. . . . . .

.Chapter 14 Multiple Integrals..

......

...1 Double Integrals, Iterated Integrals, Cross-sections

...2 Double Integrals over more general regions, Definition, Evaluation ofDouble Integrals, Properties of Double Integrals

...3 Area and Volume by Double Integration, Volume by Iterated Integrals,Volume between Two surfaces

...4 Double Integrals in Polar Coordinates, More general Regions

...5 Applications of Double Integrals, Volume and First Theorem of Pappus,Surface Area and Second Theorem of Pappus, Moments of Inertia

...6 Triple Integrals, Iterated Triple Integrals

...7 Integration in Cylindrical and Spherical Coordinates

...8 Surface Area, Surface Area of Parametric Surfaces, Surfaces Area inCylindrical Coordinates

...9 Change of Variables in Multiple Integrals, Jacobian

Matb 210 in 2015-2016

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.Chapter 14 Multiple Integrals..

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14.6 Average value of a function

14.7 Cylindrical coordinates, and Spherical Coordinates

14.7 Integration in cylindrical coordinates, and spherical coordinates.

14.7 Mass, Moments, Centroid, Moment of Inertia

Matb 210 in 2015-2016

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.. Applications of Triple Integrals.

......

Suppose an object, in region D in R3, is made of different material in which thedensity (mass per unit volume) is given by δ(x, y, z), depending on the location(x, y, z). Then the total mass of D is given approximately by the Riemann sum∑i δ(xi, yi, zi)∆Vi, which will converges to the triple integral

m =∫∫∫

Dδ(x, y, z) dV. We call it the mass of the object.

Similarly, one define.

......

the center of mass (centroid) of the object by (x, y, z) =(1m

∫∫∫D

xδ(x, y, z) dV,1m

∫∫∫D

yδ(x, y, z) dV,1m

∫∫∫D

zδ(x, y, z) dV)

.

Matb 210 in 2015-2016

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Let D be a solid object in R3, and ℓ be a straight line in R3. Then the moment

of inertia I of D around the axis ℓ is∫∫∫

Dp2δ(x, y, z) dV, where p = p(x, y, z) is

the shortest distance from the point (x, y, z) of D to the line ℓ, and δ(x, y, z) isthe density of D at the point (x, y, z).

For the coordinate axii, we have.

......

Ix = Ix-axis =∫∫∫

D(y2 + z2)δ(x, y, z) dV,

Iy = Iy-axis =∫∫∫

D(x2 + z2)δ(x, y, z) dV and

Iz = Iz-axis =∫∫∫

D(x2 + y2)δ(x, y, z)dV.

.

......

For any solid region D in R3, Define the center (x̂, ŷ, ẑ) of gyration by

x̂ =√

Ixm , ŷ =

√Iym , ẑ =

√Izm .

Matb 210 in 2015-2016

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.. Cylindrical Coordinates.

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In the cylindrical coordinate system, a point P(x, y, z) in three-dimensionalspace is represented by the ordered triple (r, θ, z), where r and θ are polarcoordinates of the projection Q(x, y, 0) of P(x, y, z) onto the xy-plane and z isthe directed distance from the xy-plane to P.

To convert from cylindrical to rectangular coordinates, we use the equationsx = r cos θ, y = r sin θ, z = z.To convert from rectangular to cylindrical coordinates, we use the equationsr =

√x2 + y2, tan θ = yx , z = z.

Matb 210 in 2015-2016

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z2 = r2 r = c (c > 0)Remark. Recall in polar coordinates, r2 = x2 + y2. Hence, the equation z2 = r2

can be expressed as z2 = x2 + y2, which is a circular cone. Another way to

understand is to look 1 = z2

x2 =∥PQ∥2∥OQ∥2 = tan

2 ∠POQ, i.e, ∠POQ = π/4, whichis a cone.

