. . . . . .

.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

Math 200 in 2011

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.Chapter 14 Multiple Integrals..

......

14.2 Properties of double integrals

14.3 Area and volume of double integration

14.4 Double integrals and iterated integral in polar coordinates

14.4 Gaussian probability distribution

Math 200 in 2011

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...1∫∫

D( f (x, y) + g(x, y) )dA =

∫∫D

f (x, y) dA +∫∫

Dg(x, y) dA.

...2∫∫

Dcf (x, y) dA = c

∫∫D

f (x, y) dA.

...3 If f (x, y) ≥ g(x, y) for all (x, y) ∈ D, then∫∫D

f (x, y) dA ≥∫∫

Dg(x, y) dA.

...4∫∫

Df (x, y) dA =

∫∫D1

f (x, y) dA +∫∫

D2

f (x, y) dA, where

D = D1 ∪ D2, and D1 and D2 donot overlap except perhaps on theirboundary.

...5∫∫

DdA =

∫∫D

1 dA = Area of D = A(D).

...6 If m ≤ f (x, y) ≤ M for all (x, y) ∈ D, then

mA(D) ≤∫∫

Df (x, y) dA ≤ MA(D).

Math 200 in 2011

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.. Double Integrals in Polar Coordinates..

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Proposition. If f (x, y) is continuous on a region D in xy-plane, and Dcan be described in polar coordinates in the following form{ (r, θ) | α ≤ θ ≤ β, g1(θ) ≤ r ≤ g2(θ) }, then∫∫

Df (x, y) dA =

∫ β

α

∫ g2(θ)

g1(θ)f (r cos θ, r sin θ) r dr dθ.

Math 200 in 2011

. . . . . .

Given a region R in xy-plane, sometimes it is rather easy to describethe boundary of R in terms of polar coordinates instead of rectangularcoordinates. An example is x2 + y2 = a2, can be easily described as{ (r, θ) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ a }. It is the reason why one needs todevelop the double integral in terms of polar coordinates, just like theone in rectangular coordinates.

Without really getting into the details, one can subdivide the planeregion R in terms of polar parameters, just like defining doubleintegral, so that we have a sectorial area ∆A = r ∆r ∆θ, so theRiemann sum is written ∑

ijf (ri cos θj, ri sin θj) ri∆r ∆θ, which will

converges to the double integral∫∫

R′f (r cos θ, r sin θ) r dr dθ, where R′

is the domain for θ and r representing the domain R in xy-plane.Math 200 in 2011

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Example. Use a double integral to find thearea enclosed by one loop of thethree-leaved rose r = sin 3θ.

Solution. The region, in polar coordinates, is{ (r, θ) | 0 ≤ θ ≤ π/3, 0 ≤ r ≤ sin 3θ },hence the area of the region D is∫ π/3

0

∫ sin 3θ

0r dr dθ =

∫ π/3

0

12

sin2(3θ)dθ =14

∫ π/3

0(1 − cos(6θ) ) dθ =

14

[θ − 1

6sin(6θ)

]π/3

0=

π

12.

Math 200 in 2011

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Example. D = { (x, y) | x2 + y2 ≤ 1, y ≥ 0 }= { (x, y) | 0 ≤ y ≤ 1, −

√1 − y2 ≤ x ≤

√1 − y2 }.

One can describe D in terms of polar coordinates as follows{ (r, θ) | − π

2 ≤ θ ≤ π2 , 0 ≤ r ≤ 1 }.

.

......

Example. For any a > 0, prove that the equation r = 2a sin θ in polarcoordinates is the circle x2 + y2 = 2ay.

Solution. The curve x2 + y2 = 2ax can be described in polarcoordinates as r2 = r2(cos2 θ + sin2 θ) = x2 + y2 = 2ay = 2ar sin θ, soone has r = 2a sin θ (0 ≤ θ ≤ π). Geometrically, it means aright-angled triangle circumscribed in a circle of diameter 2a.

Math 200 in 2011

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Example. Find the volume of the solid D bounded by the paraboloidS : z = 25 − x2 − y2 and the xy-plane.

