Mathematics schaum%27s outline of theory and problems of advanced calculus 2

(BQ) Part 2 book Advanced calculus has contents: Multiple integrals; line integrals, surface integrals, and integral theorems; infinite series; improper integrals, fourier series, fourier integrals, gamma and beta functions, functions of a complex variable.

Multiple Integrals

Much of the procedure for double and triple integrals may be

thought of as a reversal of partial differentiation and otherwise is

analogous to that for single integrals However, one complexity

that must be addressed relates to the domain of definition With

single integrals, the functions of one variable were defined on

intervals of real numbers Thus, the integrals only depended on

the properties of the functions The integrands of double and

triple integrals are functions of two and three variables,

respec-tively, and as such are defined on two- and three-dimensional

regions These regions have a flexibility in shape not possible

in the single-variable cases For example, with functions of two

variables, and the corresponding double integrals, rectangular

regions, a@ x @ b, c @ y @ d are common However, in

many problems the domains are regions bound above and below by segments of plane curves Inthe case of functions of three variables, and the corresponding triple integrals other than the regions

a@ x @ b; c @ y @ d; e @ z @ f , there are those bound above and below by portions of surfaces Invery special cases, double and triple integrals can be directly evaluated However, the systematictechnique of iterated integration is the usual procedure It is here that the reversal of partial differentia-tion comes into play

Definitions of double and triple integrals are given below Also, the method of iterated integration

Fig 9-1

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where the limit is taken so that the number n of subdivisions increases without limit and such that thelargest linear dimension of eachAkapproaches zero See Fig 9-2(a) If this limit exists, it is denoted by

ð

r

ð

and is called the double integral of F ðx; yÞ over the region r

It can be proved that the limit does exist if F ðx; yÞ is continuous (or sectionally continuous) in r.The double integral has a great variety of interpretations with any individual one dependent on theform of the integrand For example, if F ðx; yÞ ¼ ðx; yÞ represents the variable density of a flat ironplate then the double integral,Ð

A dA, of this function over a same shaped plane region, A, is the mass ofthe plate In Fig 9-2(b) we assume that F ðx; yÞ is a height function (established by a portion of a surface

z ¼ F ðx; yÞÞ for a cylindrically shaped object In this case the double integral represents a volume

ITERATED INTEGRALS

Ifr is such that any lines parallel to the y-axis meet the boundary of r in at most two points (as istrue in Fig 9-1), then we can write the equations of the curves ACB and ADB boundingr as y ¼ f1ðxÞand y ¼ f2ðxÞ, respectively, where f1ðxÞ and f2ðxÞ are single-valued and continuous in a @ x @ b In thiscase we can evaluate the double integral (3) by choosing the regions rk as rectangles formed byconstructing a grid of lines parallel to the x- and y-axes andAk as the corresponding areas Then(3) can be written

ð ð

r

F ðx; yÞ dx dy ¼

ðb x¼a

where the integral in braces is to be evaluated first (keeping x constant) and finally integrating withrespect to x from a to b The result (4) indicates how a double integral can be evaluated by expressing it

in terms of two single integrals called iterated integrals

The process of iterated integration is visually illustrated in Fig 9-3a,b and further illustrated asfollows

The general idea, as demonstrated with respect to a given three-space region, is to establish a planesection, integrate to determine its area, and then add up all the plane sections through an integrationwith respect to the remaining variable For example, choose a value of x (say, x ¼ x0Þ The intersection

of the plane x ¼ x0with the solid establishes the plane section In it z ¼ F ðx0; yÞ is the height function,and if y ¼ f1ðxÞ and y ¼ f2ðxÞ (for all z) are the bounding cylindrical surfaces of the solid, then the width

is f2ðx0Þ  f1ðx0Þ, i.e., y2 y1 Thus, the area of the section is A ¼

ðy 2

y 1

F ðx0; yÞ dy Now establish slabs

Ajxj, where for each intervalxj¼ xj xj1, there is an intermediate value xj0 Then sum these to get

an approximation to the target volume Adding the slabs and taking the limit yields

V ¼ lim

n!1

Xn j¼1

Ajxj¼

ðb a

ðg 2 ð yÞ x¼g 1 ð yÞFðx; yÞ dx

dy

If the double integral exists, (4) and (5) yield the same value (See, however, Problem 9.21.) In writing adouble integral, either of the forms (4) or (5), whichever is appropriate, may be used We call one form

an interchange of the order of integration with respect to the other form

Fig 9-3

In caser is not of the type shown in the above figure, it can generally be subdivided into regions

r1; r2; which are of this type Then the double integral over r is found by taking the sum of thedouble integrals overr1; r2;

