MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
Problem 1.
Let a and b satisfy a2 + b2 = 1 and f is a continuous function of one variable. Perform the
change of variables
u = ax + by, v = bx − ay
to show that
Z 1p
ZZ
1 − u2 · f (u) du.
f (ax + by) dx dy = 2
x2 +y 2 ≤1
−1
Solution: Solving the change of variables for x and y, we find that
au + bv
= au + bv
a2 + b2
bu − av
y= 2
= bu − av.
a + b2
The region of integration is determined by
x=
1 ≥ x2 +y 2 = (au+bv)2 +(bu−av)2 = a2 u2 +b2 v 2 +b2 u2 +a2 v 2 = (a2 +b2 )(u2 +v 2 ) = u2 +v 2
which describes the disk√bounded by√the unit circle in the uv plane. That disk is
bounded by the curves − 1 − u2 and 1 − u2 .
The determinant of the change of variables is
xu xv a b 2
2
=
yu yv b −a = | − a − b | = | − 1| = 1.
Now we change the variables in the double integral, and find that we can compute
the inner integral:
Z
1
√
Z
−1
1−u2
√
− 1−u2
Z
1
f (u) dv du =
2
p
1 − u2 · f (u) du.
−1
Problem 8, page 342.
(a) Evaluate
Z
1 Z y/2+2
2x − y dx dy
0
y/2
and sketch the region of integration in the xy plane.
Solution: The region of integration is the parallelogram with vertices (0, 0), (2, 0),
( 52 , 1), ( 21 , 1).
Z
1 Z y/2+2
Z 1
2x − y dx dy =
0
y/2
0
x=y/2+2
Z
x − yx
dy =
2
x=y/2
1
0
1
2 y 2
y
y2 y2
+ 2 − − 2y − + dy
2
2
4
2
2
MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
Z
1
=
4 dy = 4 .
0
(b) Let u = 2x − y and v = y. Find the region in the uv plane that corresponds to the region
of integration.
Solution: The region in the xy plane is bounded below by the line y = 0 and above
by y = 1. By substituting v for y, we obtain v = 0 and v = 1. The region is bounded on
the sides by the lines x = y2 and x = y2 + 2. The first of these is equivalent to
2x − y = 0 ⇔ u = 0
and the second equation is equivalent to
2x − y = 4 ⇔ u = 4.
Thus the corresponding region in the uv plane is the rectangle with vertices (0, 0),
(4, 0), (4, 1), (0, 1).
(c) Evaluate the integral using this change of variables.
Solution: The determinant of the change of variables is
2 −1 −1
1
−1
0 1 = |2| = 2 .
The integral becomes
Z 1
Z 1 2 u=4
u
u
dv =
4 dv = 4
du dv =
4 u=0
0
0
0
0 2
which is the same as the result of part (a). Good!
Z
1Z 4
Problem 10, page 342.
Evaluate
ZZ r
x+y
dA
x
− 2y
D
where D is the region in the plane bounded by the lines y = x/2, y = 0, x + y = 1.
Solution: We use the change of variables u = x + y, v = x − 2y. Making the
appropriate substitutions, the equations y = x/2, y = 0, x + y = 1 become v = 0,
u = v, u = 1, respectively. The region of integration in the uv plane is therefore the right
triangle with vertices (0, 0), (1, 0), (1, 1).
The determinant of the change of variables is
1 1 −1
1
−1
1 −2 = | − 3| = 3 .
The integral becomes
Z 1Z ur
0
0
u1
dv du =
v3
Z 1
0
2√
uv
3
v=u
Z
du =
v=0
0
1
2√ 2
u du .
3
MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
We may replace
√
3
u2 with u because u is positive in the region of integration:
2 1
Z 1
2
u
1
=
u du =
=
.
3 0
3
0 3
Problem 12, page 342.
Evaluate
(2x + y − 3)2
dx dy
2
D (2y − x + 6)
where D is the square with vertices (0, 0), (2, 1), (3, −1), (1, −2).
ZZ
Solution: We effect the change of variables u = 2y − x + 6, v = 2x + y − 3. Next,
find the equations of the lines that bound the region of integration, and obtain the corresponding equations involving u and v: x = 2y becomes u = 6. x = 2y + 5 becomes
u = 1. y = −2x becomes v = −3. y = −2x + 5 becomes v = 2.
So the corresponding region in the uv plane is the square with vertices (1, −3), (6, −3),
(6, 2), (1, 2).
The determinant of the change of variables is
−1 2 −1
1
−1
2 1 = | − 5| = 5 .
The integral becomes
6
Z 6 3 v=2
Z 6
Z 6Z 2 2
v
7
−7
35
v 1
dv du =
du =
du =
=
.
2
2
2
15u v=−3
3u 1
18
1
1 3u
1
−3 u 5
Problem 15, page 360.