Solution. The paraboloid S : z = 25 − x2 − y2

intersect the xy-plane π : z = 0 in the curve C :0 = 25 − x2 − y2, which is a circle x2 + y2 = 52.So the shadow R of the solid D after projectingonto xy-plane is given by the circular disc R =

{ (x, y) | x2 + y2 ≤ 52 }, in polar coordinates isgiven by { (r, θ) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 5}.

Then the volume of the solid D is given by∫∫R(25 − x2 − y2)dA =

∫ 2π

0

∫ 5

0(25 − r2) r dr dθ =

2π∫ 5

0(25r − r3) dr = 2π

[25r2

2− r4

4

]5

0=

6252

π.

Math 200 in 2011

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Example. Find the volume of the solid that lies under the paraboloidz = x2 + y2, above the xy-plane, and inside the cylinder x2 + y2 = 2x.

Solution. The cylinder x2 + y2 = 2x lies overthe circular disk D which can be described as{ (r, θ) | −π/2 ≤ θ ≤ π/2, 0 ≤ r ≤ 2r cos θ }in polar coordinates. The reason is that if wewrite (x, y, z) = (r cos θ, r sin θ, z) for any pointin the cylinder, then r2 = x2 + y2 ≤ 2x =

2r cos θ, i.e. r ≤ 2 cos θ. As 2 cos θ ≥ r ≥ 0,it follows that

−π/2 ≤ θ ≤ π/2. The height of the solid is the z-value of theparaboloid from the xy-plane. Hence the volume V of the solid is∫∫

D(x2 + y2) dA =

∫ π/2

−π/2

∫ 2 cos θ

0r2 · r dr dθ =

3π

2.

Math 200 in 2011

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Example. Find the volume of the solid that lies below the hemispherez =

√9 − x2 − y2, above the xy-plane, and inside the cylinder

x2 + y2 = 1.

Solution. Let R be the shadow of D after project-ing on xy-plane, then R is the circular disk cen-tered at the origin with radius 1, in polar coordi-nates { (r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π }. More-over, the top ztop =

√9 − x2 − y2 =

√9 − r2, and

zbottom = 0.

Hence the volume of the solid D is∫ 2π

0

∫ 1

0

√9 − r2 · r dr dθ =

2π

−2

∫ 1

0

√9 − r2 d(8 − r2)

= −π

[(9 − r2)3/2

3/2

]1

0

=2π

3

[93/2 − 83/2

]=

2π

3(27 − 16

√2).

Math 200 in 2011

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Example. Find the volume of the solid bounded above by theparaboloid z = 8 − x2 − y2, and below by the paraboloid z = x2 + y2.

Solution. Let P(x, y, z) be the intersection of twoparaboloids, then one has 8 − x2 − y2 = z = x2 +

y2, so x2 + y2 = 4 = 22, which is a circle. Theshadow R of the solid D is then the circular disc,in polar coordinates { (r, θ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤2π }. As 0 ≤ r ≤ 2, we have r2 ≤ 4, i.e. r2 ≤ 8− r2,so one knows that the top of the solid is given byztop = 8 − x2 − y2 = 8 − r2, and the bottom of thesolid is given by zbottom = x2 + y2 = r2.

Hence the volume of the solid D is∫ 2π

0

∫ 2

0(8 − r2 − r2) · r dr dθ = 2π

∫ 2

0(8r − r3) dr

= 2π

[4r2 − r4

4

]2

0= 24π.

Math 200 in 2011

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Example. Determine the volume of the region D common to theinteriors of the cylinders x2 + y2 = 1 and x2 + z2 = 1, in the firstoctant.

Solution. One immediately recognizes the solid D has a top given byx2 + z2 = 1, i.e. zmax(x) =

√1 − x2, and xy-plane as the bottom.

Moreover, the shadow R of the solid D is a circle disk in the 1stquadrant, so R = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤

√1 − x2 }. The volume

of D is given by∫∫

R(√

1 − x2 − 0) dA

=∫ 1

0

∫ √1−x2

0

√1 − x2 dy dx =

∫ 1

0(1 − x2)dx =

[x − x3

3

]1

0=

23

.

Math 200 in 2011

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Example. Determine the volume of the region D common to theinteriors of the cylinders x2 + y2 = 1 and x2 + z2 = 1, in the firstoctant.

