1 −
x 2 + y 2 d A ,
R
R
where R = {( x, y) : x 2 + y 2 ≤ 1} is the unit disk in R2
y
(see Figure 3.5.3). In polar coordinates ( r, θ) we know
0
that
x 2 + y 2 = r and that the unit disk R is the set
x
R′ = {( r, θ) : 0 ≤ r ≤ 1,0 ≤ θ ≤ 2 π}. Thus,
Figure 3.5.3
z =
x 2 + y 2
2 π
1
V =
(1 − r) r dr dθ
0
0
2 π
1
=
( r − r 2) dr dθ
0
0
2 π
r=1
=
r 2
dθ
2 − r 3
3
0
r=0
2 π
=
1 dθ
6
0
π
= 3
In a similar fashion, it can be shown (see Exercises 5-6) that triple integrals in cylindrical
and spherical coordinates take the following forms:
Triple Integral in Cylindrical Coordinates
f ( x, y, z) dx d y d z =
f ( r cos θ, r sin θ, z) r dr dθ dz ,
(3.25)
S
S′
where the mapping x = r cos θ, y = r sin θ, z = z maps the solid S′ in rθz-space onto the solid S in x yz-space in a one-to-one manner.
Triple Integral in Spherical Coordinates
f ( x, y, z) dx d y d z =
f ( ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ 2 sin φ dρ dφ dθ , (3.26) S
S′
where the mapping x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ maps the solid S′ in ρφθ-
space onto the solid S in x yz-space in a one-to-one manner.
3.5 Change of Variables in Multiple Integrals
123
Example 3.12. For a > 0, find the volume V inside the sphere S = x 2 + y 2 + z 2 = a 2.
Solution: We see that S is the set ρ = a in spherical coordinates, so
2 π
π
a
V =
1 dV =
1 ρ 2 sin φ dρ dφ dθ
0
0
0
S
2 π
π ρ 3 ρ= a
2 π
π a 3
=
sin φ dφ dθ =
sin φ dφ dθ
0
0
3 ρ=0
0
0
3
2 π
a 3
φ= π
2 π 2 a 3
4 πa 3
=
−
cos φ
dθ =
dθ =
.
0
3
φ=0
0
3
3
Exercises
A
1. Find the volume V inside the paraboloid z = x 2 + y 2 for 0 ≤ z ≤ 4.
2. Find the volume V inside the cone z =
x 2 + y 2 for 0 ≤ z ≤ 3.
B
3. Find the volume V of the solid inside both x 2 + y 2 + z 2 = 4 and x 2 + y 2 = 1.
4. Find the volume V inside both the sphere x 2 + y 2 + z 2 = 1 and the cone z =
x 2 + y 2.
5. Prove formula (3.25).
6. Prove formula (3.26).
7. Evaluate
sin x+ y cos x− y d A, where R is the triangle with vertices (0, 0), (2, 0) and
2
2
R
(1, 1). ( Hint: Use the change of variables u = ( x + y)/2 , v = ( x − y)/2 . )
8. Find the volume of the solid bounded by z = x 2 + y 2 and z 2 = 4( x 2 + y 2).
9. Find the volume inside the elliptic cylinder x 2
a 2 + y 2
b 2 = 1 for 0 ≤ z ≤ 2.
C
10. Show that the volume inside the ellipsoid x 2
. ( Hint: Use the change
a 2 + y 2
b 2 + z 2
c 2 = 1 is 4 πabc
3
of variables x = au, y = bv, z = cw, then consider Example 3.12. )
11. Show that the Beta function, defined by
1
B( x, y) =
tx−1(1 − t) y−1 dt , for x > 0, y > 0,
0
satisfies the relation B( y, x) = B( x, y) for x > 0, y > 0.
12. Using the substitution t = u/( u + 1), show that the Beta function can be written as
∞
ux−1
B( x, y) =
du ,
for x > 0, y > 0.
0
( u + 1) x+ y
124
CHAPTER 3. MULTIPLE INTEGRALS
3.6 Application: Center of Mass
y
Recall from single-variable calculus that for a region
R
y
= {( x, y) : a ≤ x ≤ b,0 ≤ y ≤ f ( x)} in R2 that represents
= f ( x)
a thin, flat plate (see Figure 3.6.1), where f ( x) is a con-
tinuous function on [ a, b], the center of mass of R has
R
coordinates ( ¯ x, ¯ y) given by
( ¯ x, ¯ y)
x
0
a
b
My
Mx
¯ x =
and
¯ y =
,
M
M
Figure 3.6.1
Center of mass of R
where
b ( f ( x))2
b
b
Mx =
dx ,
My =
x f ( x) dx ,
M =
f ( x) dx ,
(3.27)
a
2
a
a
assuming that R has uniform density, i.e the mass of R is uniformly distributed over the
region. In this case the area M of the region is considered the mass of R (the density is
constant, and taken as 1 for simplicity).