Evaluate
ZZ
xy
dA
− x2
D
where D is the region in the first quadrant bounded by the hyperbolas x2 − y 2 = 1, x2 − y 2 = 4
and the ellipses x2 /4 + y 2 = 1, x2 /16 + y 2 /4 = 1.
y2
Solution: We effect the change of variables u = x2 /4 + y 2 , v = x2 − y 2 . The hyperbolas become the lines v = 1 and v = 4 and the ellipses become the lines u = 1 and u = 4.
The corresponding region in the uv plane is the square with vertices (1, 1), (4, 1), (4, 4),
(1, 4).
The Jacobian is
∂(x, y) ux uy −1 x/2 2y 1
−1
=
=
2x −2y = | − 5xy| = 5xy .
∂(u, v) vx vy The integral becomes
v=4
Z 4Z 4
Z 4Z 4
Z 4
Z 4
xy 1
−1
−1
−1
dv du =
dv du =
log v
du =
log 4 du
5
5
1
1 −v 5xy
1
1 5v
1
1
v=1
=
−3
log 4 .
5
4
MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
Problem 16, page 360.
Evaluate
ZZ
(x2 + y 2 )ex
2 −y 2
dA
D
where D is the region in the first quadrant bounded by the hyperbolas x2 − y 2 = 1, x2 − y 2 = 9,
xy = 1, xy = 4.
Solution: We effect the change of variables u = xy, v = x2 − y 2 . The hyperbolas
become the lines v = 1, v = 9, u = 1, u = 4. The corresponding region in the uv plane is
the rectangle with vertices (1, 1), (4, 1), (4, 9), (1, 9).
The Jacobian is
∂(x, y) ux uy −1 y
1
x
2
2 −1
∂(u, v) = vx vy = 2x −2y = | − 2x − 2y | = 2(x2 + y 2 ) .
The integral becomes
Z 9Z 4
Z 9Z 4
Z 9
1
1 v
3 v
3 v 9
(x2 + y 2 )ev
du
dv
=
e
du
dv
=
e
dv
=
e
2(x2 + y 2 )
2
1
1
1
1 2
1 2
1
3 9
(e − e) .
2
=
Problem 16, page 342.
Evaluate by transforming to polar coordinates:
Z Z √
a2 −y 2
a
ex
−a
2 +y 2
dx dy.
0
Solution: The region of integration is the right half of the disk centered on the origin
of radius a. This is the region determined by − π2 ≤ θ ≤ π2 and 0 ≤ r ≤ a. Notice also
that x2 + y 2 = r2 .
Z
a
Z √a2 −y2
x2 +y 2
e
−a
Z
π/2
Z
dx dy =
Z
π/2
=
−π/2
r2
Z
π/2 e r drdθ =
−π/2
0
a
0
−π/2
π 2
1 a2
(e − 1) dθ = (ea − 1) .
2
2
Problem 18, page 342.
Evaluate
ZZ
D
1
p
dA
4 − x2 − y 2
where D is the disk of radius 1 centered at (0, 1)
1 r2
e
2
r=a
dθ
r=0
MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
5
Solution: We parametrize the disk in polar coordinates:
Z π Z 2 sin θ
r
√
drdθ
4 − r2
0
0
Let u = 4 − r2 and we get du = −2rdr, yielding:
−1
2
Z
π
Z
4cos2 θ
− 21
u
0
4
−1
dudθ =
2
Z π
2u
0
1
2
4 cos2 θ
4
1
dθ =
2
Z
0
π
π
1
4−4 cos θ dθ =
4θ−4 sin θ
2
0
= 2π .
Problem 6, page 355.
Let D be the region bounded on the inside by the circle of radius 1 centered at the origin and
on the outside by the square with vertices (1, 1), (1, −1), (−1, −1) (−1, 1). Find the average
value of
f (x, y) = x2 + y 2
on this region.
Solution: The area is the area of the square minus the area of the disk, i.e. 4 − π.
Now we evaluate the integral of f on D as the difference:
Z 1Z 1
Z 2π Z 1
Z 1
Z 2π 4 1
y3 1
r
2
2
3
2
x + y dydx −
r drdθ =
yx +
dx −
dθ
3
4 0
−1 −1
0
0
−1
0
−1
3
Z 1
π
2x + 2 1
π
8 π
2
2
− = −
=
2x + dx − =
3
2
3
2
3 2
−1
−1
⇒ Average =
16 − 3π
.
24 − 6π
Problem 12, page 355.
Find the center of mass of the plate that is shaped liked the region bounded by y = x2 and
y = 2x with density δ(x, y) = 1 + x + y.
Solution: We first compute the mass
Z 2
Z 2
Z 2 Z 2x
y 2 2x
x4
m=
1 + x + y dydx =
y + xy +
dx =
2x + 3x2 − x3 − dx
2 x2
2
0
x2
0
0
x4 x5 2 24
= x2 + x3 −
−
= .
