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.Chapter 14 Multiple Integrals..

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...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

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.Chapter 14 Multiple Integrals..

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14.5 Triple Integral, Upper and Lower Limits

14.5 Projections onto coordinate planes.

14.5 Mass, Moments, Centroid, Moment of Inertia

Matb 210 in 2013-2014

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Let B = [a, b] × [c, d] × [r, s] be a rectangular sol-id, and f (x, y, z) be a continuous scalar function de-fined on B. Divide B into l × m × n smaller rectan-gular solids. Label each small rectangular solid byCijk,

where 1 ≤ i ≤ l, 1 ≤ j ≤ m and 1 ≤ k ≤ n. Inside each such Cijk, pick a pointP∗

ijk = (x∗ijk, y∗ijk, z∗ijk). Denote the volume of Cijk by ∆V. Then we may form the

Riemann sum:l

∑i=1

m

∑j=1

n

∑k=1

f (P∗ijk)∆V.

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Definition. The triple integral of f over the rectangular box B is defined to∫∫∫B

f (x, y, z) dV = liml,m,n→∞

l

∑i=1

m

∑j=1

n

∑k=1

f (P∗ijk)∆V.

Remark. The triple integral over non-rectangular region can be definedsimilarly as that of double integral by summing up the smaller rectangles insidethe region in the Riemann sum, and then obtain the limit as triple integral byusing the Riemann sum.

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Remarks. In general, it is difficult to compute the triple integral of scalarfunction over a solid region in space. So one can use the same idea whichwork very well in double integral, to evaluate the triple integral via iteratedintegral. Of course, we need to more understanding of the region first.

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Fubini’s Theorem for non-rectangular region. Let D be a solid region inspace.If D = { (x, y, z) ∈ R3 | (x, y) ∈ R, and zmin(x, y) ≤ z ≤ zmax(x, y) for all(x, y) ∈ R }, where zmax(x, y) and zmin(x, y) are continuous functions definedin the region D in xy-plane. Let f (x, y, z) be a scalar function defined in the

region D, then∫∫∫

Df (x, y, z) dV =

∫∫R

(∫ zmax(x,y)

zmin(x,y)f (x, y, z)dz

)dA.

Remark. (1). The region R in xy-plane is in fact the shadow of the D under theprojection from R3 onto xy-plane, or by simply forgetting the z-coordinate. Theidea is to allow the point P(x, y, 0) varies within the region D, and then draw aline through P perpendicular to xy-plane, which will enter the solid D when zreaches zmin(x, y) and then exit the solid D when z reaches zmax(x, y).(2). The triple integral has similar properties like the double integral, here wecon’t repeat in stating these properties.(3). It is not necessary to project the solid D onto xy-plane, one can project Donto xz-plane, or yz-plane, in these cases, we should use function y = y(x, z)or x = x(y, z) respectively.

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Sketch. Sketch the region D along with its "shadow" R (vertical projection) inthe xy-plane. Label the upper and lower bounding surfaces of D and the upperand lower bounding curves of R.

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Find the z-limits of integration. Draw a line M passing through a typical point(x, y) in the shadow R parallel to the z-axis. As z increases, M enters the solidregion D at z = zmin(x, y) = f1(x, y) and leaves at z = zmax(x, y) = f2(x, y).These are the z-limits of integration

Matb 210 in 2013-2014

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Find the y-limits of integration. Draw a line L through (x, y) parallel to they-axis. As y increases, L enters R at y = ymin = g1(x) and leaves aty = ymax(x) = g2(x). These are the y-limits of integration.

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Find the x-limits of integration. Choose x-limits that include all lines through Rparallel to the y-axis (x = a and x = b in the preceding figure). These are the

x-limits of integration. The integral is∫ x=b

x=a

∫ y=g2(x)

y=g1(x)

∫ z=f1(x,y)

z=f2(x,y)F(x, y, z) dz dy dx.

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Follow similar procedures if you change the order of integration. The "shadow"of region D lies in the plane of the last two variables with respect to which theiterated integration takes place.

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Example. Evaluate triple integral

I =∫∫∫

D

dxdydz(1 + x + y + z)3 , where solid D is bounded by

the planes x + y + z = 1, x = 0, y = 0, z = 0.