In the general case where the density of a region (or lamina) R is a continuous function
δ = δ( x, y) of the coordinates ( x, y) of points inside R (where R can be any region in R2) the coordinates ( ¯ x, ¯ y) of the center of mass of R are given by
My
Mx
¯ x =
and
¯ y =
,
(3.28)
M
M
where
My =
xδ( x, y) d A ,
Mx =
yδ( x, y) d A ,
M =
δ( x, y) d A ,
(3.29)
R
R
R
The quantities Mx and My are called the moments (or first moments) of the region R about
the x-axis and y-axis, respectively. The quantity M is the mass of the region R. To see this,
think of taking a small rectangle inside R with dimensions ∆ x and ∆ y close to 0. The mass
of that rectangle is approximately δ( x∗, y∗)∆ x∆ y, for some point ( x∗, y∗) in that rectangle.
Then the mass of R is the limit of the sums of the masses of all such rectangles inside R as
the diagonals of the rectangles approach 0, which is the double integral
δ( x, y) d A.
R
Note that the formulas in (3.27) represent a special case when δ( x, y) = 1 throughout R in the formulas in (3.29).
Example 3.13. Find the center of mass of the region R = {( x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 x 2}, if the density function at ( x, y) is δ( x, y) = x + y.
3.6 Application: Center of Mass
125
y
Solution: The region R is shown in Figure 3.6.2. We have
y = 2 x 2
M =
δ( x, y) d A
R
R
1
2 x 2
x
=
( x + y) d y dx
0
1
0
0
1
y=2 x 2
Figure 3.6.2
y 2
=
x y +
dx
0
2 y=0
1
=
(2 x 3 + 2 x 4) dx
0
x 4
2 x 5 1
9
=
+
=
2
5
10
0
and
Mx =
yδ( x, y) d A
My =
xδ( x, y) d A
R
R
1
2 x 2
1
2 x 2
=
y( x + y) d y dx
=
x( x + y) d y dx
0
0
0
0
1
x y 2
y 3 y=2 x 2
1
x y 2 y=2 x 2
=
+
dx
=
x 2 y +
dx
0
2
3
2
y=0
0
y=0
1
8 x 6
1
=
(2 x 5 +
) dx
=
(2 x 4 + 2 x 5) dx
0
3
0
x 6
8 x 7 1
5
2 x 5
x 6 1
11
=
+
=
=
+
=
,
3
21
7
5
3
15
0
0
so the center of mass ( ¯ x, ¯ y) is given by
My
11/15
22
Mx
5/7
50
¯ x =
=
=
,
¯ y =
=
=
.
M
9/10
27
M
9/10
63
Note how this center of mass is a little further towards the upper corner of the region R than
when the density is uniform (it is easy to use the formulas in (3.27) to show that ( ¯ x, ¯ y) = 3 , 3
4 5
in that case). This makes sense since the density function δ( x, y) = x + y increases as ( x, y)
approaches that upper corner, where there is quite a bit of area.
In the special case where the density function δ( x, y) is a constant function on the region
R, the center of mass ( ¯ x, ¯ y) is called the centroid of R.
126
CHAPTER 3. MULTIPLE INTEGRALS
The formulas for the center of mass of a region in R2 can be generalized to a solid S in R3.
Let S be a solid with a continuous mass density function δ( x, y, z) at any point ( x, y, z) in S.
Then the center of mass of S has coordinates ( ¯ x, ¯ y, ¯ z), where
Myz
Mxz
Mxy
¯ x =
,
¯ y =
,
¯ z =
,
(3.30)
M
M
M
where
Myz =
xδ( x, y, z) dV ,
Mxz =
yδ( x, y, z) dV ,
Mxy =
zδ( x, y, z) dV , (3.31)
S
S
S
M =
δ( x, y, z) dV .
(3.32)
S
In this case, Myz, Mxz and Mxy are called the moments (or first moments) of S around the
yz-plane, xz-plane and x y-plane, respectively. Also, M is the mass of S.
Example 3.14. Find the center of mass of the solid S = {( x, y, z) : z ≥ 0, x 2 + y 2 + z 2 ≤ a 2}, if the density function at ( x, y, z) is δ( x, y, z) = 1.
z
Solution: The solid S is just the upper hemisphere inside the sphere
a
of radius a centered at the origin (see Figure 3.6.3). So since the
density function is a constant and S is symmetric about the z-axis,
( ¯ x, ¯ y, ¯ z)
then it is clear that ¯ x = 0 and ¯ y = 0, so we need only find ¯ z. We have
y
0
a
M =
δ( x, y, z) dV =
1 dV = V olume( S).
x
S
S
Figure 3.6.3
But since the volume of S is half the volume of the sphere of radius
a, which we kno