Matb 210 in 2015-2016

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Theorem. If D is a solid described in the cylindrical coordinates as{ ( r, θ, z ) | α ≤ θ ≤ β, r1(θ) ≤ r ≤ r2(θ), zmin(r, θ) ≤ z ≤ zmax(r, θ) },

and f (x, y, z) is a continuous function defined in D,

then∫∫∫

Df (x, y, z) dV =

∫ βα

∫ r2(θ)r1(θ)

∫ zmax(r, θ)zmin(r, θ)

f (r cos θ, r sin θ, z) r dz dr dθ.

Matb 210 in 2015-2016

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Definition. The average value of a function f (x, y, z) defined on a region D in

R3, to be [f ]average =

∫∫∫D f dV∫∫∫D dV

.

.

......

Example. Solid D = { (x, y, z) |√

x2 + y2 ≤ 2, y ≥ 0, 0 ≤ z ≤ 3 }. is in theshape of a half-cylinder of radius 2 and height 3. The temperature (in ◦C) at(x, y, z) is given by T(x, y, z) = 2yz2

√x2 + y2. Find the average temperature in

the tank.

Solution. We describe the tank D in cylindrical coordinates as{ (r, θ, z ) | 0 ≤ θ ≤ π, 0 ≤ r ≤ 2, 0 ≤ z ≤ 3 }.

Recall the formula [T]average =∫∫∫

D T(x,y,z) dV∫∫∫D dV

. Then∫∫∫D

T dV =∫ 2

0

∫ π0

∫ 30

2(r sin θ) · z2 · r · r dz dθ dr

= 2(∫ 2

0r3 dr

)(∫ π0

sin θ dθ)(∫ 3

0z2 dz

)= 2 × 2 × 4 × 9 = 144. Similarly,∫∫∫

DdV =

∫ 20

∫ π0

∫ 30

r dz dθ dr =(∫ 2

0r dr)(∫ π

0dθ)(∫ 3

0dz)

= 2 × π × 3 = 6π.Hence, the average temperature in the tank is 1446π =

24π .

Matb 210 in 2015-2016

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......Example. Evaluate

∫ 2−2

∫ √4−x2−√

4−x2

∫ 2√

x2+y2(x2 + y2) dzdydx.

Solution. Let D be the domain of integral in rectangular coordinates, and R beits shadow in xy-plane. Rewrite the iterated integral∫ 2

−2

∫ √4−x2−√

4−x2

∫ 2√

x2+y2(x2 + y2) dzdydx =

∫∫R

∫ 2√

x2+y2(x2 + y2) dz dAyx.

ymin(x) ymax(x)Then R = { (x, y, 0) | − 2 ≤ x ≤ 2, −

√4 − x2 ≤ y ≤

√4 − x2 }, with boundary

curves are ymin(x) = −√

4 − x2, and ymax(x) =√

4 − x2 }.Squaring both, we get a single equation y2(x) = (±

√4 − x2)2 = 4 − x2, i.e.

y2 + x2 = 4, which represents a complete circle (as −2 ≤ x ≤ 2).

Matb 210 in 2015-2016

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......Example. Evaluate

∫ 2−2

∫ √4−x2−√

4−x2

∫ 2√

x2+y2(x2 + y2) dzdydx.

Solution. The domain D of integral in rectangu-lar coordinates, and R be its shadow in xy-plane.The points of intersection of the (top) plane. Thenin cylindrical coordinates, D can be described as{ (r, θ, z) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, r ≤ z ≤ 2 }.Then by means of Fubini Theorem, and formula ofchange of variables, we can simplify iterated integral∫ 2

−2

∫ √4−x2−√

4−x2

∫ 2√

x2+y2(x2 + y2) dzdydx =

∫∫∫D(x2 + y2) dV

=∫ 2π

0

∫ 20

∫ 2r

r2 · r dz dr dθ = 2π∫ 2

0(2 − r)r3 dr

= 2π[

2r4

4− r

5

5

]20=

26π4 × 5 =

16π5

.

