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Unformatted text preview: M TOPICS IN INTEGRATION 569 x2}: dA = r4 ceszﬂ sin 9 d9 dr r=ﬂ ﬂ=ﬂ Furthermore, since the limits ef integratien with respect te 9 de net invelve r, we can
write
1 2n Ely dA : r4 (11' ceslﬂ sin 9 d9 r=ﬂ E=ﬂ
R and the denble integral in this case reduces te a precinct ef integrals. Thus we ebtain fy (1/1 = [; r5]; Hassle]? = e : WW“ a“ tr ?” Triﬂes “a! a?" a t "If rﬁt ,. W?“ “is its "
a "fft serif . a i " if??? is“?
F sites a .c .. i t ﬁﬁgﬂhﬁaﬁ a e as h a Iifagtﬁt' «'3 " F '35??? 3??” if ' “3531’
w' as as Its” trﬁttatssaameﬁatv ﬁt we “start t“ ’ r a t r ..,»..;ra afresh?! at? waist aw
Evaluate the fellewing: 4 Evaluate If sin :1th + y) dx dy ever the triangle
whese vertices are (0, 0), (2, l), (l, 2).
3 a  s s
2
(3) “Wm +y) dy d‘x (b) x y dy d‘x Sketch the dernains ef integratien ef the denble
e .
I 2 1 integrals
l 2 l 1
2 2
(c) (2): +3.» )dy dx (3) dx ﬂ
4
_..1 2 I} x MIII(1 )
1H? P
Evaluate . 2 . 2
(b) dy (ces 2y)a(1 — I: am a)dx
1 0 U
x— dx (1
y y Change the erder ef integratien, and hence evaluate
the integrals.
ever the rectangle bennded by the lines .r = 0, Wan
x = 2, y = 1 and y = 2_ Evaluate
1 l
a 2 ~ . dx
Evaluate If (a: + y ) ex rly ever the regien fer which dy 1
xE0,y;0andx+yE1. e r'rtElyU'Exn Sketch the dernain ef integratien ef the deuble
integral Sketch the demain ef integratien and evaluate 2 2.1: 1 1...:
. 2
(a) d]: Edy 2 dx (x +y) l gilII—x } x
1 x'x %J: H ﬂ , g g dJHjx
I; D H J(xIFJF)
1 a I— l
(e) div: 1 dy Express the integral in pelar ceerdinates, and hence a 2
a can?) mm '“x ‘J’ ) shew that its value is 574 VECTOR CALCULUS Figure 7.28
Threedimensional
generalization of
Green’s theorem. ' . 3787. ' Surface 5' where dS = r2 dS is the vector element of surface area and ii is a unit vector along the
normal. This generalization is called Stokes‘ theorem, and will be discussed in
Section 7.4.12 after we have formally introduced the concept of a surface integral. Evaluate the line integral i¥[sinydx+(x—cosy)dy] C taken in the anticlockwise sense, where C is the
perimeter of the triangle formed by the lines l l
y=§rcx, y=im x=0 Verify your answer using Green’s theorem in a plane. Use Green’s theorem in a piano to evaluate it [of —y) dx+ (Hy?) dy]
C as a double integral, where C is the triangle with
vertices at (0, 0), (2, O) and (2, 2) and is traversed
in the anticlockwise direction. Evaluate the line integral I = % (xydx+xdy) C where C is the closed curve consisting ofy = x2
fromx=0tox=1andy= V‘xfromx= l tox=0. 7.4.7 Surface integrals Conﬁrm your answer by applying Green‘s theorem
in the plane and evaluating I as a double integral. Use Green’s theorem in a plane to evaluate the line
integral final 3;?) dx + (ey+ 4x2) dy] C where C is the circle x2 +y2 : 4. (Hint: use polar
coordinates to evaluate the double integral.) Evaluate a 2a—x
dx % d},
0 1 4a + { y + x)'
using the transformation of coordinates u = x + y,
v=x—y Using the transformation
x + y = u, — = 12 show that 1 2—y 4 2 l
dej x 2y6x+ydx=J clqu e“dv=e2—l
0 y x o 0 _________.___d.__________________________________m_._______________...___________._____H.u____________m..______________.____..._________________ The extensions of the idea of an integral to line and double integrals are not the only
generalizations that can be made. We can also extend the idea to integration over a
general surface S. Two types of such integrals occur: M Topics IN INTEGRATION 531 Evaluate the area of the surface .s = 2 — x2 — y: lying Evaluate the surface integral ﬁg U(x, y, 2:) d3, 11. above the (x, y) plane. (Hint: Use polar coordinates I?” i where S is the surface of the paraboloid
 I to evaluate the double integral.) a = 2 — (xii + ya) above the (x, y) plane and
U(x, y, s) is given by  Evaluate (a) ffﬂxi + yi)dS, where S is the surface area of . . . . .
the plane 23; + y + 23 = 6 cut me by the planes Give a physical interpretation in each case.