Solution. The last 3 equations are the coordinate planes: yz, xz and xy-planes.The oblique plane S : x + y + z = 1 and the xy-plane z = 0 intersect at points(x, y, z) with z = 0 and 1 = x + y + 0 = x + y. so their intersection is a lineℓ = { (x, y, 0) | x + y = 1 }.And xy-plane z = 0 and xz-plane y = 0 meet the point (x, y, z) with y = z = 0,i.e. their intersection is the x-axis= { (x, 0, 0) | x ∈ R }.Similarly, xy-plane z = 0 and yz-plane x = 0 meet at the y-axis.

So (yellow) shadow R on xy of solid D is bounded line ℓ, x-axis and y-axis.The shadow R of D on xy-plane is given by

R = { (x, y, 0) | x + y ≤ 1, x ≥ 0, y ≥ 0},

and

D = { (x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y }.

Matb 210 in 2013-2014

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Example. Evaluate triple integral

I =∫∫∫

D

dxdydz(1 + x + y + z)3 , where solid D is bounded by

the planes x + y + z = 1, x = 0, y = 0, z = 0.

Solution.Let R = { (x, y ) | x + y ≤ 1, x ≥ 0, y ≥ 0} be the shadow of D on xy-plane.Then

D = { (x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y }.

I =∫∫∫

D

dxdydz(1 + x + y + z)3 =

∫∫R

(∫ 1−x−y

0

dz(1 + x + y + z)3

)dxdy

=∫∫

R

[−1

2(1 + x + y + z)2

]1−x−y

z=0dxdy =

12

∫∫R

[1

(1 + x + y)2 − 14

]dxdy

=12

∫ 1

0

∫ 1−y

0

dx(1 + x + y)2 dy − 1

16= · · · = 1

2ln 2 − 5

16.

Matb 210 in 2013-2014

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Example. Evaluate∫∫∫

D

√x2 + z2 dV, where

D is the solid region bounded by the paraboloidS : y = x2 + z2 and the plane π : y = 4.

Solution. First determine the intersection ofthese surfaces π and S, take any intersection point P(x, y, z), sox2 + z2 = y = 4, so the intersection is a circle C : { (x, 4, z) | x2 + z2 = 22 } onthe plane π. We project D onto xz-plane, then its shadow R is bounded theimage C′ of C, i.e. x2 + z2 ≤ 22 ♡.For any point Q(x, y, z) in D, then (x, 0, z) is in R, so it follows from ♡ that

ybottom(x, z) = x2 + z2 ≤ 22 = 4 = ytop(x, z),this means that the plane π lies above the paraboloid S over the D, the solidD = { (x, y, z) | 0 ≤ x2 + z2 ≤ 4, x2 + z2 ≤ y ≤ 4 }. So

∫∫∫D

√x2 + z2 dV

=∫∫

R

∫ ymin(x,z)=4

ymin(x,z)=x2+z2

√x2 + z2 dy dAxz =

∫∫R(4 − x2 − z2)

√x2 + z2 dAxz =∫ 2π

0

∫ 2

0(4 − r2)

√r2 · rdrdθ = 2π

∫ 2

04r2 − r4dr = 2π

[4r3

3− r5

5

]2

0=

128π

15.

Matb 210 in 2013-2014

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Example. Determine the volume of the solid D bounded above by the planez = y + 2 and bounded below by the paraboloid z = x2 + y2.

Solution. Let P(x, y, z) be any point in the intersection of theplane and the paraboloid, we have x2 + y2 = z = x + 2,i.e. x2 − x + y2 = 2, or equivalently in more familiar formx2 + (y − 1

2 )2 = 9

4 , which is a circle C in xy-plane.Then the shadow R of D on xy-plane bounded by the circleC is given by { (x, y, 0) | x2 + (y − 1

2 )2 ≤ 9

4 }.

Then for any (x, y, 0) ∈ R, we have x2 + (y − 12 )

2 ≤ 94 i.e. x2 + y2 ≤ x + 2, so

zmax(x, y) = y + 2, and zmin(x, y) = x2 + y2.

D = { (x, y, z) | x2 + (y − 12)2 ≤ (3/2)2, x2 + y2 ≤ z ≤ y + 2 }.

The volume of D is given by∫∫∫

DdV =

∫∫R

∫ zmax(x,y)

zmin(x,y)1 dzdA

=∫∫

R(zmax(x, y)− zmin(x, y)) dzdA =

∫∫R(y + 2 − x2 − y2) dA.

Remark. We will skip the detailed calculation in the double integral.