Matb 210 in 2015-2016

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......

Example. A solid T is bounded by the conez =

√x2 + y2 and the plane z = 2. The mass

density at any point of the solid T is proportionalto the distance between the axis of the cone andthe point. Find the mass of T.

Solution. Since the density of the solid at a point P(x, y, z) is proportional to thedistance from the z-axis to the point P(x, y, z), we see that the density functionδ(x, y, z) = k

√x2 + y2 for some constant k. Moreover, T can be described as{ (x, y, z) | 0 ≤ x2 + y2 ≤ 4,

√x2 + y2 ≤ z ≤ 2 }.

In terms of cylindrical coordinates, T may be given as{ (r, θ, z) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, r ≤ z ≤ 2 }.

So the mass of the solid T is∫∫∫T

δ(x, y, z) dV =∫∫∫

Tk√

x2 + y2 dV = k∫ 2π

0

∫ 20

∫ 2r

r · r dz dr dθ

= 2kπ∫ 2

0(2 − r)r2 dr = 2kπ

[2r3

3− r

4

4

]20= 2kπ(

163

− 4) = 8kπ3

.

Matb 210 in 2015-2016

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......

Example. Find the volume of the solid region D bounded below by theparaboloid z = x2 + y2, and above by the plane z = 2x.

Solution. First we sketch the graphs of the paraboloid S1 and the plane S2, andwe want to find the common intersection of S1 and S2. Let P(x, y, z) be anypoint of the common intersection, i.e. they satisfy both 2x = z = x2 + y2, so(x − 1)2 + y2 = 1, which represents a cylinder in space. In fact, the intersectionis a curve obtained by the intersection of the plane and the cylinder above,which is a bounded closed curve. Inside the cylinder, we have(x − 1)2 + y2 ≤ 1, i.e. x2 + y2 ≤ 2x, or equivalently the graph of S1 is below thegraph of the plane S2. Hence the solid region D = { (x, y, z) | x2 + y2 ≤ 2x, andx2 + y2 ≤ z ≤ 2x }. Then one can switch to cylindrical coordinates to describeD as { (r, θ, z) | 0 ≤ r ≤ 2 cos θ, − π2 ≤ θ ≤

π2 and r

2 ≤ z ≤ 2r cos θ }. The

volume of D =∫∫∫

D1 dV =

∫∫x2+y2≤2x

(∫ 2xx2+y2

1 dz)

dA =∫ π/2−π/2

∫ 2 cos θ0

(∫ 2r cos θr2

1 dz)

r drdθ.

Matb 210 in 2015-2016

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Example. Find the volume of the region D that lies inside both the spherex2 + y2 + z2 = 4 and the cylinder x2 + y2 − 2x = 0.

Solution. The cylinder S is given by (x − 1)2 + y2 = 1, so any point P(x, y, z)inside S is given by (x − 1)2 + y2 ≤ 1. In terms of cylindrical coordinates, it canbe described by (r, θ, z) by r2 cos2 θ + r2 sin2 θ − 2r cos θ ≤ 0, i.e. r ≤ 2 cos θ, itfollows that the region can be described in terms of cylindrical coordinates as{ (r, θ, z) | − π2 ≤ θ ≤

π2 , r ≤ 2 cos θ and −

√4 − r2 ≤ z ≤

√4 − r2 }. It follows

from the definition of triple integral that the volume of the region

D =∫∫∫

D1 dV =

∫ π/2−π/2

∫ 2 cos θ0

(∫ √4−r2−√

4−r2dz

)r drdθ =

2∫ π/2−π/2

∫ 2 cos θ0

√4 − r2 · rdrdθ = 2

∫ π/2−π/2

[(4 − r2)3/2

3

]2 cos θ0

dθ = · · · .