e=ﬂ,z=2,y=0,y=3; (1)) HS: d8, where Sis the surface area of the
hemisphere x2 + y2 + a2 = 1 (2 1‘s 0) cut off
by the cylinder 3:2 — x + y2 = 0. (a) 1 (lirr).:ri+r2 (c): Determine the surface area of the plane
2x+y+2z=16cutoffbyx=0,y=0
andx2+y2= 64. Show that the area of that portion of the surface
of the paraboloid x2 + y2 = 4: included between
(a) ti: (xy, 11:2, .1: + s) andS is the part of the planes: =1 ands = 3 is ‘13? 1:04 — 3'2}. the plane 2x + 2y + s = 6 included in the ﬁrst octant; Evaluate the surface integral in Example 7.26 using
03) v = (3y, 2x3, .33) and S is the surface of the cylindrical polar coordinates. cylinders:2 +y2 =1, 0 1:: a < 1. Evaluate I f 31: ~ d3, where If F = yr? + (x — 23:21,? — xyk, evaluate the surface
Show that ﬂgsi d5 = in, where Sis the surface of integral ff3(cur1F) d5, where S is the surface of
the sphere is2 +y2 + a2 =1, e E U. the sphere is: +y2 + a2 = erg, s w 0. 7.4.9 Volume integrals In Section 7.4.? we deﬁned the integral of a function over a curved surface in three dimensions. This idea can be extended to deﬁne the integral of a function of three
variables through a region T of threedimensional Space by the limit . _ . ' _  ' I _ .  . '  . '  a _  . _   . ._“':da_.'_r.n .c..,.' .H'.'.'_a':":'.
'.r_£':'eh ae'nihi'l'iﬁ'haﬁﬁ of _'._ :14— ..o.. where AV]. (E = 1, . . . , n) is a partition of T into n elements of volume, and (iii$1.51.) is
a point in AK. as illustrated in Figure 7.36. In terms of rectangular cartesian coordinates the triple integral can, as illustrated in
Figure 7.37, be written as b gzix} hgixiy}
f (x: y... 2) fl V = dx dy for, y, z) (12' (7.32) :1 Elite} alter]
1" Note that there are six different orders in which the integration in (7.32) can be
carried out. .5; As we saw for double integrals in (7.28), the expression for the element of volume
" dV = dx dy dz under the transformation a: = x(u, v, w), y = y(u, u. w), z' = z(u, I}, w) may
be obtained using the Jacobian 584 VECTOR CALCULUS ________________._.._—...—.————~——‘— In this example it is natural to use spherical polar coordinates, so that r = 3M3 (r2 sin29) r2 sin9 dr d6 c1¢ 41! a
sphere H II
= i”; r4dr sin36 d6
41m 0 0 2
=§Ma W Evaluate the triple integrals i 2 3 (MJ de dyJ x2};de
0 0 I
2 3 4 (mi j [ xyzzdzdydx
(l l 2 Show that I z .HZ
J, £12]de (x+y+z)dy=0
7] (l x—z Evaluate III sin (x + y + z) (1.: dy dz over the
portion of the positive octant cut off by the plane
x + y + z 2 1:. Evaluate LUV xyz dx d y dz, where V is the region
bounded by the planes )5 = 0, y = 0, z = 0 and
x + y + z = 1. Sketch the region contained between the parabolic
cylinders y z x2 and x 2 y2 and the planes 2 : 0 and
x + y + z = 2. Show that the volume of the region
may be expressed as the triple integral 1 :1 Z—x—y
J J J dz dy dx
0 x2 0 and evaluate it. Use spherical polar coordinates to evaluate “’eroc2 + y2 + 22) dx dydz V 2 0 (up = 3M,(§a5)<§)(2n) 411a where Vis the region in the ﬁrst octant lying within
the sphere x2 +322 +22 = 1. Evaluate Hf xzyzziix + y + 2) dx dy dz thr0ughout
the region deﬁned byx+y+z $1,): a 0,y 2 0,
z B 0. Showthatifx+y+z=u,y+2=uuaridz=uvw
then Hence evaluate the triple integral exp[~4(x + y + 2):] dx dy dz V where Vis the volume of the tetrahedron bounded
bytheplanesx=0,y=0,z=0andx+y+z= 1. Evaluate HIV yz dx dydz taken throughout the
prism with sides parallel to the z axis, whose base
is the triangle with vertices at (0, 0, O), (1, 0, 0),
(0, 1, 0) and whose top is the triangle with vertices
at (0, 0, 2), (1, O, 1), (0, 1, 1). Find also the position
of the centroid of this prism. Evaluate HI 2 dx dy dz thmughout the region
deﬁnedbyx2+y2 S 22,x2+yz+z2 S1,z > 0. Using spherical polar coordinates, evaluate
III x dx dy d2 throughout the positive octant of
the sphere x2 +y2 + 22 = a2. M ...
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 Fall '10
 ChinC.Lee

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