Vector Calculus by Michael Corral - HTML preview

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