Matb 210 in 2013-2014

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Example. Determine the volume of the solid D bounded above by the planez = y + 2 and bounded below by the paraboloid z = x2 + y2.

Solution. If we project the solid D onto yz-plane, then itsshadow T in yz-plane is bounded by z = y2 and the linez = y + 2. In this case, in yz-plane, we haveT = { (y, z) | − 1 ≤ y ≤ 2, y2 ≤ z ≤ y + 2 }.

For any point P(0, y, z) ∈ T, the line through P(0, y, z), parallel to x-axis meetthe paraboloid z = x2 + y2 at Q(xmax, y, z) and Q′(xmax, y, z), soxmax(y, z) =

√z − y2, and xmin(y, z) = −

√z − y2. The volume of D is

Vol(D) =∫∫∫

DdV =

∫∫T

∫ xmax

xmin

dx dAzy =∫ 2

−1

∫ y+2

y2

∫ √z−y2

−√

z−y2dx dz dy

= 2∫ 2

−1

∫ y+2

y2

√z − y2 dzdy = 2

∫ 2

−1

[23

(z − y2

)3/2]y+2

y2dy

=43

∫ 2

−1(2 + y − y2)3/2dy ∗

=43

∫ 3/2

−3/2

(94− u2

)3/2du

♡=

274

∫ π/2

−π/2cos4 θ dθ =

8132

π.

Remarks. In (*) we use u = y − 12 , and in (♡), we use u = 3

2 sin θ.

Matb 210 in 2013-2014

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Example. Find the volume of the solid region D enclosed by the surfacesz = x2 + 3y2 and z = 8 − x2 − y2.

Solution. First sketch region D. Graphsurfaces (on the left) intersect at(x, y, z) with x2 + 3y2 = z = 8− x2 − y2

i.e. x2 + 2y2 = 4, z > 0.The boundary points of shadow R ofD, on xy-plane is given by (x, y, 0) with

x2 + 2y2 = 4.”Upper” boundary of R is the curvey =

√(4 − x2)/2, ”lower" boundary is

y = −√(4 − x2)/2, with −2 ≤ x ≤ 2.

Next is to find the z upper- and lowerlimits of integration. The line Mthrough a typical point (x, y, 0) in R parallel to the z-axis enters D atz = x2 + 3y2 and leaves at z = 8 − x2 − y2.

D = { (x, y, z) | − 2 ≤ x ≤ 2, −√(4 − x2)/2 ≤ y ≤

√(4 − x2)/2,

x2 + 3y2 ≤ z ≤ 8 − x2 − y2 }.

Matb 210 in 2013-2014

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Example. Find the volume of the solid region D enclosed by the surfacesz = x2 + 3y2 and z = 8 − x2 − y2.

Solution. The volume of the solid D is∫∫∫D

dV =∫ 2

−2

∫ ymax=√

(4−x2)/2

ymin=−√

(4−x2)/2

∫ zmax=8−x2−y2

zmin=x2+3y2dz dy dx

=∫ 2

−2

∫ √(4−x2)/2

−√

(4−x2)/2(8 − 2x2 − 4y2) dy dx

=∫ 2

−2

[(8 − 2x2)y − 4

3y3]y=

√(4−x2)/2

y=−√

(4−x2)/2dx

=∫ 2

−2

(2(8 − 2x2)

√4 − x2

2− 8

3

(4 − x2

2

)3/2)dx

=∫ 2

−2

[8(

4 − x2

2

)3/2

− 83

(4 − x2

2

)3/2]dx

=4√

23

∫ 2

−2(4 − x2)3/2dx ∗

= 8π√

2. (* after plugging x = 2 sin u.)

Matb 210 in 2013-2014

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Example. Set up the limits of integration in dydzdx, for evaluating the tripleintegral of a function F(x, y, z) over the tetrahedron D with vertices(0, 0, 0), (1, 1, 0), (0, 1, 0), and (0, 1, 1).

Solution. We sketch D along with its "shad-ow" R in the xz-plane. The upper (right-hand)bounding surface of D lies in the plane y = 1.The lower (left-hand) bounding surface lies inthe plane y = x + z. The upper boundary ofR is the line z = 1 − x. The lower boundaryis the line z = 0. R = { (x, z) | 0 ≤ x ≤1, 0 ≤ z ≤ 1 − x }. It remains to find the y-limits of integration. A line through a typicalpoint (x, z) in R parallel to the y-axis enters Dat y = x + z and leaves at y = 1.

