Solutions for PSet 9
1. (11.9:8) Using Fubini’s Theorem (we assumed that the double integral exists):
tx
t t txy
e
ey
dx dy =
dx dy =
3
3
1
0 y
[0,t]×[1,t] y
t ty2
t
tx t
e −1
y
dy =
dy =
y −3 e y
t
ty 2
x=0
1
1
t
1 ty2
1
1
1
1
1 2
− 3e +
= 2 − − 3 et + 3 et
ty y=1
t
t t
t
t
2. (11.15:2)
0
1
S
(1 + x) sin y d x d y =
(1 + x)(1 − cos(1 + x))d x =
0
1
x
0
1+x
(1 + x) sin y d y d x =
(x + 2) − (x + 1) sin(x + 1) − cos(x + 1)
2
3
=
+ cos 1 + sin 1 − cos 2 − 2 sin 2
2
3. (11.15:6) The volume can be computed as the double integral of the function
6 − x − 2y
f (x, y) =
over region S = {(x, y)|0 ≤ x ≤ 6, 0 ≤ y ≤ (6 − x)/2}:
3
6 6−x
2
6 − x − 2y
6 − x − 2y
dy dx =
dy dx =
3
3
S
0
0
6
6−x
6
6
6−x
y2 2
(6 − x)2
(6 − x)3
y−
dx = −
=6
dx =
3
3 y=0
12
36
0
0
0
4. (11.15:13) The domain we integrate over is given as
S = {−6 ≤ x ≤ 2,
1
x2 − 4
≤ y ≤ 2 − x}
4
1
0
Observe the points of intersection of the two functions of x are at (−6, 8) and
(2, 0). Integrating in x√first will require dividing the domain into two regions,
as√on 0 ≤ y ≤ 8,
√ − 4 + 4y ≤ x ≤ 2 − y while on −1 ≤ y ≤ 0 we see
− 4 + 4y ≤ x ≤ 4 + 4y.
Therefore we can evaluate our integral
2 2−x
0 √4y+4
f (x, y) d y d x =
f (x, y) d x d y +
√
2
−6
x −4
4
−1
− 4y+4
0
8
2−y
√
− 4y+4
f (x, y) d x d y
5. (11.18:10) Place the coordinate system so that the sides of the rectangle be­
come parallel to the axis and A = (0, 0), B = (0, b), C = (a, b) and D = (a, 0).
The side AB then is along the y axis and the side AD is along the x axis. The
rectangle can be described as Q = {0 ≤ x ≤ a, 0 ≤ y ≤ b}. The distances
of any point (x, y) from segment AB and AD are x and y respectively. Thus,
density f (x, y) and mass m(Q) can be defined as:
f (x, y) = x × y
2
ab
m(Q) =
f (x, y) d y d x =
2
Q
Then the coordinates of the center of mass can be computed as:
1
2
x =
x(xy) d y d x = a
3
m(Q)
Q
2
1
y =
y(xy) d y d x = b
m(Q)
3
Q
6. Let fS , fR represent the density functions for S, R respectively. We define
⎧
⎨ fR (x) if x ∈ R
fS (x) if x ∈ S
(1)
fR∪S (x) =
⎩
0
otherwise
Then
d
x
d
y
d
x
d
y
+
xf d x d y
xf
xf
∪
R
S
R
S S
xT = R∪S
.
= R
dx dy
f dx dy +
f dx dy
f
R∪S R∪S
R R
S S
2
xf
d
d
y
=
x
f
d
x
d
y
=
x
m(R)
and
xfS d x d y =
Now
observe
that
x
R
R
R
R
R
R
S
xS
f
d
x
d
y
=
x
m(S).
Thus
S
S S
xT =
xR m(R) + xS m(S)
.
m(R) + m(S)
A similar argument works for y T and the result follows immediately.
3
MIT OpenCourseWare
http://ocw.mit.edu
18.024 Multivariable Calculus with Theory
Spring 2011
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