TRIPLE INTEGRALS

The above results are easily generalized to closed regions in three dimensions For example,consider a function F ðx; y; zÞ defined in a closed three-dimensional region r Subdivide the regioninto n subregions of volumeVk, k ¼ 1; 2; ; n Letting ðk; k; kÞ be some point in each subregion,

we form

lim

n!1

Xn k¼1

ðb x¼a

ðg 2 ðxÞ y¼g 1 ðxÞ

ðf 2 ðx;yÞ z¼f 1 ðx;yÞF ðx; y; zÞ dz

Extensions to higher dimensions are also possible

Fig 9-4

TRANSFORMATIONS OF MULTIPLE INTEGRALS

In evaluating a multiple integral over a regionr, it is often convenient to use coordinates other thanrectangular, such as the curvilinear coordinates considered in Chapters 6 and 7

If we let ðu; vÞ be curvilinear coordinates of points in a plane, there will be a set of transformationequations x ¼ f ðu; vÞ; y ¼ gðu; vÞ mapping points ðx; yÞ of the xy plane into points ðu; vÞ of the uv plane

In such case the regionr of the xy plane is mapped into a region r0of the uv plane We then have

is the Jacobian of x and y with respect to u and v (see Chapter 6)

Similarly if ðu; v; wÞ are curvilinear coordinates in three dimensions, there will be a set of mation equations x ¼ f ðu; v; wÞ; y ¼ gðu; v; wÞ; z ¼ hðu; v; wÞ and we can write

is the Jacobian of x, y, and z with respect to u, v, and w

The results (9) and (11) correspond to change of variables for double and triple integrals

Generalizations to higher dimensions are easily made

THE DIFFERENTIAL ELEMENT OF AREA IN POLAR COORDINATES, DIFFERENTIALELEMENTS OF AREA IN CYLINDRAL AND SPHERICAL COORDINATES

Of special interest is the differential element of area, dA, for polar coordinates in the plane, and thedifferential elements of volume, dV, for cylindrical and spherical coordinates in three space With these

in hand the double and triple integrals as expressed in these systems are seen to take the following forms.(See Fig 9-5.)

The transformation equations relating cylindrical coordinates to rectangular Cartesian onesappeared in Chapter 7, in particular,

x ¼ cos ; y ¼  sin ; z ¼ zThe coordinate surfaces are circular cylinders, planes, and planes (See Fig 9-5.)

At any point of the space (other than the origin), the set of vectors @r

In the cylindrical case r ¼ cos i þ  sin j þ zk and the set is

The transformation equations relating spherical and rectangular Cartesian coordinates are

x ¼ rsin cos ; y ¼ rsin sin ; z ¼ rcos

In this case the coordinate surfaces are spheres, cones, and planes (See Fig 9-5.)

Fig 9-5

Following the same pattern as with cylindrical coordinates we discover that

Of course, the order of these integrations may be adapted to the geometry

The coordinate surfaces in spherical coordinates are spheres, cones, and planes If r is heldconstant, say, r ¼ a, then we obtain the differential element of surface area

Solved Problems

DOUBLE INTEGRALS

9.1 (a) Sketch the regionr in the xy plane bounded by y ¼ x2; x ¼ 2; y ¼ 1

(b) Give a physical interpreation to

ð ð

r

ðx2þ y2Þ dx dy

(c) Evaluate the double integral in (b)

(a) The required regionr is shown shaded in Fig 9-6 below

(b) Since x2þ y2is the square of the distance from any point ðx; yÞ to ð0; 0Þ, we can consider the doubleintegral as representing the polar moment of inertia (i.e., moment of inertia with respect to the origin) ofthe regionr (assuming unit density)

We can also consider the double integral as representing the mass of the regionr assuming adensity varying as x2þ y2

.(c) Method 1: The double integral can be expressed as the iterated integral

ð2 x¼1

ðx 2 y¼1

ðx2þ y2Þ dy dx ¼

ð2 x¼1

ðx 2 y¼1

ðx2þ y2Þ dy

dx ¼

ð2 x¼1

x2y þy33

x 2

y¼1dx

¼

ð2 x¼1

x4þx6

3  x21

3

!

dx ¼1006105The integration with respect to y (keeping x constant) from y ¼ 1 to y ¼ x2corresponds formally

to summing in a vertical column (see Fig 9-6) The subsequent integration with respect to x from x ¼ 1

to x ¼ 2 corresponds to addition of contributions from all such vertical columns between x ¼ 1 and

x ¼2

Method 2: The double integral can also be expressed as the iterated integral

ð4 y¼1

ð2 x¼ pffiffiyðx2þ y2Þ dx dy ¼

ð4 y¼1

ð2 x¼ pffiffiyðx2þ y2Þ dx

dy ¼

ð4 y¼1

x3

3 þ xy2 2

x¼ pffiffiydy

¼

ð4 y¼1

In this case the vertical column of regionr in Fig 9-6 above is replaced by a horizontal column as

in Fig 9-7 above Then the integration with respect to x (keeping y constant) from x ¼pffiffiffiy

to x ¼ 2corresponds to summing in this horizontal column Subsequent integration with respect to y from

y ¼1 to y ¼ 4 corresponds to addition of contributions for all such horizontal columns between y ¼ 1and y ¼ 4