The required integral is∫∫∫

DF(x, y, z) dV =

∫ 1

0

∫ 1−x

0

∫ 1

x+zF(x, y, z) dy dz dx.

Matb 210 in 2013-2014

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Example. Set up the limits of integration in dz dy, dx, for evaluating the tripleintegral of a function F(x, y, z) over the tetrahedron D with vertices(0, 0, 0), (1, 1, 0), (0, 1, 0), and (0, 1, 1).

The required integral is∫∫∫

DF(x, y, z) dV =

∫ 1

0

∫ 1

x

∫ y−x

0F(x, y, z) dz dy dx.

Matb 210 in 2013-2014

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Example. Determine the volume of the solid region D in the first octantbounded by the coordinate planes and the surface z = 4 − x2 − y

Solution. The level surface z = 4 − x2 − y intersectsthe xy-plane at a curve C : 4 = x2 + y. It follows thatthe shadow R of D on the xy-plane is given by R =

{ (x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 4 − x2 }. In this case,the top is given by zmax(x, y) = 4 − x2 − y, and thebottom is given by zmin(x, y) = 0.

Then the volume of D is given by∫∫∫

D1 dV =

∫∫R

∫ zmax=4−x2−y

zmin=01 dz, dAyx

=∫∫

R(4 − x2 − y) dAyx =

∫ 2

0

∫ 4−x2

0(4 − x2 − y) dy dx

=∫ 2

0

[(4 − x2)y − y2

2

]y=4−x2

y=0=∫ 2

0

[(4 − x2)2 − (4 − x2)2

2

]dx

=12

∫ 2

0(4 − x2)2 dx =

12

∫ 2

0(16 − 8x2 + x4) dx =

12× (16 × 2 − 8

3× 23 +

25

5)

= 16 − 323

+165

= 16 × (1 − 23+

15) =

12815

.

Matb 210 in 2013-2014

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Example. Determine the volume of the solid region D in the first octantbounded by the coordinate planes, the plane x + y = 4, and the cylinderS : y2 + 4z2 = 16.

Solution. Cylinder S : y2 + 4z2 = 16 is a surface inspace given by { (x, y, z)|x ∈ R, and y2 + 4z2 = 16 }.The shadow R of D on the yz-plane is given by R =

{ (y, z) | x ≥ 0, y ≥ 0, and y2 + 4z2 ≤ 16 }= { (y, z) | 0 ≤ z ≤ 2, 0 ≤ y ≤ 2

√4 − z2 }. In this

case, the top is given by xmax(y, z) = 4 − y, and thebottom is given by xmin(y, z) = 0.

Then the volume of D is given by∫∫∫

R1 dV =

∫∫R

∫ xmax=4−y

xmin=0dx dAyz

=∫ 2

0

∫ 2√

4−z2

0(4 − y) dy dz =

∫ 2

0

[4y − y2

2

]2√

4−z2

0dz

=∫ 2

0(8√

4 − z2 − 2(4 − z2) ) dz

= 8 × 14× π × 22 − 8 × 2 +

23× 23 = 8π − 32

3.

Matb 210 in 2013-2014

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Example. Given∫∫∫

DdV =

∫ 1

−1

∫ 1

x2

∫ 1−y

0dzdydx, rewrite the integral as an

equivalent iterated integral in the order (a) dy dz dx

Solution. For any point P(x, 0, 0), the line through P parallel to y-axis meet thesolid D at the curve y = x2 at Q, so Q(x, x2, 0). Then the line through point Qparallel to z-axis meets the top y + z = 1 before leaving the solid D at R. HenceR(x, x2, 1 − x2). The shadow S of D on the xz-plane isS = { (x, 0, z) | − 1 ≤ x ≤ 1, 0 ≤ z ≤ 1 − x2 }.

(Continue..)Matb 210 in 2013-2014

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Example. Given∫∫∫

DdV =

∫ 1

−1

∫ 1

x2

∫ 1−y

0dzdydx, rewrite the integral as an

equivalent iterated integral in the order (a) dy dz dx

Solution. For any point P(x, z, 0) in S, the line through P first passes throughthe cylinder (side) y = x2 before entering the solid D, and then meets the plane(top) y + z = 1 before leaving the solid D, it follows that ymin = x2, and

ymax = 1 − z. Hence, we have∫∫∫

DdV =

∫ 1

−1

∫ 1−x2

0

∫ 1−z

x2dy dz dx.

