. . . . . .

.Chapter 14 Multiple Integrals..

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

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

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

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

...5 Applications of Double Integrals, Volume and First Theorem ofPappus, 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, SurfacesArea in Cylindrical Coordinates

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

. . . . . .

.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

. . . . . .

Let B = [a, b] × [c, d] × [r, s] be a rectangu-lar solid, and f (x, y, z) be a continuous scalarfunction defined on B. Divide B into l × m × nsmaller rectangular solids. Label each smallrectangular solid by Cijk,

where 1 ≤ i ≤ l, 1 ≤ j ≤ m and 1 ≤ k ≤ n. Inside each such Cijk, picka point P∗

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.

.

......

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

to∫∫∫

Bf (x, y, z) dV = lim

l,m,n→∞

l

∑i=1

m

∑j=1

n

∑k=1

f (P∗ijk)∆V.

Remark. The triple integral over non-rectangular region can bedefined similarly as that of double integral by summing up the smallerrectangles inside the region in the Riemann sum, and then obtain thelimit as triple integral by using the Riemann sum.

. . . . . .

.

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Remarks. In general, it is difficult to compute the triple integral ofscalar function over a solid region in space. So one can use the sameidea which work very well in double integral, to evaluate the tripleintegral via iterated integral. Of course, we need to moreunderstanding of the region first.

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Fubini’s Theorem for non-rectangular region. Let D be a solid regionin space.If D = { (x, y, z) ∈ R3 | (x, y) ∈ R, andzmin(x, y) ≤ z ≤ zmax(x, y) for all (x, y) ∈ R }, where zmax(x, y) andzmin(x, y) are continuous functions defined in the region D inxy-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 Dunder the projection from R3 onto xy-plane, or by simply forgettingthe z-coordinate. The idea is to allow the point P(x, y, 0) varies withinthe region D, and then draw a line through P perpendicular toxy-plane, which will enter the solid D when z reaches zmin(x, y) andthen exit the solid D when z reaches zmax(x, y).(2). The triple integral has similar properties like the double integral,here we con’t repeat in stating these properties.(3). It is not necessary to project the solid D onto xy-plane, one canproject D onto xz-plane, or yz-plane, in these cases, we should usefunction y = y(x, z) or x = x(y, z) respectively.

. . . . . .

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Sketch. Sketch the region D along with its "shadow" R (verticalprojection) in the xy-plane. Label the upper and lower boundingsurfaces of D and the upper and lower bounding curves of R.

. . . . . .

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Find the z-limits of integration. Draw a line M passing through atypical point (x, y) in the shadow R parallel to the z-axis. As zincreases, M enters the solid region D at z = zmin(x, y) = f1(x, y) andleaves at z = zmax(x, y) = f2(x, y). These are the z-limits of integration

. . . . . .

.

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

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.

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Find the x-limits of integration. Choose x-limits that include all linesthrough R parallel to the y-axis (x = a and x = b in the precedingfigure). 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.

.

......

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

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Example. Evaluate the triple integral

I =∫∫∫

D

dxdydz(1 + x + y + z)3 , where the solid

D is bounded by the planesx + y + z = 1, x = 0, y = 0, z = 0.

Solution. Let R = { (x, y ) | x + y ≤ 1, x ≥ 0, y ≥ 0} be theprojection image of D onto xy-plane. ThenD = { (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.

. . . . . .

.

......

Example. Evaluate∫∫∫

D

√x2 + z2 dV,

where D is the solid region bounded bythe paraboloid S : y = x2 + z2 and theplane π : y = 4.

Solution. First determine the intersection of these surfaces E and S,take any intersection point P(x, y, z), so x2 + z2 = y = 4, so theintersection is a circle C : { (x, 4, z) | x2 + z2 = 22 } in the plane π. Weproject D onto xz-plane, then its shadow R is bounded the image C′

of C. For any point Q(x, y, z) in D, then (x, 0, z) is in R, so we haveybottom(x, z) = x2 + z2 ≤ 22 = 4 = ytop(x, z), this means that the planeπ 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 =

∫D

∫ ymin(x,z)=4

ymin(x,z)=x2+z2

√x2 + z2 dy dAxz =∫

D(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.

. . . . . .

.

......

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

Solution. Let P(x, y, z) be any point in the intersectionof the plane and the paraboloid, we have x2 + y2 =

z = x + 2, i.e. x2 + (y − 12 )

2 = 94 . Then shadow R

of D on xy-plane is given by { (x, y, 0) | x2 + (y −12 )

2 = 94 }. It follows that the volume of D is given by∫∫∫

DdV =

∫∫R(ztop − zbottom) dzdA

=∫∫

R(y + 2 − x2 − y2) dA.

Remark. We will complete the calculation in the following example.

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.

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

Solution. If we project the solid D onto yz-plane, thenits shadow T in yz-plane is bounded by z = y2 andthe line z = 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-axismeet the 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

given by∫∫∫

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.

