Thomas Calculus 13th [Solutions]

Section 15.2 Double Integrals Over General Regions 1097
82. One way would be to partition R into two triangles with
the line y 1. The integral of f over R could then be
written as a sum of integrals that could be evaluated by
integrating first with respect to x and then with respect
to y:
f ( x, y)
dA
R
1 2 ( y/2) 2 2 ( y/2)
f ( x, y) dx dy
f ( x, y) dx dy.
0 2 2y
1 y 1
Partitioning R with the line x 1 would let us write the
integral of f over R as a sum of iterated integrals with
order dy dx.
83.
b b 2 2 b b 2 2 b 2 b 2 b 2 b 2
x y y x y x x y
e dx dy e e dx dy e e dx dy e dx e dy
b b b b b b b b
b 2
2
2
2
2
2
x b x b x
e dx 2 e dx 4 e dx ; taking limits as b gives the stated result.
b 0 0
84.
1 3 2 2 3 1
x
3 1 3
1 1
3 dy
dy dx x dx dy x dy
0 0
2/3
0 0
2/3
0
2/3 3 3 0
2/3
( y 1) ( y 1) ( y 1) 0
( y 1)
3
1 1
3 1/3 1/3
3 lim b dy
0
2/3 3
lim dy
lim ( 1) b
y lim ( y 1)
2/3
b 1 ( y 1) b 1 b ( y 1) b 1 0 b 1
b
1/3 1/3 1/3 1/3 3 3
lim ( b 1) ( 1) lim ( b 1) (2) (0 1) 0 2 1 2
b 1 b 1
85-88. Example CAS commands:
Maple:
f: (x,y) - 1/x/y;
q1: Int( Int (f(x,y) y 1..x), x 1..3);
evalf(q1);
value(q1);
evalf( value(q1) );
89-94. Example CAS commands:
Maple:
f: (x,y) - exp(x^2);
c,d: 0,1;
g1: y - 2*y;
g2: y - 4;
q5: Int( Int( f(x,y), x g1(y)..g2(y) ),y c..d );
value( q5);
plot3d(0, (x g1(y)..g2(y), y c..d, color pink, style patchnogrid, axes boxed, orientation [-90,0]
scaling constrained, title "#89(Section 15.2)" );
r5 : Int( Int( f(x,y), y 0..x / 2), x 0..2 ) Int( Int( f(x, y 0..1), x 2..4);
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