9.2 Find the volume of the region bound by the elliptic paraboloid z ¼ 4  x21

4y2and the plane

ð2 ffiffiffiffiffiffiffiffi4x 2 p

0

4  x21

4y2

dy dx ¼4

ð2 04y  x2y 1

4

y33

!2 ffiffiffiffiffiffiffiffi4x 2 p

0dx

¼ 16

Hint: Use trigonometric substitutions to complete the integrations

9.3 The geometric model of a material body is a plane region R bound by y ¼ x2and y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi

2  x2

p

onthe interval 0@ x @ 1, and with a density function  ¼ xy (a) Draw the graph of the region.(b) Find the mass of the body (c) Find the coordinates of the center of mass (See Fig 9-8.)

(a)

Fig 9-8

M ¼

ðb a

ðf 2

f 1

 dy dx ¼

ð1 0

ð ffiffiffiffiffiffiffiffip 2x 2

x 2

yx dy dx ¼

ð1 0

y22

" # ffiffiffiffiffiffiffiffip 2x 2

x 2

x dxðbÞ

¼

ð1 0

0

¼ 724(c) The coordinates of the center of mass are defined to be

xx ¼ 1M

ðb a

ðf 2 ðxÞ

f 1 ðxÞx dy dx and
yy ¼ 1

M

ðb a

ðf 2 ðxÞ

f 1 ðxÞy dy dxwhere

M ¼

ðb a

ðf 2 ðxÞ

f 1 ðxÞ dy dxThus,

M
xx ¼

ð1 0

ð ffiffiffiffiffiffiffiffip 2x 2

x 2

x xy dy dx ¼

ð1 0

x2 y22

" # ffiffiffiffiffiffiffiffip 2x 2

x 2

dx ¼

ð1 0

M
yy ¼

ð1 0

9.4 Find the volume of the region common to the intersecting cylinders x2þ y2¼ a2 and

ð ffiffiffiffiffiffiffiffiffiffip a 2 x 2 y¼0

z dy dx

¼ 8

ða x¼0

ð ffiffiffiffiffiffiffiffiffiffip a 2 x 2 y¼0

9.5 Find the volume of the region bounded by

z ¼ x þ y; z ¼ 6; x ¼ 0; y ¼ 0; z ¼ 0

Required volume ¼ volume of region shown in Fig 9-10

¼

ð6 x¼0

ð6x y¼0f6  ðx þ yÞg dy dx

¼

ð6 x¼0

ð6  xÞy 1

2y2 6x

y¼0dx

¼

ð6x¼0

1

2ð6  xÞ2dx ¼36

In this case the volume of a typical column (shown darkly shaded) corresponds to f6  ðx þ yÞg dy dx.The limits of integration are then obtained by integrating over the region r of the figure Keeping xconstant and integrating with respect to y from y ¼ 0 to y ¼ 6  x (obtained from z ¼ 6 and z ¼ x þ yÞcorresponds to summing all columns in a slab parallel to the yz plane Finally, integrating with respect to xfrom x ¼ 0 to x ¼ 6 corresponds to adding the volumes of all such slabs and gives the required volume

TRANSFORMATION OF DOUBLE INTEGRALS

9.6 Justify equation (9), Page 211, for changing variables in a double integral

In rectangular coordinates, the double integral of Fðx; yÞ over the region r (shaded in Fig 9-11) is

ð ð

r

F ðx; yÞ dx dy We can also evaluate this double integral by considering a grid formed by a family of u and

vcurvilinear coordinate curves constructed on the regionr as shown in the figure

Fig 9-11

Let P be any point with coordinates ðx; yÞ or ðu; vÞ, where x ¼ f ðu; vÞ and y ¼ gðu; vÞ Then the vector rfrom O to P is given by r ¼ xi þ yj ¼ f ðu; vÞi þ gðu; vÞj The tangent vectors to the coordinate curves u ¼ c1and v ¼ c2, where c1and c2are constants, are@r=@v and @r=@u, respectively Then the area of region r ofFig 9-11 is given approximately by @r

@u

r

@vBut

1

4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u2þ v2pAnother method: Solve the given equations for x and y in terms of u and v and find the Jacobian directly

9.8 Find the polar moment of inertia of the region in the xy plane bounded by x2 y2¼ 1,

x2 y2¼ 9, xy ¼ 2; xy ¼ 4 assuming unit density

Under the transformation x2 y2¼ u, 2xy ¼ v the required region r in the xy plane [shaded in Fig.9-12(a)] is mapped into region r0

of the uv plane [shaded in Fig 9-12(b)] Then:

