Sc. (Calculus of Several Variables) (S-III)
4.107
Multiple Integrals
2.
1 оу
9. The integral S S ſ dz dy dx =
0 -1 0
2
1 1-X 1-Z
1-x
1
1-2
(a) S S S dx dy dz
(b)
V1-z
S sh dy dz
0-V1-
-1 0 o
2.
Х
-V1-z x²
dx
1 1
1.-V2
گو
sss dy dx dz
1 1 1 -√2
(d) ſ I ſ dz dx dy
0 0 - 1
0 0 - 1
Answers
Answer:
Problem 1 (16.1.15). Use symmetry to evaluate Z Z
R
x
3
dA for R = [−4, 4] × [0, 5].
Solution. For each y, the x-integral goes from −4 to 4, so is symmetric about the x-axis, while the
function x
3
is odd, meaning that (−x)
3 = −x
3
. This means that the negative and positive portions
of the integral cancel, so each x-integral is 0, hence the entire area integral is 0.
Problem 2 (16.1.17). Use symmetry to evaluate Z Z
R
sin x dA for R = [0, 2π] × [0, 2π].
Solution. For each y, the x-integral is the integral of sin x over a whole period, so is 0. Thus the
entire area integral is 0.
Problem 3 (16.1.24). Evaluate Z 1
−1
Z π
0
x
2
sin y dx dy.
Solution.
Z y=1
y=−1
Z x=π
x=0
x
2
sin y dx dy =
Z y=1
y=−1
x
3
3
sin y
x=π
x=0
dy
=
Z y=1
y=−1
π
3
3
sin y dy
=
π
3
3
[− cos y]
y=1
y=−1
=
π
3
3
(cos(−1) − cos 1)
= 0.
Problem 4 (16.1.31). Evaluate Z 2
1
Z 4
0
dy dx
x + y
.
Solution.
Z x=2
x=1
Z y=4
y=0
dy dx
x + y
=
Z x=2
x=1
[ln(x + y)]y=4
y=0 dx
=
Z x=2
x=1
(ln(x + 4) − ln(x)) dx
= [(x + 4) ln(x + 4) − (x + 4)]x=2
x=1 − [x ln x − x]
x=2
x=1
= [(6 ln 6 − 6) − (5 ln 5 − 5)] − [(2 ln 2 − 2) − (1 ln 1 − 1)]
= 6 ln 6 − 5 ln 5 − 2 ln 2.
3