Matb 210 in 2015-2016

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Remark. In the calculation above, one needs to pay attention to the following :(sin2 θ)3/2 = | sin3 θ| ̸= sin3 θ for all θ ∈ [−π2 ,

π2 ].∫ π/2

−π/2

[(4 − r2)3/2

3

]2 cos θ0

dθ

=∫ π/2−π/2

[(4 − 4 cos2 θ)3/2 − (4 − 4 cos2 0)3/2

3

]dθ

=∫ π/2−π/2

(4 sin2 θ)3/2

3dθ

=13

∫ π/2−π/2

|2 sin θ|3 dθ

=23

∫ π/20

|2 sin θ|3 dθ

=163

∫ π/20

sin3 θ dθ.

Matb 210 in 2015-2016

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Example. Find the volume and the centroid of solid D, with same density,bounded by paraboloid S : z = b(x2 + y2) (b > 0) and plane π : z = h (h > 0).

Solution. The intersection of plane π : z = h and the paraboloidS : z = b(x2 + y2) is { (x, y, h) | x2 + y2 =

(√h/b

)2 }.D is described as { (x, y, z) | 0 ≤ x2 + y2 ≤ hb , b(x

2 + y2) ≤ z ≤ b }, in polarcoordinates as { (r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤

√hb , br

2 ≤ z ≤ b }.

The volume of the solid D is given by∫∫∫

DdV =

∫ 2π0

∫ √ hb

0

∫ hbr2

r dz dr dθ

= 2π∫ √ h

b

0(hr − br3) dz dr = 2π

(12

ha2 − 14

ba4)=

πh2

2b.

It follows from rotational symmetry of the region D about z-axis, and constantdensity δ = 1, (x, y) = (0, 0). Then

z =1V

∫∫∫D

z dV =2b

πh2

∫ 2π0

∫ √ hb

0

∫ hbr2

rz dz dr dθ =4bh2

∫ √ hb

0

(h2r2

− b2r5

2

)dr

=4bh2

(h2

2× 1

2(

√hb)2 − b

2

2× 1

6× (√

hb)6)

=23

h.

Matb 210 in 2015-2016

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.. Spherical Coordinates

ρ =√

x2 + y2 + z2 = ∥−→OP∥ =length of vector

−→OP. ϕ = angle be-

tween the vector−→OP and the z-axis,

0 ≤ ϕ ≤ π.θ = angle between the shadow ofvector

−→OP onto xy-plane and the x-

axis, 0 ≤ θ ≤ 2π.x = ρ sin ϕ cos θ,y = ρ sin ϕ sin θ,z = ρ cos ϕ.

Matb 210 in 2015-2016

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Matb 210 in 2015-2016

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......Example. Find a spherical coordinate equation for the cone z =√

x2 + y2.

Solution. Substitute for x, y, and z with spherical coordinates:ρ cos ϕ = z =

√x2 + y2 =

√(ρ sin ϕ cos θ)2 + (ρ sin ϕ sin θ)2 = ρ sin ϕ i.e.

cos ϕ = sin ϕ, so tan ϕ = 1. As 0 ≤ ϕ ≤ π, we have ϕ = π/4..

......

Example. Find a spherical coordinate equation for the spherex2 + y2 + (z − a)2 = a2.

Solution. Substitute for x, y, and z with spherical coordinates:a2 = x2 + y2 + (z − a)2 = (ρ sin ϕ cos θ)2 + (ρ sin ϕ sin θ)2 + (ρ cos ϕ − a)2 =(ρ sin ϕ)2 + (ρ cos ϕ)2 − 2aρ cos ϕ + a2 = ρ2 − 2aρ cos ϕ + a2, i.e. ρ2 = 2aρ cos ϕ,hence ρ = 2a cos ϕ.Remark. One should note that x2 + y2 = ρ2 sin2 ϕ..

......Example. Rewrite the sphere x2 + y2 + z2 = 2az (a > 0) in sphericalcoordinates.

