2 to 7) x^2 y dy dx
$= Integral(1 to 5) x^2 {Integral(2 to 7) y dy} dx$
The inner integral has no dependency at all on x, in its limits or in its integrand. It can therefore be extracted as a “constant” factor from inside the outer integral.
$= {Integral(1 to 5) x^2 dx} {Integral(2 to 7) y dy}$
= [x^3 / 3](1 to 5) [y^2 / 2](2 to 7) = 62*15
Therefore mass = 930 kg.
rectangle (1,2) to (5,7),
vertical strips

rectangle (1,2) to (5,7),
horizontal strips
OR
We can choose to sum horizontally first:
mass = Integral(2 to 7)(1 to 5) x^2 y dx dy
$= Integral(2 to 7) y {Integral(1 to 5) x^2 dx} dy$
The inner integral has no dependency at all on y, in its limits or in its integrand. It can therefore be extracted as a “constant” factor from inside the outer integral.
$= {Integral(1 to 5) x^2 dx} {Integral(2 to 7) y dy}$
which is exactly the same form as before, leading to the same value of 930 kg.

Example 2:

Evaluate Integral(over R) (6x + 2y^2) dA
where R is the region enclosed by the parabola
x = y² and the line x + y = 2.

The upper boundary changes form at x = 1.
The left boundary is the same throughout R.
The right boundary is the same throughout R.
Therefore choose horizontal strips.
I = Integral(-2 to 1)(y^2 to 2-y) (6x + 2y^2) dx dy
= Integral(-2 to 1) [3x^2 + 2xy^2](x = y^2 to 2-y) dx dy
= Integral(-2 to 1) (3(2-y)^2 + 2(2-y)y^2 - (3y^4 + 2y^4)) dy
= Integral(-2 to 1)
((12 - 12y + 3y^2) + (4y^2 - 2y^3) - 5y^4) dy
= Integral(-2 to 1) (12 - 12y + 7y^2 - 2y^3 - 5y^4) dy
= [12y - 6y^2 + (7/3)y^3 - (1/2)y^4 - y^5](-2 to 1)
region between parabola and line,
horizontal strips

= (12 - 6 + (7/3) - (1/2) - 1)
- (-24 - 24 - (56/3) - 8 + 32)
Therefore I = 99/2

6.2 Double Integrals (Plane Polar and Other Coordinates)

Plane polar coordinates (r, θ) are related to Cartesian coordinates (x, y) by the equations:
x = r cos θ , y = r sin θ .
The differential for area is
dA = dx dy = Jacobian dr dt
= r dr dt
If the boundary of the region over which one wishes to integrate consists of circular arcs and/or radial lines, then a transformation into plane polar coordinates may simplify the integration.

Example 3:

Find the centre of mass for a plate of uniform density, whose boundary is the portion of the circle x² + y² = a² that is inside the first quadrant.

Use plane polar coordinates.

Boundaries:

The positive x-axis is the line θ = 0.
The positive y-axis is the line θ = π /2 .
The circle is r² = a² which is r = a.
region is quarter circle,
in the first quadrant

Mass:
Surface density σ is a constant.
Dm = s dA = s dx dy = s r dr dt
m = Integral(0 to pi/2)(0 to a) s r dr dt
m = s Integral(0 to pi/2) 1 dt * Integral(0 to a) r dr
m = s [t](0 to pi/2) * [r^2 / 2](0 to a)
= s pi a^2 / 4
[or, directly, mass = density × (area of quarter-circle).]

Moments about the x-axis:
DMx = y Dm = (r sin t) s r dr dt
Mx = Integral(0 to pi/2)(0 to a) (r sin t) s r dr dt
Mx = s Integral(0 to pi/2) sin t dt * Integral(0 to a) r^2 dr
Mx = s [-cos t](0 to pi/2) * [r^3 / 3](0 to a)
= s 1 a^3 / 3
Mx = m yBar implies yBar = Mx / m = (s a^3 / 3) / (s pi a^2 / 4)
= 4a / (3pi)
By symmetry, xBar = yBar
Therefore the centre of mass is located at

(4a/(3pi), 4a/(3pi))

Example 4:

Find the area enclosed by one loop of the polar curve r = cos 2θ.

