[PDF] Triple Integrals in Cylindrical Coordinates Many applications involve

View - Download Triple Integrals in Cylindrical Coordinates Many applications involve

Previous PDF

Next PDF

Triple Integrals in Cylindrical Coordinates

Many applications involve densities for solids that are best expressed in non- Cartesian coordinate systems. In particular, there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example, suppose thatf(r;)g(r;)in polar coordinates and that U(x;y;z)is a continuous function. IfSis the solid betweenz=f(x;y)and z=g(x;y)over a regionRin thexy-plane, then Z Z Z S

U(x;y;z)dV=ZZ

R"

Zf(x;y)

g(x;y)U(x;y;z)dz# dA Let"s suppose now that inpolar coordinates,Ris bounded by=; =; r=p();andr=q():SincedA=rdrdin polar coordinates, a change of variables intocylindrical coordinatesis given by Z Z Z S

U(x;y;z)dV=Z

Z q() p()Z g(r;) f(r;)U(rcos;rsin();z)r dzdrd(1) In practice, however, it is often more straightforward to simply evaluate the ...rst integral inzand then transform the resulting double integral into polar coordinates.

EXAMPLE 1Evaluate the following

Z 1 0Z p1x2 0Z 2xy

0x2+y2dzdydx

Solution:Rather than employ (1) directly, let"s ...rst evaluate the integral inz. That is, Z 1 0Z p1x2 0Z 2xy

0x2+y2dzdydx=Z

1 0Z p1x2

0zx2+zy22xy

0dydx Z 1 0Z p1x2 0

2xyx2+y2dydx

Z Z R

2xyx2+y2dA

1 whereRis the quarter of the unit circle in the1stquadrant. In polar coordinates,Ris bounded by= 0,==2; r= 0;and r= 1:Thus, Z 1 0Z p1x2 0Z 2xy

0x2+y2dzdydx=Z

=2 0Z 1 0

2rcos()rsin()r2rdrd

Z =2 0Z 1 0 sin(2)r5drd Z =2 0 sin(2)r66 1 0 d Z =2 016 sin(2)d 16 Check your Reading:Where did thercome from in (1)?

Triple Integrals in Spherical CoordinatesIfU(r;;z)is given in cylindrical coordinates, then the spherical transformation

z=cos(); r=sin() transformsU(r;;z)intoU(sin();;cos()):Similar to polar coordinates, we have@(z;r)@(;)=dd so that a triple integral in cylindrical coordinates becomes Z Z q() p()Z g(r;) f(r;)U(r;;z)r dzdrd=Z Z q() p()Z g(sin();) f(sin();)U(sin();;cos())r dd d

However,r=sin();which leads to the following:

2 Triple Integrals in Spherical Coordinates:IfSis a solid bounded in spherical coordinates by=,=; =p(); =q(); =f(;);and=g(;);and ifU(;;)is contin- uous onS;then Z Z Z S

U(;;)dV=Z

Z q() p()Z g(;) f(;)U(;;)2sin()ddd (2)

In particular, it is important to notice that

dV=2sin()ddd and it is acceptable to usedV=rdddsincer=sin():It is also important to remember the relationships given by the two right triangles relating (x;y;z)to(;;): In particular,x2+y2=r2impliesx2+y2=2sin2()andr2+z2=2implies that x

2+y2+z2=2(3)

3 Also remember thatranges over[0;2];whileranges over[0;]. EXAMPLE 2Use spherical coordinates to evaluate the triple in- tegral Z 1 1Z p1x2 p1x2Z p1x2y2 p1x2y2px

2+y2+z2dzdydx

Solution:To begin with, we notice that this iterated integral re- duces to Z Z Z unit spherepx

2+y2+z2dV

As a result, in spherical coordinates it becomes

Z 2 0Z 0Z 1 0p

22sin()ddd

sincex2+y2+z2=2:Thus, we have Z 2 0Z 0Z 1 0

3sin()ddd=Z

2 0Z 0Z 1 0

3sin()ddd

Z 2 0Z 0 44
1 0 sin()dd 14 Z 2 0Z 0 sin()dd 14 Z 2 0 cos()j 0d 4 EXAMPLE 3Find the volume of the solid above the conez2= x

2+y2and below the planez= 1:

Solution:The conez2=x2+y2corresponds to==4in spher- ical. Moreover,z= 1corresponds tocos() = 1;or= sec():Thus,

V=Z Z Z

S dV=Z 2 0Z =4 0Z sec() 0

2sin()ddd

Z 2 0Z =4 0 33
sec() 0 sin()dd 13 Z 2 0Z =4 0 sec3()sin()dd

However,sec()sin() = tan();so that

V=13 Z 2 0Z =4 0 tan()sec2()dd

Thus,u= tan(); du= sec2()d,u(0) = tan(0) = 0and

u(=4) = tan(=4) = 1yields V=13 Z 2 0Z 1 0 udud= 13 Z 2 012 d 3 5 Check your Reading:Why does the conez2=x2+y2correspond to==4?