Solution. ρ2 = x2 + y2 + z2 = 2az = 2aρ cos ϕ, so ρ = 2a cos ϕ (0 ≤ ϕ ≤ π).

Matb 210 in 2015-2016

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......

Theorem. If D is a solid described in the spherical coordinates as{ (ρ, ϕ, θ) | α ≤ θ ≤ β, ϕ1(θ) ≤ ϕ ≤ ϕ2(θ), ρ1(ϕ, θ) ≤ ρ ≤ ρ2(ϕ, θ) },

and f (x, y, z) is a continuous function defined in D, then∫∫∫D

f (x, y, z) dV

=∫ β

α

∫ ϕ2(θ)ϕ1(θ)

∫ ρ2(ϕ, θ)ρ1(ϕ, θ)

f (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) ρ2 sin ϕ dρ dϕ dθ.

Remark. In general, it is hard to see if a solid D can be expressed in the formstated above. However, solid ball, and cone, or their intersection are typicalexamples that one can think of.

Matb 210 in 2015-2016

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......

∫∫∫D

f (x, y, z) dV

=∫ β

α

∫ ϕ2(θ)ϕ1(θ)

∫ ρ2(ϕ, θ)ρ1(ϕ, θ)

f (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) ρ2 sin ϕ dρ dϕ dθ.

...1 Rotate the green door with points(ρ, ϕ, θ) fixed by θ = θ0 about z-axis fromθmin = α to θmax = β to trap the solid D.

...2 In the green plane set by θ, bend the redray OP with ∠ZOP = ϕ away from z-axisfrom ϕmin = ϕ1(θ) to ϕmax = ϕ2(θ), likeopening a Chinese paper fan.

...3 Extend the red ray in polar coordinate(ρ, ϕ, θ) with fixed ϕ and θ from ρ fromρmin = ρ1(ϕ, θ) and ρmax = ρ2(ϕ, θ).

Matb 210 in 2015-2016

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......Example. Prove that the volume of a solid sphere D of radius a is43 πa

3.

Solution. Let D be described in spherical coordinates as{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π, 0 ≤ ρ ≤ a }, then the volume of the solid

sphere D is∫∫∫

D1 dV =

∫ 2π0

∫ π0

∫ a0

ρ2 sin ϕ dρ dϕ dθ

= 2π∫ π

0

[ρ3

3

]a0

sin ϕ dϕ = 2π × a3

3×∫ π

0sin ϕ dϕ =

43

πa3.

Matb 210 in 2015-2016

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......

Example. Evaluate∫∫∫

De(x

2+y2+z2)3/2dV, where

D = { (x, y, z) | x2 + y2 + z2 ≤ a } is the solid ball of radius a.

Solution. We use spherical coordinates to describe the solid ball D as{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π, 0 ≤ ρ ≤ a } in spherical coordinatesystem, then

e(x2+y2+z2)3/2 = e(ρ

2)3/2 = eρ3,

and∫∫∫

De(x

2+y2+z2)3/2 dV =∫ 2π

0

∫ π0

∫ a0

eρ3ρ2 sin ϕ dρ dϕ dθ

= 2π[∫ π

0sin ϕ dϕ

] [∫ a0

eρ3ρ2dρ

]= 2π[1 − (−1)]×

[eρ

3

3

]a0

=4π3(ea

3 − 1).

Matb 210 in 2015-2016

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......

Example. A solid D is cut from a solid ball of radius 1 centered at the origin bythe inequalities z ≥ 0 and y ≥ x. The mass density of D at a point (x, y, z) isgiven by the function δ(x, y, z) = 30z2 kg /m3.(a) Find the total mass of S. (b) Find the average mass density of S.