Boundaries:

0 < r < cos(2 theta);
-pi/4 < theta < +pi/4

Area = Integral(-pi/4 to +pi/4)(0 to cos(2t)) 1 r dr dt

loop, -pi/4 < theta < +pi/4,
0 < r < cos(2 theta)

Area = [sin(4t)/16 + t/4](-pi/4 to +pi/4)
= (0 + pi/16) - (0 - pi/16)
Therefore

Area = pi/8

In general, in plane polar coordinates,

general polar region general polar integral

In another coordinate system (u, v), use the Jacobian to transform the double integral:

Integral(D) f dA = Integral(over u & v) f J du dv

Some examples can be found in Problem Set 10.

6.3 Triple Integrals

The concepts for double integrals (surfaces) extend naturally to triple integrals (volumes).
The element of volume, in terms of the Cartesian coordinate system (x, y, z) and another coordinate system (u, v, w), is

dV = Jacobian(x, y, z; u, v, w) du dv dw
and
Integral(D) f dV = Integral(over u, v & w) f J du dv dw

The most common choices for non-Cartesian coordinate systems in real ³ are:

Cylindrical Polar Coordinates:

x = r cos phi ,
y = r sin phi,
z = z
for which the differential volume is
dV = r dr d_phi dz

Spherical Polar Coordinates:

x = r sin theta cos phi ,
y = r sin theta sin phi,
z = r cos theta
for which the differential volume is
dV = r^2 sin theta dr d_theta d_phi

Example 5:

The density of an object is equal to the reciprocal of the distance from the origin.
Find the mass and the average density inside the sphere r = a .

Use spherical polar coordinates.
density = 1/r
mass = Integral(0 to 2pi)(0 to pi)(0 to a) (1/r) r^2 sin t dr dt df
mass = Integral(0 to a) r dr * Integral(0 to pi) sin t dt
* Integral(0 to 2pi) 1 df
mass = [r^2 / 2](0 to a) * [-cos t](0 to pi) * [f](0 to 2pi)
Therefore

mass = 2 pi a^2

average density = mass / volume = 2 pi a^2 / (4/3 pi a^3)
Therefore

average density = 3/(2a)

[Note that the mass is finite even though the density is infinite at the origin!]

Example 6:

Find the proportion of the mass removed, when a hole of radius 1, tangent to a diameter, is bored through a uniform sphere of radius 2.

Cross-section at right angles to the axis of the hole:

Use cylindrical polar coordinates, with the z-axis aligned parallel to the axis of the cylindrical hole.

The plane polar equation of the boundary of the hole is then
r = 2 cos θ

The entire circular boundary is traversed once for
-pi/2 < theta < +pi/2
hole, radius 1,
bored through sphere, radius 2;
view from above

Cross-section parallel to the axis of the hole:

At each value of r , the distance from the equatorial plane to the point where the hole emerges from the sphere is
z = sqrt(2^2 - r^2)
The element of volume for the hole is therefore
dV = 2z dA
= 2 sqrt(2^2 - r^2) r dr dt
V =
Integral(-pi/2 to pi/2)(0 to 2 cos t) 2 sqrt(4 - r^2) r dr dt
We cannot separate the two integrals,
because the upper limit of the inner integral,
(r = 2 cos θ), is a function of the
variable of integration in the outer integral.

The geometry is entirely symmetric about θ = 0
V = 4 *
Integral(0 to pi/2)(0 to 2 cos t) sqrt(4 - r^2) r dr dt
hole, radius 1,
bored through sphere, radius 2;
view from side

V =
4 Integral(0 to pi/2) [(4 - r^2)^(3/2) / (-3)](0 to 2 cos t) dt
= (4/3) Integral(0 to pi/2) [((4 - 4 cos^2 t)^(3/2)
- (4 - 0)^(3/2)) / (-3)] dt = (4/3) Integral(0 to pi/2) [4^(3/2) - (4 sin^2 t)^(3/2)] dt
= (32/3) Integral(0 to pi/2) (1 - sin^3 t) dt

$= (32/3) { Integral(0 to pi/2) (1) dt - Integral(0 to pi/2) (1 - cos^2 t) sin t dt$
Let u = cos θ , then du = – sin θ dθ .
θ = 0 implies u = 1 and θ = π/2 implies u = 0
V = (32/3) (pi/2 - Integral(1 to 0) (1 - u^2) (- du))
= (32/3) (pi/2 - [u - (u^3 /3)](0 to 1))
V = 16pi/3 - 64/9
The density is constant throughout the sphere. Therefore
ratio of masses = ratio of volumes = 1/2 - 2/(3pi)
Therefore the proportion of the sphere that is removed is