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

Solution. To find the limits of in-tegration for evaluating the inte-gral, we first sketch the regionD. The surfaces (as shownabove) intersect on the ellipti-cal cylinder x2 + 3y2 = 8− x2 −y2 or x2 + 2y2 = 4, z > 0.The boundary of the shadow Rof D, after the projection ontoxy-plane, is an ellipse with thesame equation: x2 + 2y2 = 4.The ”upper” boundary of R is the curve y =

√(4 − x2)/2. The lower

boundary is the curve y = −√(4 − x2)/2. In this case, one can see

that −2 ≤ x ≤ 2. It remains to find the z-limits of integration. The lineM passing through a typical point (x, y) in R parallel to the z-axisenters D at z = x2 + 3y2 and leaves at z = 8 − x2 − y2.

. . . . . .

.

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Example. Find the volume of the solid region D enclosed by thesurfaces z = 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.)

. . . . . .

.

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

Solution. We sketch D along with its"shadow" R in the xz-plane. The up-per (right-hand) bounding surface ofD lies in the plane y = 1. The low-er (left-hand) bounding surface liesin the plane y = x + z. The upperboundary of R is the line z = 1 − x.The lower boundary is the line z = 0.R = { (x, z) | 0 ≤ x ≤ 1, 0 ≤ z ≤1 − x }. It remains to find the y-limitsof integration. A line through a typi-cal point (x, z) in R parallel to the y-axis enters D at y = x + z and

leaves at y = 1. The required integral is∫∫∫D

F(x, y, z) dV =∫ 1

0

∫ 1−x

0

∫ 1

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

. . . . . .

.

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

The required integral is∫∫∫D

F(x, y, z) dV =∫ 1

0

∫ 1

x

∫ y−x

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

. . . . . .

.

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Example. Determine the volume of the solid region D in the firstoctant bounded by the coordinate planes and the surfacez = 4 − x2 − y

Solution. The level surface z = 4− x2 − y inter-sects the xy-plane at a curve C : 4 = x2 + y. Itfollows that the shadow R of D on the xy-planeis given by R = { (x, y) | 0 ≤ x ≤ 2, 0 ≤y ≤ 4 − x2 }. In this case, the top is given byzmax(x, y) = 4 − x2 − y.

Then the volume of D is given by∫∫

R(4 − x2 − y) dA

=∫ 2

0

∫ 4−x2

0(4 − x2 − y) dy dz =

∫ 2

0

[(4 − x2)y − y2

2

]4−x2

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 − 83× 23 +

25

5

)= 16 − 32

3+

165

= 16 × (1 − 23+

15) =

12815

.

. . . . . .

.

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Example. Determine the volume of the solid region D in the firstoctant bounded by the coordinate planes, the plane x + y = 4, andthe cylinder y2 + 4z2 = 16.

Solution. The shadow R of D on the yz-planeis given by R = { (y, z) | x ≥ 0, y ≥ 0, andy2 + 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.

Then the volume of D is given by∫∫R(4 − y) dA =

∫ 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.

. . . . . .

.

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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-axismeet the solid D at the curve y = x2 at Q, so Q(x, x2, 0). Then the linethrough point Q parallel to z-axis meets the top y + z = 1 beforeleaving the solid D at R. Hence R(x, x2, 1 − x2). The shadow S of D onthe xz-plane is S = { (x, 0, z) | − 1 ≤ x ≤ 1, 0 ≤ z ≤ 1 − x2 }.

(Continue..)

. . . . . .

.

......

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 passesthrough the cylinder (side) y = x2 before entering the solid D, andthen meets the plane (top) y + z = 1 before leaving the solid D, itfollows that ymin = x2, and ymax = 1 − z. Hence, we have∫∫∫

DdV =

∫ 1

−1

∫ 1−x2

0

∫ 1−z

x2dy dz dx.

. . . . . .

.

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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 line through A parallel to y-axis meets thecylinder y =

√x at a point B(x,

√x, 0). Then the

line through B parallel to z-axis meets the planez = 1 − y at C(x,

√x, 1 −

√x). It follows that the

shadow R of solid D onto xz-plane is given byR = { (x, z) | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 −

√x }.

For any point P(x, 0, z) in the shadow, the line through P parallel toy-axis meets 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 that ymin =

√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.

. . . . . .

.

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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 byR = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x }. Note that the top of the solidD is the graph z = 1 − x2, and the bottom of D is xy-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.

. . . . . .

.

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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. For any P(x, y, z) in the intersec-tion curve C (in red) of the two surfacesz = 1 − x2, and y = 1 − x. Then P(x, y, z)satisfies the equations: z = 1 − x2, andy = 1 − x. It follows that P(x, 1 − x, 1 − x2).

After projecting the point P(x, 1 − x, 1 − x2) of the curve C ontoyz-plane, we have Q(0, 1 − x, 1 − x2), set y = 1 − x, and z = 1 − x2,i.e. z = (1 − x)(1 + x) = −(1 − x)(−1 − x) = −y(−2 + y) = 2y − y2

for all 0 ≤ y ≤ 1. Let R1 = { (y, z) | 0 ≤ y ≤ 1, 0 ≤ z ≤ −y(2 + y) }and R2 = { (y, z) | 0 ≤ y ≤ 1, − y(2 + y) ≤ z ≤ 1 }. It follows that theintegral is given by∫ 1

0

∫ 2y−y2

0

∫ √1−z

0dx dz dy +

∫ 1

0

∫ 1

2y−y2

∫ 1−y

0dx dz dy.