Applications in Spherical and Cylindrical CoordinatesTriple integrals in spherical and cylindrical coordinates occur frequently in ap-

plications. For example, it is not common for charge densities and other real- world distributions to havespherical symmetry,which means that the density is a function only of the distance. ( Note: Scientists and engineers useboth to denote charge density and also to denote distance in spherical coordinates. The context in whichappears will indicate how it is being used). EXAMPLE 4The charge density for a certain charge cloud con- tained in a sphere of radius10cm centered at the origin is given by (x;y;z) = 100px

2+y2+z2Ccm

3 What is the total charge contained within a sphere? (C= micro- coulombs )

Solution:If

denotes the solid sphere of radius10cm centered at the origin, then the total charge is

Q=Z Z Z

100px

2+y2+z2dV

However,x2+y2+z2=2leads to

Q= 100Z

2 0Z 0Z 10 0 2sin()ddd 6 :2500sin(i.e., charge density is proportional to). Evaluation of the integral leads to

Q= 100Z

2 0Z 0 44
10 0 sin()ddd = 100 Z 2 0Z 0

2500sin()dd

= 100 Z 2 0 5000d
= 1;000;000 C which isQ=coulombs. Triple integrals in spherical and cylindrical coordinates are common in the study of electricity and magnetism. In fact, quantities in the ...elds of electricity and magnetism are often de...ned in spherical coordinates to begin with. EXAMPLE 5The power emitted by a certain antenna has a power density per unit volume of p(;;) =P0

2sin4()cos2()

whereP0is a constant with units in Watts. What is the total power within a sphere of radius10m?

Solution:The total powerPwill satisfy

P=Z Z Z

P 0

2sin4()cos2()dV

Z 2 0Z 0Z 10 0P 0

2sin4()cos2()2sin()ddd

=P0Z 2 0Z 0Z 10 0 sin4() sin() cos2()ddd = 10P0Z 2 0Z

01cos2()2sin() cos2()dd

Let us now letu= cos(); du=cos()d; u(0) = 1;andu() =

1:Then

P=10P0Z

2 0Z 1

11u22cos2()dud

= 10P0Z 2 01615
cos2d 7

However,2cos2() = cos(2) + 1;so that

P=8025

P0Z 2 0 (cos(2) + 1)d=325

P0Watts

Check your Reading:Why does the conez2=x2+y2correspond to==4? The Inverse Square LawSuppose two point masses with massesmandMrespectively are located a dis- tancerapart. Sir Isaac Newton"sinverse square lawstates that the magnitude jFjof the gravitational force between the two point masses is jFj=GMmr 2(4) whereGis the universal gravitational constant. However, as Newton realized and struggled with for some time, objects in the real world are not point-masses and instead, the law (4) might need to be modi...ed. In particular, let"s suppose that one of the bodies is not a "point-mass,"but instead is a sphere of radiusRwith uniform mass density:Forr > Rconstant, let"s suppose that the sphere is centered at(0;0;r):If the other body is a point- mass "satellite"of massmlocated at the origin, then the gravitational force is directed along thez-axis. Suppose now that a small "piece"of the sphere is located at a point(;;) (in spherical coordinates), and suppose that it has a small massdM:Then the 8 distance between the small piece and the origin is: so that by (4) the "small"magnitudedjFjof the gravitational force between the small "piece"and the satellite is djFj=Gm dM 2(5) The amount ofdjFjin the vertical direction is then given bycos()djFj(see above). Thus, the total gravitational force in the vertical direction is jFj=Z Z Z S cos()djFj=Z Z Z

SGmcos()

2dM whereSis the sphere corresponding to the "planet". IfdVdenotes the volume of a small "piece"of the sphere, thendM=dV;which leads to jFj=GmZ Z Z

Scos()

2dV(6)

In Cartesian coordinates, the sphereSis given by

x

2+y2+ (zr)2=R2or x2+y2+z22rz+r2=R2

In spherical coordinates this becomes

22rcos() +r2R2= 0

which by the quadratic formula leads to =rcos()pR

2r2(1cos2())

=rcos()qR

2r2sin2()

9

Thus, the sphere is contained between

1=rcosqR

2r2sin2()and 2=rcos+qR

2r2sin2()

Let us also note thatranges from0tosin1(R=r)whileranges over[0;2]:

Evaluating (6) in spherical coordinates leads to

jFj=GmZ 2 0Z sin1(R=r) 0Z 2

1cos()

22sin()ddd

=GmZ 2 0Z sin1(R=r) 0Z 2

1cos()sin()ddd

=GmZ 2 0Z sin1(R=r) 0 (21)cos()sin()ddd

Since21= 2R2r2sin2()1=2;this in turn leads to

jFj=2GmZ 2 0Z sin1(R=r)

0R2r2sin2()1=2sin()cos()dd

If we letu() =R2r2sin2();then the limits of integration become u(0) =R2and u sin 1Rr =R2r2R2r 2 = 0 10

Moreover,du=2r22sin()cos()d;so that

jFj=Gmr 2Z 2 0Z sin1(R=r)

0R2r2sin2()1=22r2sin()cos()dd

Gmr 2Z 2 0Z 0 R

2u1=2du d

Gmr 2Z 2 0u

3=23=2

0 R 2d Gmr 2Z 2 02R33 d Gmr 24R33
However, the volume of the sphere isV= 4R3=3;so that the mass of the sphere isM=V=4R3=3:Thus, we have shown that jFj=GMmr 2 That is, a uniformly-dense spherical "planet"of massMand a point-mass of massMat the center of the sphere have the same gravitational attraction on a "satellite"point mass outside the sphere. Since the electromagnetic force also satis...es an inverse square law, this result also says that the electromagnetic force between spheres with uniform charge density is equivalent to the electromagnetic force between point-charges.

Exercises

Convert to cylindrical coordinates and evaluate:

quotesdbs_dbs17.pdfusesText_23

[PDF] triumph paris nord moto aulnay sous bois

[PDF] trivia questions pdf

[PDF] trouble du langage oral définition

[PDF] trouble langage oral définition

[PDF] true polymorphism

[PDF] truth table generator exclusive or

[PDF] tugboat specifications

[PDF] turk amerikan konsoloslugu new york

[PDF] tutorialkart flutter

[PDF] two characteristics of oral language

[PDF] two dimensional array example

[PDF] two dimensional array in c++

[PDF] two factors that affect the brightness of a star

[PDF] two letter country codes

[PDF] two main parts of scratch window