Thomas Calculus 13th [Solutions]

Chapter 15 Additional and Advanced Exercises 1151
11.
e e
b xy b xy b t xy
b
dx e dy dx e dx dy lim e dx dy lim e dy
0 x
0 a a 0 a t 0 a t
y
0
ax bx xy t
b
lim 1 e
yt b
1
b
dy dy ln y ln b
a t
y y a y a a
12. (a) The region of integration is sketched at the right
(b)
2 2
asin a y 2 2
ln x y dx dy
0 y cot
a r 2
ln r dr d ;
0 0
2
u r 1
a
2
2
2
0 0 ln u du d
du r dr
2 2
1 a 1 2 2 2
u ln u u d 2a ln a a lim t ln t d a (2 ln a 1) d a ln a 1
2 0 0 2 0 t 0
2 0
2
2 2
a cos (tan ) x 2 2 a a x 2 2
ln x y dy dx ln x y dy dx
0 0 a cos 0
13.
14.
15.
16.
x u m( x t) ( ) ( )
( ) x x m x t
( ) x m x t
e f t dt du e f t du dt ( x t) e f ( t) dt;
also
0 0 0 t
0
x v u m( x t) ( ) ( )
( ) x x v m x t
( ) x x m x t
e f t dt du dv e f t du dv dt ( v t) e f ( t)
dv dt
0 0 0 0 t t 0 t
x
2
1 2 ( ) ( ) ( )
0 2 ( ) m x t
( ) x x x t m x t
v t e f t dt e f
0 2
( t ) dt
t
1 x
1 x
1 1
f ( x) g( x y) f ( y) dy dx g( x y) f ( x) f ( y) dy dx g( x y) f ( x) f ( y)
dx dy
0 0 0 0 0 y
1 1
f ( y) g( x y) f ( x) dx dy;
0 y
1 1 1 1 1
0 0 (| |) ( ) ( ) x
g x y f x f y dx dy g
0 0 ( x y ) f ( x ) f ( y ) dy dx g
0 x
( y x ) f ( x ) f ( y ) dy dx
1 1 1 1
g
0 ( x y ) f ( x ) f ( y
y
) dx dy g
0 x
( y x ) f ( x ) f ( y ) dy dx
1 1 1 1 1 1
g( x y) f ( x) f ( y) dx dy g( x y) f ( y) f ( x) dx dy 2 g( x y) f ( x) f ( y) dx dy,
0 y 0 y 0 y
and the statement now follows.
simply interchange x and y variable names
2
2 3 x / a
3 3 4 4 a
y 2
x x x x a
2 6 2 6
( ) a x/ a 2 2 a 2 a
1 2
Io
a x y dy dx x y dx dx a
0 0 0 3 0 3 4 12 0
4 12
;
0
a a a a
1 1 3 4 1 4 1 1
o ( ) 2 6 0 3 3 .
4
I a a a a a Since I
1 1
4
3
o ( a ) a
2 2
0, the value of a does provide a
minimum for the polar moment of inertia Io
( a).
2 4 2 2 2 2
3
14 64
0 2 (3) 3 0
4 x
x
3 3
104
I x y dy dx x dx
o
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