Calculus III
Practice Problems 8: Answers
1. What is the mass of the lamina bounded by the curves y
is δ x y
xy?
3x and y
6x
x 2 where the density function
Answer. Let R represent the region covered by the lamina.
x
PSfrag replacements
3
From the figure we see that R is a type 1 domain, bounded above by y 6x
the range 0 x 3. (To see that, solve 6x x 2 3x for x). Thus the mass is
Mass R
0
x 2 and below by y
3x over
6x x2
3
δ dA
xydy dx
3x
The inner integral is
6x x2
xydy
x
3x
y2
2
x 12x which is 81 2 21 20
42 525.
Thus
Mass
1
2
3
5
x
36x2
2
6x x2
3x
4
12x3
1 x6
2 6
27x3 dx
0
x
4
9x2
12x5
5
27x4
4
1 5
x
2
12x4
81 9
2 6
3
0
27x3
36
5
27
4
2. A lamina filled with a homogeneous material (the density is identicaly equal to 1) is in the shape of the
region 0 x π 0 y sin x. Find its center of mass.
Solution.
π
Mass
0
sin x
dydx
0
π
sin xdx
0
1
ydydx 2 π 8 and the center of mass is π 4 π 8 .
π
Thys ȳ
0
0
π
sin x
0
π 2. To find ȳ, we calculate
Now, by symmetry, the x-coordinate of the center of mass is x̄
Momy
2
0
sin2 xdx
π
4
3. The surface H, given in cylindrical coordinates by z 2θ is a helicoid. What is the volume of the region
R bounded above by H 0 θ 2π , below by the plane z 0 and lying over the disc r 1?
Answer. Here we use polar coordinates. The volume is
zdA
R
R
2θ rdrd θ
2π
1
2
0
0
rθ rdrd θ
2
1
rdr
0
2π
θ dθ
0
2π 2
4. A beach B is shaped in the form of a crescent. We model this on the area between the circle of radius 1, centered at the origin, and the circle of radius 3/4 centered at the point (3/4,0), where the units are in miles. Suppose that the human density σ decreases as we move from the beach according to σ x y
1000 x 2 y2 2
people per square mile.What is the population on that beach?
Answer. We move to polar coordinates r θ Then the crescent is the domain bounded by the circles r 1
and r
3 2 cos θ . These curves intersect when cos θ 2 3, let α represent those angles. Then B is given
by the relations α θ α 1 r
3 2 cos θ , and the population is the integral of the population density
on this beach:
3 cos θ
α
2
103
Population
σ dA
rdrd θ
r4
α 1
The inner integral is
3
2
cos θ
1
Then
Since cos α
2 3 tan α
2
1 r 2
1 4 sec θ dθ
9
103
dr
r3
103
2
Population
2
103
α
2
α
3
2
103
1
2
cos θ
1
103
θ
2
4
sec2 θ
9
4
tan θ
9
α
5, and the population is 397.5 people.
4
103 tan α
9
α
5. The curve z
x 1 2 0 z 1 is rotated about the z-axis, enclosing, together with the xy-plane, a
3-dimensional region R. R is filled with a substance whose density is inversely proportional to the distance
from the z-axis. Find the total mass of this object.
r 1
Answer. The region is that under the curve (using polar coordinates) z
r 1 on the xy-plane. The density is δ k r. Thus, the mass is
R
zδ drd θ 2π
0
6. As u v runs through the region u 2
r 1
1
0
v2
2k
r
r 1
2π k
rdr d θ
3
3
1
0
2
above the disc R
0 2π k
3
1, the vector function
u v I u v J
2
Xuv
2
2
2
uvK
describes a surface S in three space. Write down the double integral which must be calculated to find the
surface area of S.
Answer. We have to integrate dS
Xu
X
2uI
u
Xv d udv:
2uJ
vK
Xv
2vI
2vJ
uK
2v 2u J 4uvK and the ensuing calculation leads to
X X ! 2 2 u v Thus
S 2 2 u v dudv where R is the unit disc. We now switch to polar coordinates: u v
r cos θ sin θ r cos2θ , so
cos 2θ r drdθ 2 cos 2θ dθ 4 2 " cos 2θ dθ S 2 2 2
The last equality comes from the observation that the integral of cos 2θ around the full circle is 8 times the
integral of cos 2θ through π 4 radians. Thus
1 2
S 4 2 #
2
2 2
Xu
2v 2
Xv
2u2 I
u
2
2
2
v
2
2
2
R
2
2π
1
0
2
2
2π
3
0
2
2
2
π 4
0
0
$ 7. Find the volume of the region lying above the disc x 2
z sin π x2 y2 2 .
y2
1 in the xy-plane, and below the surface
sin π r 2 lying
Answer. Switching to polar coordinates, this is the volume bounded above by the surface z
above the disc r 1. Thus
V
2π
zdA
R
0
This integral we calculate by parts;
1
0
1
0
sin π r 2 rdrd θ
2π
1
0
r sin π r 2 dr
2 cos π r 2 dr
π
2 2
2 2
1
sin π r 2
π
π
π 1 π % r sin π r 2 dr
2
r cos π r 2
π
1
1
0
0
1
0
8. Find the mass of the lamina of the region R lying between the ellipses x 2
the density function is δ x y
x 2 y2 .
4y2
1 and x2
4y2
4, where
Answer. Make the change of coordinates u x v 2y. Then R corresponds to the region S in uv-space
bounded by the circles of radius 1 and 2. Writing x and y in terms of u and v by x u y v 2, we calculate
the Jacobian:
∂ xy
1
1 0
det
0 12
∂ uv
2
& '
Thus
∂ x y
Mass δ dA x y dxdy x y dudv
∂ u v
1 v
dudv u
2
4
Now, we switch to polar coordinates in u v space, obtaining
1
r cos θ 1 r sin θ rdrdθ Mass
2
R
2
2
R
S
2
2
S
2π
2
0
2
1
2
2
2
4
2
2
Integrating first with respect to θ , since
2π
0
cos2 θ d θ
we finally obtain
2π
0
sin2 θ d θ
π
5π 2 3
5π 16 1 75π
r dr
8 1
8
4
32
We note that we could have made just one coordinate change, directly from x y to r θ :
Mass
x
r cos θ
r sin θ
2
y
but doing it in two steps is conceptually clearer and computationally easier.
9. Find the area of the region R in the first quadrant bounded by the curves y 2
2x y2
5x x2
4y x2
Answer. Make the change of coordinates
y2
x
u
x2
y
v
u 5 4 v 10. Then
( dxdy ( ∂ x y dudv Area R ∂ u v To calculate the Jacobian, we have to solve for x y in terms of u v. After a little bit of algebra, we find
x u " v " y u " v " so that R corresponds to the region S in uv-space given by the inequalities 2
R
1 3 2 3
S
2 3 1 3
Now, computing the partial derivatives, we find
" " u " v "
" " u " v " '
∂ x y
1 Area R ∂ u v dudv 3
∂ xy
∂ uv
Finally
det
&
1
2 3 2 3
v
3u
2 1 3 1 3
u
v
3
2
1 3 1 3
3
1 2 3 2 3
3
1
9
5
S
2
4
9
13 10
dudv
4
6
10y.