Matb 210 in 2013-2014

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Example. The figure shows the region of integration for the integral∫ 1

0

∫ 1√

x

∫ 1−y

0f (x, y, z) dz dy dx. Rewrite this integral as an equivalent iterated

integral in the five other orders.

Solution. For any point A(x, 0, 0) on the x-axis, the linethrough A parallel to y-axis meets the cylinder y =

√x

at a point B(x,√

x, 0). Then the line through B parallelto z-axis meets the plane z = 1 − y at C(x,

√x, 1 −

√x).

It follows that the shadow R of solid D onto xz-plane isgiven by R = { (x, z) | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 −

√x }.

Matb 210 in 2013-2014

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Example. The figure shows the region of integration for the integral∫ 1

0

∫ 1√

x

∫ 1−y

0f (x, y, z) dz dy dx. Rewrite this integral as an equivalent iterated

integral in dy dz dx.

Solution. For any point A(x, 0, 0) on the x-axis, the linethrough A parallel to y-axis meets the cylinder y =

√x

at a point B(x,√

x, 0). Then the line through B parallelto z-axis meets the plane z = 1 − y at C(x,

√x, 1 −

√x).

It follows that the shadow R of solid D onto xz-plane isgiven by R = { (x, z) | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 −

√x }.

For any point P(x, 0, z) in the shadow, the line through P parallel to y-axismeets the cylinder y =

√x at the point (x,

√x, z) before entering the solid D

and and meets the plane z = 1 − y at the point (x, 1 − z, z), it follows thatymin =

√x, and ymax = 1 − z. It follows that

D = { (x, y, z) | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 −√

x,√

x ≤ y ≤ 1 − z }. Then we have∫ 1

0

∫ 1√

x

∫ 1−y

0f (x, y, z) dz dy dx =

∫ 1

0

∫ 1−√

x

0

∫ 1−z√

xdy dz dx.

Matb 210 in 2013-2014

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Example. The figure shows the solid region D of integration for the integral∫ 1

0

∫ 1−x2

0

∫ 1−x

0f (x, y, z) dy dz dx. Rewrite this integral as an equivalent iterated

integral in the dz dy dx order.

Solution. The shadow R of D onto xy-plane is the triangle given by

R = { (x, y, 0) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x }.

Note that the top of the solid D is the graph z = 1 − x2, and the bottom of D isxy-plane, i.e. zmax(x, y) = 1 − x2, and zmin(x, y) = 0. The desired iterated

integral is∫ 1

0

∫ 1−x

0

∫ 1−x2

0f (x, y, z) dz dy dx.

Matb 210 in 2013-2014

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Example. The figure shows the solid region D of integration for the integral∫ 1

0

∫ 1−x2

0

∫ 1−x

0f (x, y, z) dy dz dx. Rewrite this integral as an equivalent iterated

integral in the dx dz dy order.

Solution. For any P(x, y, z) in the intersection curveC of the two surfaces z = 1 − x2, and y = 1 − x.Then P(x, y, z) satisfies the equations: z = 1 − x2, andy = 1 − x, so P(x, 1 − x, 1 − x2) in C. After projectingP(x, 1− x, 1− x2) of the curve C onto yz-plane, we haveQ(0, 1 − x, 1 − x2) in the red curve. Set y = 1 − x, andz = 1 − x2 with 0 ≤ x ≤ 1, and then eliminate x as

follows: z = 1 − x2 = 1 − (1 − y)2 = 2y − y2 for all 0 ≤ y ≤ 1.Let R1 = { (0, y, z) | 0 ≤ y ≤ 1, 0 ≤ z ≤ −y(2 + y) } be the lower part andR2 = { (0, y, z) | 0 ≤ y ≤ 1, − y(2 + y) ≤ z ≤ 1 } be the upper part of squareon yz-plane. It follows that the integral is given by∫∫

R1

∫ 1−y

0f (x, y, z) dz dA +

∫∫R2

∫ √1−z

0f (x, y, z) dz dA =∫ 1

0

∫ 2y−y2

0

∫ 1−y

0f (x, y, z) dx dz dy +

∫ 1

0

∫ 1

2y−y2

∫ √1−z

0f (x, y, z) dx dz dy.

Matb 210 in 2013-2014