4
10 0
5
Next,
Z
Z 2 Z 2x
Z 2
xy 2 2x
2
xδ(x, y)dA =
x(1 + x + y) dydx =
dx
xy + x y +
2 x2
D
0
x2
0
3
Z 2
x5
2x
3x4 x5 x6 2 28
2
3
4
=
2x + 3x − x − dx =
+
−
−
= .
2
3
4
5
12 0
5
0
6
MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
Similarly,
Z
y(1 + x + y) dydx =
yδ(x, y)dA =
0
D
Z
2
=
0
Z 2
2 Z 2x
Z
x2
0
y 2 + xy 2 y 3
+
2
3
2x
dx
x2
3
14x3 x4 + x5 x6
2x
7x4 x5 x6 x7 2 324
2x +
=
−
− dx =
+
−
−
−
3
2
3
3
6
10 12 21 0
35
2
Therefore,
x=
7
41
,y =
.
6
21
Problem 6.
Find the mass of the flat elliptical plate
4x2 + 9y 2 ≤ 1,
with density δ = y 2 .
Solution: Use new coordinates u = 2x, v = 3y, or equivalently
v
u
y= .
x=
2
3
Under this change of coordinates, the ellipse corresponds to the unit disc u2 + v 2 ≤ 1.
The mass equals
2
Z
ZZ
1
v
2
y dA =
dA.
6 u2 +v2 ≤1 3
4x2 +9y 2 ≤1
The latter integral will be evaluated using polar coordinates:
Z 2π Z 1
Z 2π
Z 2π
4 2 1
1
1
1
1 θ sin(2θ) 2π
3
2
2
r sin θ drdθ =
r sin θ 0 dθ =
sin θ dθ =
−
54 0
216 0
216 0
216 2
4
0
0
=
π
.
216
Problem 7.
Using the change of coordinates
y
,
x
find the polar moment of inertia in the region bounded by xy = 1, the x-axis, and the two lines
x = 1 and x = 2 with density δ = 1
u = xy, v =
Solution: We need to evaluate the integral
Z 2Z 1
x
(x2 + y 2 )xy dydx.
1
0
The Jacobian is
∂(u, v) y
=
y
∂(x, y) − x2
x 2y
= 2v.
1 =
x
x
MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
7
We have
u 2
, y = uv.
v
Under the change of coordinates, the original region is mapped to a triangle. Specifically, the x axis is mapped to the origin, the hyperbola xy = 1 becomes the line u = 1,
the two vertical lines x = 1 and x = 2 become u = v and u = 4v respectively. The
integral becomes
v=u
Z Z 1Z u Z 1
1
1 1 u
13
3 3u2
u
− + uv
+ uv
dv du =
+
du =
.
du =
2v
2 0
v
8
8
0
u/4 v
0 2
v=u/4
x2 =
Problem 8.
Find the center of mass of the region with density δ = xy bounded by the parabolas y = x2
and x = y 2 .
Solution: First we compute the mass:
√x
Z Z
Z 1 Z √x
1
1 x3 x6 1
1 1 2
1 1
2
5
= .
xy
x − x dx =
−
m=
dx =
xy dydx =
2 0
2 0
2 3
6 0 12
0
x2
x2
Next,
Z
Z
xδ(x, y) dA =
1Z
x
1
x y dydx =
2
2
x2
0
D
√
=
Similarly,
Z
Z
yδ(x, y) dA =
0
D
1Z
√
x2
x
Z 1
2 2
√ x
x y
0
x2
1
dx =
2
Z
0
1
1 x4 x7 1
x −x dx =
−
2 4
7 0
3
6
3
9
⇒x=
.
56
14
1
xy 2 dydx =
3
Z 1
xy
0
3
√x
x2
1
dx =
3
Z
0
1
7
1 2x 2 x8 1
x −x dx =
−
3 7
8 0
5
2
7
3
9
⇒y=
.
56
14
This last computation could have been avoided, by making use of the symmetry of the
problem.
=
Problem 9.
(i) Draw the gradient fields for
f (x, y) = lnr and g(x, y) = ax + by.
Solution: We have
x
y
∇f =
,
.
x2 + y 2 x2 + y 2
This the radial field pointing away from the origin whose length decreases quadratically
in the radius.
Next, ∇g = (a, b) is a constant vector field. All arrows point in the same direction,
and have same magnitude.
8
MATH 52, FALL 2007 - SOLUTIONS TO PROBLEM SET 2.
(ii) Draw the field
yi − xj
.
r
Solution: The vector field consists of vectors tangent to the circle which “rotate” the
circle clockwise.
F=
(iii) Write down a the vector field whose vector at (x, y) is tangent to the circle through (x, y)
with center at the origin, clockwise direction and whose magnitude is r12 .
Solution: By the above example, we have:
yi − xj
G=
.
r3