1 Triple Integrals

Mass problem. Find the mass

M of a solid G whose density

(the mass per unit volume)

is a continuous nonnegative

function δ(x, y, z).

1. Divide the box enclosing G into

subboxes, and exclude all those

subboxes that contain points

outside of G. Let n be the number

of all the subboxes inside G, and

let ∆Vk = ∆xk∆yk∆zk be the volume of the k-th subbox.

2. Choose any point (x∗k, y∗k, z∗k) in the k-th subbox. The mass of the

k-th subbox is ∆Mk ≈ δ(x∗k, y∗k, z∗k)∆Vk. Thus,

M ≈n∑k=1

∆Mk =

n∑k=1

δ(x∗k, y∗k, z∗k)∆Vk =

n∑k=1

δ(x∗k, y∗k, z∗k)∆xk∆yk∆zk

This sum is called the Riemann sum.

3. Take the sides of all the subboxes to 0, and therefore the number

of them to infinity, and get

M = limn→∞

n∑k=1

δ(x∗k, y∗k, z∗k)∆Vk =

˚G

δ(x, y, z) dV

The last term is the notation for the limit of the Riemann sum,

and it is called the triple integral of δ(x, y, z) over G.

1

Properties of triple integrals

1. If f, g are continuous on G, and c, d are constants, then

˚G

(c f (x, y, z) + d g(x, y, z)

)dV = c

˚G

f (x, y, z) dV

+ d

˚G

g(x, y, z) dV

2. If G is divided into two solids G1 and G2, then

˚G

f (x, y, z) dV =

˚G1

f (x, y, z) dV +

˚G2

f (x, y, z) dV

The mass of the entire solid

is the sum of the masses of

the solids G1 and G2.

2

Evaluating triple integrals over rectangular boxes

a ≤ x ≤ b , c ≤ y ≤ d , k ≤ z ≤ l

˚G

f (x, y, z) dV =

ˆ b

a

ˆ d

c

ˆ l

k

f (x, y, z) dz dy dx

or any permutation, e.g.

˚G

f (x, y, z) dV =

ˆ d

c

ˆ l

k

ˆ b

a

f (x, y, z) dx dz dy

Example. Find the mass of the box

1

3≤ x ≤ 1

2, 0 ≤ y ≤ π , 0 ≤ z ≤ 1

if its density is

δ(x, y, z) = xz sin(xy)

Answer : M = 112 +

√3

4π −12π ≈ 0.0620106

3

Evaluating triple integrals over simple xy-, xz-, yz-solids

A solid G is called a simple xy-solid

if it is bounded above by a surface

z = g2(x, y), below by a surface

z = g1(x, y), and its projection

on the xy plane is a region R.

˚G

f (x, y, z) dV =

¨R

[ˆ g2(x,y)

g1(x,y)

f (x, y, z) dz

]dA

Example. Find the mass of the solid G defined by the inequalities

π

6≤ y ≤ π

2, y ≤ x ≤ π

2, 0 ≤ z ≤ xy

if its density is

δ(x, y, z) = cos(z/y)

Answer : M = 5π12 −

√32 ≈ 0.442972

4

Volume of G =

˚G

dV

Example. Find V of G bounded by the surfaces

y = x2 , y + z = 4 , z = 0

Answer : V = 25615

Integration in other orders

A simple xz-solid˚G

f (x, y, z) dV =

¨R

[ˆ g2(x,z)

g1(x,z)

f (x, y, z) dy

]dA

A simple yz-solid˚G

f (x, y, z) dV =

¨R

[ˆ g2(y,z)

g1(y,z)

f (x, y, z) dx

]dA

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2 Centre of gravity and centroid

Mass, centre of gravity and centroid of a lamina

Recall that a lamina is an idealised flat object that is thin enough to

be viewed as a 2-d plane region R.

The mass M of a lamina with density δ(x, y) is

M =

¨R

δ(x, y) dA

The centre of gravity of

a lamina is a unique point (x, y)

such that the effect of gravity

on the lamina is equivalent to

that of a single force acting

at the point (x, y)

x =1

M

¨R

x δ(x, y) dA , y =1

M

¨R

y δ(x, y) dA

Example. Find the centre of gravity of a lamina with density

δ(x, y) = x + 1 bounded by x2 + (y + 1)2 = 1

Answer : M = π , x = 1/4 , y = −1

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For a homogeneous lamina with δ(x, y) = const,

the centre of gravity is called the centroid of the lamina

or the centroid of the region R

because it does not depend on δ(x, y) = const.

x =1

A

¨R

x dA , y =1

A

¨R

y dA , A =

¨R

dA

Example. Find the centroid of a region bounded by

(x− 1)2 + y2 = 1 , and (x− 2)2 + y2 = 4

Answer : A = 3π , x = 7/3 , y = 0

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Mass, centre of gravity and centroid of a solid

The centre of gravity of a solid G with density δ(x, y, z) is a unique

point (x, y, z) such that the effect of gravity on the solid is equivalent

to that of a single force acting at the point (x, y, z)

x =1

M

˚G

x δ(x, y, z) dV , y =1

M

˚G

y δ(x, y, z) dV

z =1

M

˚G

z δ(x, y, z) dV , M =

˚G

δ(x, y, z) dV

Example. Find the centre of gravity of a solid G bounded by the

surfaces x2 + y2 = 1 and x2 + y2 = 4, above by the surface z =

5− x2 − y2, and below by the surface z = 0 with the density

δ(x, y, z) = e5−x2−y2−z

Answer : M = π(e4 − e− 3) ≈ 153.561 , z = 2e4−2e−212(e4−e−3)

≈ 0.85

For a homogeneous solid with δ(x, y) = const, the centre of gravity

is called the centroid of the solid.

x =1

V

˚G

x dV , y =1

V

˚G

y dV

z =1

V

˚G

z dV , V =

˚G

dV

Example. Find the centroid of the solid below z = 10 − x2 − y2,

inside of x2 + y2 = 1, and above z = 0

Answer : V = 19π/2 , x = 0 , y = 0 , z = 271/57 ≈ 4.75439

8

3 Triple integrals in cylindrical and spherical coordi-

nates

Cylindrical coordinates

Cylindrical wedge or

cylindrical element of volume

is interior of intersection of

two cylinders: r = r1 , r = r2

two half-planes: θ = θ1 , θ = θ2

two planes: z = z1 , z = z2

The dimensions: θ2−θ1 , r2−r1 , z2−z1 are called the central angle,

thickness and height of the wedge.

Divide G by cylindrical wedges

˚G

f (r, θ, z) dV = limn→∞

∞∑n=1

f (r∗k, θ∗k, z∗k)∆Vk

∆Vk = [ area of base ] · [ height ]

= r∗k∆rk∆θk∆zk

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Theorem.

Let G be a solid whose upper suface is

z = g2(r, θ) and whose lower suface is

z = g1(r, θ) in cylindrical coordinates.

If projection of G on the xy-plane is

a simple polar region R, and

if f (r, θ, z) is continuous on G, then

˚G

f (r, θ, z) dV =

¨R

[ˆ g2(r,θ)

g1(r,θ)

f (r, θ, z) dz

]dA

=

ˆ θ2

θ1

ˆ r2(θ)

r1(θ)

ˆ g2(r,θ)

g1(r,θ)

f (r, θ, z) r dz dr dθ

Example. V and centroid of G bounded above by

z =√

25− x2 − y2, below by z = 0, and laterally by x2 + y2 = 9.

Answer : V = 122π/3 , z = 1107/488

Converting triple integrals from rectangular to cylindrical

coordinates

˚G

f (x, y, z) dV =

˚

appropriatelimits

f (r cos θ, r sin θ, z) r dz dr dθ

Example.

ˆ 3

−3

ˆ √9−x2−√9−x2

ˆ 9−x2−y2

0

x2 dz dy dx =243

4π

10

Spherical coordinates

Spherical wedge or

spherical element of volume

is interior of intersection of

two spheres: ρ = ρ1 , ρ = ρ2

two half-planes: θ = θ1 , θ = θ2

nappes of two right circular

cones: φ = φ1 , φ = φ2

The numbers:

θ2 − θ1 , ρ2 − ρ1 , φ2 − φ1are the dimensions of the wedge.

Divide G by spherical wedges

˚G

f (r, θ, φ) dV = limn→∞

∞∑n=1

f (r∗k, θ∗k, φ

∗k)∆Vk

∆Vk = (ρ∗k)2 sinφ∗k∆ρk∆φk∆θk

˚G

f (r, θ, φ) dV =

˚

appropriatelimits

f (r, θ, φ) ρ2 sinφ dρ dφ dθ

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Example. V and centroid of G bounded above by

x2 + y2 + z2 = 16 and below by z =√x2 + y2.

Answer : V = 64(2−√

2)π/3 > 0 , z = 3/2(2−√

2) ≈ 2.56

Converting triple integrals from rectangular to spherical

coordinates

˚G

f (x, y, z) dV =

˚

appropriatelimits

f (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) ρ2 sinφ dρ dφ dθ

Example.

ˆ 2

−2

ˆ √4−x2−√4−x2

ˆ √4−x2−y2

0

z2√x2 + y2 + z2 dz dy dx =

64

9π

12