Required polar moment of inertia ¼

ð ðr

ð8 v¼4

du dv ¼8

where we have used the results of Problem 9.7

Note that the limits of integration for the regionr0

can be constructed directly from the regionr in the

xyplane without actually constructing the regionr0

In such case we use a grid as in Problem 9.6 Thecoordinates ðu; vÞ are curvilinear coordinates, in this case called hyperbolic coordinates

[Fig 9-13(b) below]

Since@ðx; yÞ

@ð; Þ¼ , it follows that

ð ðr

We can also write the integration limits forr0

immediately on observing the regionr, since for fixed ,

 varies from  ¼ 2 to  ¼ 3 within the sector shown dashed in Fig 9-13(a) An integration with respect to

 from  ¼ 0 to  ¼ 2
then gives the contribution from all sectors Geometrically,  d d represents thearea dA as shown in Fig 9-13(a)

9.10 Find the area of the region in the xy plane bounded by the lemniscate2¼ a2cos 2

Here the curve is given directly in polar coordinates ð; Þ By assigning various values to  and findingcorresponding values of, we obtain the graph shown in Fig 9-14 The required area (making use ofsymmetry) is

4

ð
=4

¼0

ða ffiffiffiffiffiffiffiffifficos 2

a ffiffiffiffiffiffiffiffifficos 2

(c) Evaluate the triple integral in (b)

(a) The required regionr is shown in Fig 9-15

(b) Since x2þ y2þ z2is the square of the distance from any point ðx; y; zÞ to ð0; 0; 0Þ, we can consider thetriple integral as representing the polar moment of inertia (i.e., moment of inertia with respect to theorigin) of the regionr (assuming unit density)

We can also consider the triple integral as representing the mass of the region if the density varies

as x2þ y2þ z2

(c) The triple integral can be expressed as the iterated integral

ða x¼0

ðax y¼0

ðaxy z¼0

ðx2þ y2þ z2Þ dz dy dx

¼

ða x¼0

ðax y¼0

x2z þ y2z þz

33axy

z¼0

dy dx

¼

ða x¼0

ðax y¼0

x2ða  xÞ  x2y þ ða  xÞy2 y3þða  x  yÞ3

y¼0dx

¼

ða 0

dx

¼

ða 0

20

The integration with respect to z (keeping x and y constant) from z ¼ 0 to z ¼ a  x  y sponds to summing the polar moments of inertia (or masses) corresponding to each cube in a verticalcolumn The subsequent integration with respect to y from y ¼ 0 to y ¼ a  x (keeping x constant)corresponds to addition of contributions from all vertical columns contained in a slab parallel to the yzplane Finally, integration with respect to x from x ¼ 0 to x ¼ a adds up contributions from all slabsparallel to the yz plane

corre-Although the above integration has been accomplished in the order z; y; x, any other order isclearly possible and the final answer should be the same

9.12 Find the (a) volume and (b) centroid of the region r bounded by the parabolic cylinder

z ¼4  x2and the planes x ¼ 0, y ¼ 0, y ¼ 6, z ¼ 0 assuming the density to be a constant

The regionr is shown in Fig 9-16

Fig 9-16

ðaÞ Required volume ¼

ð ð ðr

dx dy dz

¼

ð2x¼0

ð6y¼0

ð4x2 z¼0

dz dy dx

¼

ð2 x¼0

ð6 y¼0ð4  x2Þ dy dx

¼

ð2 x¼0ð4  x2Þy

6

y¼0dx

¼

ð2 x¼0ð24  6x2Þ dx ¼ 32

(b) Total mass ¼

ð2 x¼0

ð6 y¼0

ð4x 2 z¼0

dz dy dx ¼ 32 by part (a), since is constant Then

xx ¼Total moment about yz plane

ð2 x¼0

ð6 y¼0

ð4x 2 z¼0

x dz dy dxTotal mass ¼24

32 ¼

34

yy ¼Total moment about xz plane

Ð2 x¼0

Ð6 y¼0

Ð4x2z¼0 y dz dy dxTotal mass ¼96

32 ...

y2< /sup>¼ x; y2< /small>¼ 8x; x2< /small>¼ y; x2< /small>¼ 8y about the x-axis is 27 9
=2 [Hint: Let y2< /small>¼ ux; x2< /small>¼ vy.]

9.35 Find the area of. .. by the spheres x2< /small>ỵ y2< /small>ỵ z2< /small>ẳ a2< /small>

and

x2< /sup>ỵ y2< /small>ỵ z2< /small>ẳ b2< /small>

where... xẳ0f2xịx2< /small>ị  x2< /small>g dx ỵ fx ỵ x2< /small>ị2< /small>g dx2< /small>ị ẳ

1 02x3ỵ x2< /small>ỵ 2x5