Andrew Rosen
AT X)
Chapter 13 - Multiple Integration (L
E
13.1 - Double Integration over Rectangular Regions
f
is continuous on a rectangle with
¨
ˆ
Fubini's Theorem: If
b
ˆ
ˆ
d
d
ˆ
c
a
f (x, y) dx dy
a
c
´
then
b
f (x, y) dy dx =
f (x, y) dA =
D
[a, b] × [c, d] = D,
2
−2x − y + 6 dy , then it becomes −2xy − y2 + 6y + C )
2
−1 ´ 2 −27
´ 2 ´ −1
´29 2
y2 −27
dy
→
(9x
+
y)
dx
dy
→
x
+
xy
+
y
dy
→
y
+
2
2
2
2 → −25
0
0 −2
0
−2
Example: (Note - if it's
0
The average value of an integrable function
˜
f¯ =
f
over a region
˜
f (x, y) dA
D
=
Area of D
D
D
is:
f (x, y) dA
˜
dA
D
Remember: If you are looking for the bounds on an integral with respect to
dx,
the lower bound is the lower
function and the upper bound is the higher function. If you are looking for the bounds on an integral with
respect to
dy ,
the lower bound is the left-most function and the upper bound is the right-most function,
assuming that this is on a typical
xy
plane.
13.2 - Double Integrals over General Regions
With integrals over nonrectangular regions, the order of integration cannot be simply switched. A correct
x1 = a, x2 = b, y1 = c,
statement would be the following where the bounds are
¨
ˆ
b
ˆ
f (x, y) dy dx
a
D
ˆ
g(x)
d
ˆ
h(y)
f (x, y) dA =
f (x, y) dx dy
c
D
y2 = d:
h(x)
f (x, y) dA =
¨
and
g(y)
1
How to nd the bounds of a double integral :
1. Determine with which variable the inner integral is with respect to:
dx
or
dy
2. The bounding curves determine the limits of integration for the variable determined in the rst step
3. The bounds of the remaining variable is the projection of the region on that axis
1 It
is important to be able to identify the standard quadric surfaces
1
Example: Double integral with
dx
on the inside and
dy
Compute the iterated integral of the region bounded by
and
dy
on the outside
y = 14 − x, y = 3,
and
x=4
with
dx
on the inside
and
x=4
with
dy
on the inside
on the outside
Step 1: Plot the region
Step 2: Determine the bounds for the
dx
integral
dy
integral
4 ≤ x ≤ −y + 14
Step 3: Determine the bounds for the
3 ≤ y ≤ 10
Step 4: Write the iterated integral:
ˆ
10
ˆ
−y+14
f (x, y) dx dy
3
Example: Double integral with
dy
4
on the inside and
dx
Compute the iterated integral of the region bounded by
and
dx
on the outside
y = 14 − x, y = 3,
on the outside
Step 1: Plot the region (see previous plot)
Step 2: Determine the bounds for the
dy
integral
dx
integral
3 ≤ y ≤ 14 − x
Step 3: Determine the bounds for the
4 ≤ x ≤ 11
Step 4: Write the iterated integral:
ˆ
11
ˆ
14−x
f (x, y) dy dx
4
3
g(x, y) ≥ f (x, y) is:
¨
V =
(g(x, y) − f (x, y)) dA
The volume between two surfaces where
D
If
D
is a region in the
xy
plane then the area of that region is:
¨
AreaD =
dA
D
2
2
13.3 - Double Integrals in Polar Coordinates
Before doing double integration in polar coordinates, it is essential to recall the following identities:
p
r=
x2 + y 2 → r 2 = x2 + y 2
x = r cos(θ), y = r sin(θ)
As long as
β − α ≤ 2π
while
α≤θ≤β
polar rectangular region for a function,
and 0 ≤ a ≤ r ≤ b,
f (x, y):
¨
ˆ
ˆ
θ=β
then the following is the double integral over a
r=b
f [r cos(θ), r sin(θ)] r dr dθ
f (r, θ) dA =
r=a
θ=α
D
With double integrals over non-rectangular polar regions, the order of integration cannot be switched directly.
A more general expression for the double integral over a general polar rectangular region where
r ≤ h(θ)
and
β − α ≤ 2π
with
α≤θ≤β
¨
f (x, y),
for a function,
ˆ
θ=β
ˆ
0 ≤ g(θ) ≤
is:
r=h(θ)
f (r, θ) dA =
f [r cos(θ), r sin(θ)] r dr dθ
θ=α
D
r=g(θ)
Note: Don't forget to multiply the integrand by a factor of
r!
Area of Polar Regions:
¨
A=
ˆ
θ=β
ˆ
r=h(θ)
dA =
r dr dθ
θ=a
D
r=g(θ)
The following trigonometric identities are crucial to integration of squared trigonometric functions:
2 It
sin2 (θ) =
1
(1 − cos(2θ))
2
cos2 (θ) =
1
(1 + cos(2θ))
2
is essential to be able to recall the typical trigonometric values for sine, cosine, and tangent
3
Example:
Set up an equation to nd the area of 1 leaf of the rose,
r = cos(2θ)
1) Plot the function:
2) Write the bounds
0 ≤ r ≤ cos(2θ)
and
− π4 ≤ θ ≤
π
4
Note: When writing the bounds of integration for
θ, make sure that it is going from a lower value to a higher
value and make sure it is the correct region of the boundary. In this case,
π
7π
4 to 4 would be incorrect since
it'd sweep more than one leaf.
3) Set up double integral
ˆ
ˆ
π
4
cos(2θ)
r dr dθ
A=
−π
4
0
ˆ
h(x)
13.4 - Triple Integrals
˚
ˆ
b
ˆ
H(x,y)
f (x, y, z) dV =
f (x, y, z) dz dy dx
a
D
g(x)
G(x,y)
Note: Five other orders of integration could be set up based on Fubini's Theorem
How to Find the Bounds of Integration:
1. Imagine that you really, really hate calculus, and triple integrals make you want to stab things. This
3
attitude will help you greatly!
2. Your rst set of limits can be gured out if you imagine (fatally) stabbing the three-dimensional
boundary in the direction of the axis of the variable you're integrating. Find where the knife (or spear,
rusty lance, extra sharp pencil, etc.)
enters the three-dimensional boundary for the rst time and
where it exits. Basically, nd the entrance and exit wounds. Write this as an inequality (eg:
where the integral is with respect to
a ≤ z ≤ b,
dz ).
3. