Multiple Integrals

1 Double Integrals

Definite integrals appear when one solves

Area problem. Find the area A of the region R bounded above by

the curve y = f (x), below by the x-axis, and on the sides by x = a

and x = b.

A =

ˆ b

a

f (x) dx = limmax ∆xi→0

n∑k=1

f (x∗k)∆xk

Mass problem. Find the mass M of a rod of length L whose linear

density (the mass per unit length) is a function δ(x), where x is the

distance from a point of the rod to one of the rod’s ends.

M =

ˆ L

0

δ(x) dx

1

Double integrals appear when one solves

Volume problem. Find the volume

V of the solid G enclosed between

the surface z = f (x, y) and a region R

in the xy-plane where f (x, y) is

continuous and nonnegative on R.

Mass problem. Find the mass M of

a lamina (a region R in the xy-plane)

whose density (the mass per unit area)

is a continuous nonnegative function

δ(x, y) defined as

δ(x, y) = lim∆A→0

∆M

∆A

where ∆M is the mass of the small

rectangle of area ∆A

which contains (x, y).

2

Let us consider the volume problem.

1. Divide the rectangle enclosing R into subrectangles, and exclude

all those rectangles that contain points outside of R. Let n be the

number of all the rectangles inside R, and let ∆Ak = ∆xk∆yk be

the area of the k-th subrectangle.

2. Choose any point (x∗k, y∗k) in the k-th subrectangle. The volume

of a rectangular parallelepiped with base area ∆Ak and height

f (x∗k, y∗k) is ∆Vk = f (x∗k, y

∗k)∆Ak. Thus,

V ≈n∑k=1

∆Vk =

n∑k=1

f (x∗k, y∗k)∆Ak =

n∑k=1

f (x∗k, y∗k)∆xk∆yk

This sum is called the Riemann sum.

3. Take the sides of all the subrectangles to 0, and therefore the num-

ber of them to infinity, and get

V = limmax ∆xi,∆yi→0

n∑k=1

f (x∗k, y∗k)∆Ak =

¨R

f (x, y) dA

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

and it is called the double integral of f (x, y) over R.

3

In what follows we identify

limmax ∆xi,∆yi→0

n∑k=1

· · · ≡ limn→∞

n∑k=1

· · ·

If f is continuous but not nonnegative on R then the limit represents

a difference of volumes – above and below the xy-plane. It is called

the net signed volume between R and the surface z = f (x, y), and

it is given by the limit of the corresponding Riemann sum that is the

double integral of f (x, y) over R

¨R

f (x, y) dA = limn→∞

n∑k=1

f (x∗k, y∗k)∆Ak = lim

n→∞

n∑k=1

f (x∗k, y∗k)∆xk∆yk

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

M =

¨R

δ(x, y) dA

4

Properties of double integrals

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

¨R

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

)dA = c

¨R

f (x, y) dA + d

¨R

g(x, y) dA

2. If R is divided into two regions R1 and R2, then

¨R

f (x, y) dA =

¨R1

f (x, y) dA +

¨R2

f (x, y) dA

The volume of the entire solid

is the sum of the volumes of

the solids above R1 and R2.

5

Double integrals over rectangular regions

The symbolsˆ b

a

f (x, y) dx and

ˆ d

c

f (x, y) dy

where in the first integral y is fixed while in the second integral x is

fixed, denote partial definite integrals.

Examples.

(i)

ˆ 2

1

sin(2x− 3y) dx , (ii)

ˆ 1

0

sin(2x− 3y) dy .

We can then integrate the resulting functions of y and x with respect

to y and x, respectively. This two-stage integration process is called

iterated or repeated integration.

Notation ˆ d

c

ˆ b

a

f (x, y) dx dy =

ˆ d

c

[ˆ b

a

f (x, y) dx

]dy

ˆ b

a

ˆ d

c

f (x, y) dy dx =

ˆ b

a

[ˆ d

c

f (x, y) dy

]dx

These integrals are called iterated integrals.

Example.

(i)

ˆ 1

0

ˆ 2

1

sin(2x− 3y) dx dy , (ii)

ˆ 2

1

ˆ 1

0

sin(2x− 3y) dy dx

6

Theorem. Let R be the rectangle a ≤ x ≤ b, c ≤ y ≤ d. If f (x, y)

is continuous on R then¨R

f (x, y) dA =

ˆ d

c

ˆ b

a

f (x, y) dx dy =

ˆ b

a

ˆ d

c

f (x, y) dy dx

Example. Use a double integral

to find V under the surface

z = 3πex sin y + e−x

and over the rectangle

R = {(x, y) : 0 ≤ x ≤ ln 3 , 0 ≤ y ≤ π}

V =38

3π ≈ 39.7935 > 0

7

2 Double integrals over nonrectangular regions

ˆ b

a

ˆ d

c

f (x, y) dy dx =

ˆ b

a

[ˆ d

c

f (x, y) dy

]dx

Replace c→ g1(x), d→ g2(x). Then

ˆ b

a

ˆ g2(x)

g1(x)

f (x, y) dy dx =

ˆ b

a

[ˆ g2(x)

g1(x)

f (x, y) dy

]dx

a x b

c

d

x

y

a x b

y = g2(x)

y = g1(x)

x

y

Figure 1: The rectangle becomes a type I region: g2(x) ≥ g1(x).

c→g1, d→g2−−−−−−→

Theorem. If R is a type I region on which f (x, y) is continuous, then

¨R

f (x, y) dA =

ˆ b

a

ˆ g2(x)

g1(x)

f (x, y) dy dx

Example. Find the volume V of the solid enclosed by the surfaces

z = 0, y2 = x, and x + z = 1.

V =8

15

8

Similarly, inˆ d

c

ˆ b

a

f (x, y) dx dy =

ˆ d

c

[ˆ b

a

f (x, y) dx

]dy

we replace a→ h1(y), b→ h2(y). Then

ˆ d

c

ˆ h2(y)

h1(y)

f (x, y) dx dy =

ˆ d

c

[ˆ h2(y)

h1(y)

f (x, y) dx

]dy

a b

y

c

d

x

y

d

y

cx = h2(y)x = h1(y)

x

y

Figure 2: The rectangle becomes a type II region: h2(y) ≥ h1(y).

a→h1, b→h2−−−−−−→

Theorem. IfR is a type II region on which f (x, y) is continuous, then

¨R

f (x, y) dA =

ˆ d

c

ˆ h2(y)

h1(y)

f (x, y) dx dy

Example. Find the volume V of the solid enclosed by the surfaces

z = 0, y2 = x, and x + z = 1.

9

Some regions belong to both type I and II, e.g. the region from the

previous example, a disc and a triangle.

Reversing the order of integration. Example.ˆ 4

0

ˆ 2

√y

ex3dx dy =

1

3

(e8 − 1

)

Area calculated as a double integral

The solid enclosed between

the plane z = 1, and a region R

is a right cylinder with base R,

height h = 1, and with

cross-sectional area A equal

to the area of R.

Its volume is

V = Ah = A =

¨R

1 dA

Thus

Area of R =

¨R

1 dA =

¨R

dA

Example. Find A of R enclosed between

y = xm and y = xn.

y = xm

y = xn

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

10

3 Double integrals in polar coordinates

Polar coordinates

Figure 3: x = r cos θ , y = r sin θ , r2 = x2 + y2 , tan θ = yx .

We identify (r, θ + 2πn) ∼ (r, θ) , n = ±1 , ±2 , . . .

and (−r, θ) ∼ (r, θ + π)

because they give the same x, y coordinates.

The graph of r = a is the circle of radius a centred at O.

The graph of θ = α, r ≥ 0 is the ray making an angle of α with the

polar axis.

θ=α

θ=β

r=a

r=b

0.0 0.5 1.0 1.5 2.0

0.0

0.5

1.0

1.5

2.0

A polar rectangle is a region enclosed

between two rays, θ = α, θ = β, and

two circles r = a, r = b.

Its area is

A = 12b

2(β − α)− 12a

2(β − α)

= 12(b + a)(b− a)(β − α)

= r∆r∆θ, where r = 12(b + a),

∆r = b− a is the radial thickness, ∆θ = β −α is the central angle.

11

A simple polar region

is a region enclosed between

two rays, θ = α, θ = β, and

two continuous polar curves

r = r1(θ), r = r2(θ) which satisfy

(i) α ≤ β,

(ii) β − α ≤ 2π,

(iii) 0 ≤ r1(θ) ≤ r2(θ).

Examples.

θ=αθ=β

r = r2(θ)

-1.0 -0.5 0.0 0.5 1.0 1.5 2.0

0.0

0.5

1.0

1.5

2.0

y

Figure 4: r1(θ) = 0 and β − α < 2π

r1(θ) = 0 and β − α = 2π r1(θ) 6= 0 and β − α = 2π

12

Let us consider the volume problem in polar coordinates.

1. Divide the rectangle enclosing R into polar subrectangles, and

exclude all those rectangles that contain points outside of R. Let

n be the number of all the rectangles inside R, and let

∆Ak = rk ∆rk ∆θk be the area of the k-th polar subrectangle.

2. Choose any point (r∗k, θ∗k) in the k-th subrectangle. The volume

of a right cylinder with base area ∆Ak and height f (r∗k, θ∗k) is

∆Vk = f (r∗k, θ∗k)∆Ak. Thus,

V ≈n∑k=1

∆Vk =

n∑k=1

f (r∗k, θ∗k)∆Ak =

n∑k=1

f (r∗k, θ∗k) rk ∆rk ∆θk

This sum is called the polar Riemann sum.

3. Take the sides of all the subrectangles to 0, and therefore the num-

ber of them to infinity, and get

V = limn→∞

n∑k=1

f (r∗k, θ∗k)∆Ak =

¨R

f (r, θ) dA

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

and it is called the polar double integral of f (r, θ) over R.

13

The limit of the polar Riemann sum is the same for any choice of

points (r∗k, θ∗k). Let us choose (r∗k, θ

∗k) to be the centre of the k-th polar

rectangle, that is r∗k = rk, θ∗k = θk = 1

2(θk−1 + θk). Then, the polar

double integral is given by

¨R

f (r, θ) dA = limn→∞

n∑k=1

f (rk, θk) rk ∆rk ∆θk

This formula is similar to the one for the double integral in rectangular

coordinates, and it is valid for any region R.

Theorem. If R is a simple polar region enclosed between two rays,

θ = α, θ = β, and two continuous polar curves r = r1(θ), r = r2(θ),

and if f (r, θ is continuous on R, then

¨R

f (r, θ) dA =

ˆ β

α

ˆ r2(θ)

r1(θ)

f (r, θ) r dr dθ

Example. Find the volume of the solid below z = 1−x2− y2, inside

of x2 + y2 − x = 0, and above z = 0. Answer : V = 5π/32

Area of R =

¨R

dA =

ˆ β

α

ˆ r2(θ)

r1(θ)

r dr dθ =1

2

ˆ β

α

(r2(θ)2−r1(θ)2

)dθ

Example. Find A of R in the first quadrant that is outside r = 2

and inside the cardioid r = 2(1 + cos θ). Answer : A = 4 + π/2

Example. Find A enclosed by the three-petaled rose r = cos 3θ.

Answer : A = π/4

14

Converting double integrals from rectangular to

polar coordinates

¨R

f (x, y) dA =

¨

appropriatelimits

f (r cos θ, r sin θ) r dr dθ

It is especially useful if f (x, y) = g(x2 + y2) = g(r)

or f (x, y) = g(y/x) = g(tan θ)

Example. ˆ 1

−1

ˆ √1−x2

0

(x2 + y2)3/2 dy dx =π

5

Example.

ˆ 1

0

ˆ √4−x2

√1−x2

x√x2 + y2

dy dx +

ˆ 2

1

ˆ √4−x2

0

x√x2 + y2

dy dx =3

2

Example. ˆ ∞−∞

e−x2dx =

√π

15

4 Parametric surfaces

A curve: x = x(t) , y = y(t) , z = z(t) , t0 ≤ t ≤ t1

orientationa

tb x

y

z

Pa

Pb

orientation

A surface: x = x(u, v) , y = y(u, v) , z = z(u, v) , (u, v) ∈ R

16

Example. z = 4− x2 − y2

Rectangular coordinates: x = u, y = v, z = 4− u2 − v2

Polar coordinates: x = r cos θ, y = r sin θ, z = 4− r2

17

1 2 3 4 5 6x

1

2

3

4

5

6

z

Polar coordinates are useful for

surfaces of revolution

generated by revolving a curve

z = f (x), x ≥ 0 in the xz-plane,

or, equivalently, a curve

z = f (y), y ≥ 0 in the yz-plane

about the z-axis.

These surfaces are graphs of

functions z = f (√x2 + y2),

and can be parametrised as

x = r cos θ, y = r sin θ, z = f (r)

Example. A right cone of height h and base radius a oriented along

the z-axis, with vertex pointing up, and with the base located at z = 0.

x = r cos θ , y = r sin θ , z = h(1− r

a) , 0 ≤ r ≤ a , 0 ≤ θ ≤ 2π

18

Cylindrical coordinates

x = r cos θ , y = r sin θ , z

r ≥ 0 , 0 ≤ θ ≤ 2π

Plane: z = const

Circular cylinder: r = const

Half-plane: θ = const

1 2 3 4 5 6x

1

2

3

4

5

6

zCylindrical coordinates are useful for

surfaces of revolution generated

by revolving a curve x = f (z)

in the xz-plane or, equivalently,

a curve y = f (z) in the yz-plane

about the z-axis.

These surfaces are parametrised as

x = f (ζ) cos θ, y = f (ζ) sin θ, z = ζ

Example. A right cone of height h

and base radius a oriented along the z-axis, with vertex pointing up,

and with the base located at z = 0.

x = (1−ζh

)a cos θ , y = (1−ζh

)a sin θ , z = ζ , 0 ≤ ζ ≤ h , 0 ≤ θ ≤ 2π

19

Spherical coordinates

x = ρ sinφ cos θ ,

y = ρ sinφ sin θ ,

z = ρ cosφ ,

ρ ≥ 0 , 0 ≤ θ ≤ 2π ,

0 ≤ φ ≤ π .

Sphere: r = const

Half-plane: θ = const

Cone: φ = const

Example. x2 + y2 + z2 = 9 ⇒ ρ = 3

x = 3 sinφ cos θ , y = 3 sinφ sin θ , z = 3 cosφ ,

0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π .

The constant φ-curves are the lines of latitude.

The constant θ-curves are the lines of longitude.

Example. z =√

x2+y2

3 ⇒ φ = π3

x =√

32 ρ cos θ , y =

√3

2 ρ sin θ , z = 12ρ ,

ρ ≥ 0 , 0 ≤ θ ≤ 2π .

The constant ρ-curves are circles.

The constant θ-curves are half-lines.

20

1 2 3 4x

0.5

1.0

1.5

2.0

2.5

3.0

zSphere is an example of a

surface of revolution generated by

revolving a parametric curve

x = f (t), z = g(t) or, equivalently,

a parametric curve y = f (t), z = g(t)

about the z-axis.

Such a surface is parametrised as

x = f (t) cos θ , y = f (t) sin θ , z = g(t) .

Example. Torus.

Similarly, we can get surfaces of revolution by

revolving a parametric curve

y = f (u), x = g(u) or, equivalently,

a parametric curve z = f (u), x = g(u)

about the x-axis.

They are parametrised as

x = g(u) ,

y = f (u) cos v ,

z = f (u) sin v

Finally, we can get surfaces of

revolution by revolving a parametric curve z = f (u), y = g(u) or,

equivalently, a parametric curve x = f (u), y = g(u) about the y-axis.

They are parametrised as

x = f (u) sin v , y = g(u) , z = f (u) cos v

21

Vector-valued functions of two variables

The vector form of parametric eqs for a surface

~r = ~r(u, v) = x(u, v)~i + y(u, v)~j + z(u, v)~k

~r(u, v) is a vector-valued functions of two variables.

Partial derivatives

~ru =∂~r

∂u=∂x

∂u~i +

∂y

∂u~j +

∂z

∂u~k

~rv =∂~r

∂v=∂x

∂v~i +

∂y

∂v~j +

∂z

∂v~k

Tangent planes to parametric surfaces

Let σ be a parametric surface in 3-space.

Definition. A plane is said to be tangent to σ at P0 provided a line

through P0 lies in the plane if and only if it is a tangent line at P0 to

a curve on σ.

σ : ~r(u, v) , P0(a, b, c) ∈ σa = x(u0, v0) , b = y(u0, v0) , c = z(u0, v0)

If ∂~r∂u 6= 0 then it is tangent to the constant v-curve.

If ∂~r∂v 6= 0 then it is tangent to the constant u-curve.

Thus, if ∂~r∂u ×

∂~r∂v 6= 0 at (u0, v0) then it is orthogonal to both tangent

vectors and is normal to the tangent plane and the surface at P0.

22

~n =∂~r∂u ×

∂~r∂v∣∣∂~r

∂u ×∂~r∂v

∣∣is called the principle normal vector

to the surface ~r = ~r(u, v) at (u0, v0).

Thus, the tangent plane equation is

~n · (~r − ~r0) = 0

Example. Tangent plane to

x = uv , y = u , z = v2 at (2,−1)

Figure 5: Whitney’s umbrella.

Answer : x + y + z = 1

23

Surface area

Let ~r = x(u, v)~i + y(u, v)~j + z(u, v)~k be a smooth parametric

surface on a region R of the uv-plane, that is ∂~r∂u ×

∂~r∂v 6= 0 on R, and

therefore there is a tangent plane for every (u, v) ∈ R.

Rk : ∆Ak = ∆uk∆vk

∆Sk ≈ Area of parallelogram =

∣∣∣∣∂~r∂u∆uk ×∂~r

∂v∆vk

∣∣∣∣=

∣∣∣∣∂~r∂u × ∂~r

∂v

∣∣∣∣∆uk∆vk =

∣∣∣∣∂~r∂u × ∂~r

∂v

∣∣∣∣∆Ak

Thus,

S ≈n∑k=1

∣∣∣∣∂~r∂u × ∂~r

∂v

∣∣∣∣∆Ak

In the limit n→∞

S =

¨R

∣∣∣∣∂~r∂u × ∂~r

∂v

∣∣∣∣ dAExamples. Surfaces of revolution: Sphere, cone, torus

24

Surface area of surfaces of the form z = f (x, y)

x = u , y = v , z = f (u, v) , ~r = u~i + v~j + z ~k

∂~r

∂u=~i +

∂z

∂u~k ,

∂~r

∂v= ~j +

∂z

∂v~k

∂~r

∂u× ∂~r

∂v= ~k − ∂z

∂v~j − ∂z

∂u~i

∣∣∣∣∂~r∂u × ∂~r

∂v

∣∣∣∣ =

√1 +

(∂z

∂u

)2

+

(∂z

∂v

)2

S =

¨R

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dA

Example. S of z = x2 + y2 below z = 1.

Answer : π6(5√

5− 1) ≈ 5.330

25

5 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.

26

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.

27

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 −1

2π ≈ 0.0620106

28

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 −

√3

2 ≈ 0.442972

29

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

30

6 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

31

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

32

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−21

2(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

33

7 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

34

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

35

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 − φ1

are 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

36

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

37