Solution. One immediately recognizes the solid D has a top given byx2 + z2 = 1, i.e. zmax(x) =

√1 − x2, and xy-plane as the bottom.

Moreover, the shadow R of the solid D is a circle disk in the 1stquadrant, so R can be described in terms of polar coordinates as{ (r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2 }. The volume of D is given by∫ π/2

0

∫ 10

√1 − r2 cos2 θ rdrdθ

∗= 2

3 .(*): the answer 2

3 is given by integration via rectangular coordinates.

Math 200 in 2011

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Example. Show that the volume of the solid region E bounded by thethree cylinders x2 + y2 = 1, y2 + z2 = 1 and x2 + z2 = 1 is 16 − 8

√2.

Solution. The solid is symmetric with re-spect to the 3 coordinate axes. Thus, it suf-fices to compute the volume of the portion Dwith nonnegative x, y, z-coordinates. In otherwords, D lies in the first octant of the coordi-nate system.

Math 200 in 2011

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Example. Show that the volume of the solid region E bounded by thethree cylinders x2 + y2 = 1, y2 + z2 = 1 and x2 + z2 = 1 is 16 − 8

√2.

Solution. The solid is symmetric with re-spect to the 3 coordinate axes. Thus, it suf-fices to compute the volume of the portion Dwith nonnegative x, y, z-coordinates. In otherwords, D lies in the first octant of the coordi-nate system.For the solid D, it is bounded on top by the graphs of z =

√1 − x2 and

z =√

1 − y2. On its side and bottom, it is bounded by the cylinderx2 + y2 = 1 and the three coordinate planes. Furthermore, the graphsof z =

√1 − x2 and z =

√1 − y2 intersect along a curve on the plane

x = y. Thus the solid is under the graph of z =√

1 − x2 over theregion D in xy-plane described as { (r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/4 }in the polar coordinates. Hence, the volume of original solid E= 16

∫∫D

√1 − x2 dA = 16

∫ π/40

∫ 10

√1 − r2 cos2 θrdrdθ =

16∫ π/4

0

[− (1−r2 cos2 θ)3/2

cos θ

]r=1

r=0dθ = 16

∫ π/40

1−sin3 θ3 cos2 θ

dθ =

163 [tan θ − sec θ − cos θ]π/4

0 = 16 − 8√

2.Math 200 in 2011

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.

......Example. Prove with the polar-coordinates that

∫ +∞

0e−x2

dx =

√π

2.

Solution. As∫ +∞

0e−x dx = lim

M→+∞

∫ M

0e−xdx = lim

M→+∞

(1 − e−M

)= 1,

it follows from comparison test with 0 ≤ e−x2 ≤ e−x that the improper

integral∫ +∞

0e−x2

dx converges. Let IM =∫ M

0e−x2

dx. Need to prove

that limM→+∞

IM =

√π

2. It suffices to show that

π

4= lim

M→+∞I2M.

By Fubini theorem, we have

I2M =

(∫ M

0e−x2

dx)·(∫ M

0e−y2

dy)=

∫ M

0

∫ M

0e−x2

e−y2dy dx =∫∫

RM

e−x2−y2dA, where RM = [0, M]× [0, M] is a square of length M.

Continue...

Math 200 in 2011

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......Example. Prove with the polar-coordinates that

∫ +∞

0e−x2

dx =

√π

2.

Solution. Let DM be quarter circular disk of radius M centered atorigin. By means of polar coordinates, we have

JM =∫∫

DM

e−x2−y2dA =

∫ π/2

0

∫ M

0e−r2

r dr dθ

=π

2

∫ M

0e−r2 × −1

2d(−r2) =

π

4(1 − e−M2

).

It follows that limM→+∞

JM =π

4.

As f (x, y) = e−x2−y2 ≥ 0, and DM ⊂ RM ⊂ DM√

2, it follows fromproperty of double integral that JM ≤ I2

M ≤ JM√

2. The result followsfrom sandwich theorem thatπ

4= lim

M→+∞JM ≤ lim

M→+∞I2M ≤ lim

M→+∞JM

√2 = lim

M→+∞JM =

π

4.