Solution. (a) D is described in spherical coordinates by{ (ρ, θ, ϕ) | 0 ≤ ϕ ≤ π/2, π/4 ≤ θ ≤ 5π/4 0 ≤ ρ ≤ 1 }.The total mass of D is∫∫∫

Dδ(x, y, z) dV =

∫ π/20

∫ 5π/4π/4

∫ 10

30(ρ cos ϕ)2ρ2 sin ϕ dρ dθ dϕ

= 30(∫ π/2

0cos2 ϕ sin ϕ dϕ

)(∫ 5π/4π/4

dθ)(∫ 1

0ρ3 dρ

)= 2π.

(b) The volume of D is∫∫∫

DdV =

∫ π/20

∫ 5π/4π/4

∫ 10

ρ2 sin ϕ dρ dθ dϕ

=

(∫ π/20

sin ϕ dϕ)(∫ 5π/4

π/4dθ)(∫ 1

0ρ2 dρ

)= 1 × π × 1

3=

π

3.

The average mass of D is 2ππ/3 = 6.

Matb 210 in 2015-2016

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......

Example. Use spherical coordinates to find the volume of the solid that liesabove the cone z =

√x2 + y2 and below the sphere x2 + y2 + z2 = z.

Solution. The sphere x2 + y2 + z2 = z can be written as ρ2 = ρ cos ϕ, i.e.ρ = cos ϕ. Since the upper solid cone bounded by z =

√x2 + y2 can be best

described in spherical coordinate by 0 ≤ ϕ ≤ π4 . D can be parameterized by{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/4, 0 ≤ ρ ≤ cos ϕ }

in terms of spherical coordinates. For any point P(x, y, z) in D, the distancefrom the point P to the xy-plane is z = ρ cos ϕ. The volume of D is∫∫∫

DdV =

∫ 2π0

∫ π/40

∫ cos ϕ0

ρ2 sin ϕ dρ dϕ dθ = · · · = 5π9√

2.

Matb 210 in 2015-2016

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Example. Let D be a region bounded above by the sphere x2 + y2 + z2 = 32

and below by the cone z =√

x2 + y2. The mass density at any point in D is itsdistance from the xy-plane. Find the total mass m of D.

Solution. Since the cone z =√

x2 + y2 can be best described in sphericalcoordinate by 0 ≤ ρ ≤ π4 . In terms of spherical coordinates, D can beparameterized by { (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/4, 0 ≤ ρ ≤ 3 }. For anypoint P(x, y, z) in D, the distance from the point P to the xy-plane is z = ρ cos ϕ.

The total mass m =∫∫∫

DzdV =

∫ 2π0

∫ π/40

∫ 30

ρ cos ϕ · ρ2 sin ϕ dρdϕdθ

= 2π∫ 3

0ρ3dρ ·

∫ π/40

sin ϕ cos ϕdϕ = 2π × 34

4× sin

2(π/4)2

=81π

8.

Matb 210 in 2015-2016

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......

Example. Let D be ice-cream solid bounded above by the spherex2 + y2 + z2 = 2az and below by the cone z =

√3(x2 + y2).

Find the volume of D.

Solution. The sphere x2 + y2 + z2 = 2az is written in spherical coordinates asρ2 = x2 + y2 + z2 = 2az = 2aρ cos ϕ, i.e. ρ = 2a cos ϕ.

The cone z =√

3(x2 + y2) is written in spherical coordinates as

ρ cos ϕ =√

3ρ2 sin2 ϕ, i.e. tan ϕ = 1√3

and so ϕ = π/6.In terms of spherical coordinates, D can be described as

{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/6, 0 ≤ ρ ≤ 2a cos ϕ }.

The volume of D is given by∫∫∫

D1 dV

=∫ 2π

0

∫ π/60

∫ 2a cos ϕ0

ρ2 sin ϕ dρdϕdθ

= 2π × 8π3

∫ π/60

a3 cos3 ϕ sin ϕ dϕ =16πa3

3

[−1

4cos4 ϕ

]π/60

=712

πa3.

Matb 210 in 2015-2016