Math 200 in 2011

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.Applications of Double Integrals..

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Suppose a (planar) object, in region R, is made of different material inwhich the density (mass per unit area) is given by δ(x, y), dependingon the location (x, y). Then the total mass of R is given approximatelyby the Riemann sum ∑i δ(xi, yi)∆Ai, which will converges to the

double integral m =∫∫

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

.

......

Similarly, one can define the center of mass (centroid) of the object by

(x, y) =(

1m

∫∫D

xδ(x, y) dA,1m

∫∫D

yδ(x, y) dA)

.

Math 200 in 2011

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Example. A lamina R is made of the part of circular disc of radius a in1st quadrant. Its density is proportional to the distance from theorigin. Determine the position of its centroid.

Solution. In terms of polar coordinates, D can be described as{ (r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ π/2 }. Let δ(x, y) = k

√x2 + y2 = kr.

Then the mass m =∫ π/2

0

∫ a

0kr · r dr dθ =

π

2· ka3

3=

kπa3

6.

As all the conditions on the lamina R are symmetric in x and y, so

x = y =1m

∫ π/2

0

∫ a

0(r sin θ) · kr · r dr dθ =

km

(∫ π/2

0sin θdθ

)·(∫ a

0r3 dr

)=

kkπa3/6

× 1 × a4

4=

3a2π

.

Math 200 in 2011

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.First Theorem of Pappus: Volume of Revolution..

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Suppose that a plane region R is revolved around an axis in its planegenerating a solid of revolution with volume V. Assume that the axisdoes not intersect the interior of R. Then the volume V of the solid is

V = A · d,

where A is the area of R and d is the distance traveled by the centroidof R.

Remark. Cut the region R into vertical strips, and each vertical stripafter rotating will form a ring, which contributes ∆Vi = 2πx∗i ∆Ai,where Ai is the area of the vertical strip. It follows from Riemann sumthat

V ≈n

∑i=1

∆Vi =n

∑i

2πx∗i ∆Ai.Math 200 in 2011

. . . . . .

.Volume of a Sphere........Prove that the volume of a sphere is 4

3 πa3.

Solution. Let R be a region bounded by the up-per semi-circle x2 + y2 = a2, y ≥ 0 and x-axis.

In polar coordinates, R can be described as{ (r, θ) | 0 ≤ θ ≤ π, 0 ≤ r ≤ a }. Using polar coordinates, the area of

R is =∫ π

0

∫ a

0r dr dθ =

a2

2× π = πa2/2. Similarly, y of R is given by

1πa2/2

∫∫R

ydA =1

πa2/2

∫ π

0

∫ a

0r sin θ · r dr dθ

=1

πa2/2

(∫ π

0sin θ dθ

)( ∫ a

0r2 dr

)=

1πa2/2

× [− cos θ ]π0 ×[

r3

3

]a

0

=2

πa2 × 2 × a3

3=

4a3π

. By Pappus’s theorem, the volume of a sphere

of radius a = 2πy× Area of R = 2π · 4a3π

· πa2

2=

43

πa3.

Math 200 in 2011

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.Second Theorem of Pappus: Surface Area of Revolution..

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Suppose that a plane curve C is revolved around an axis in its planethat does not intersect the curve. Then the area A of the surface ofrevolution generated is

A = s · d,

where s is the arc-length of C and d is the distance traveled by thecentroid of C.

Math 200 in 2011

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Let R be a plane lamina and ℓ be a straight line that may or may notlie in xy-plane. Then the moment of inertia I of R around the axis ℓ is∫∫

Rp2δ(x, y)dA, where p = p(x, y) is the shortest distance from the

point (x, y) of R to the line ℓ, and δ(x, y) is the density of R at the point(x, y).

For the coordinate axii, we have.

......

Ix = Ix-axis =∫∫

R(y2)δ(x, y)dA, Iy = Iy-axis =

∫∫R(x2)δ(x, y)dA and

Iz = Iz-axis =∫∫

R(x2 + y2)δ(x, y)dA.

.

......

For any plane lamina R, Define the center (x, y, z) of gyration by

x =√

Ixm , y =

√Iym , z =

√Izm .

Math 200 in 2011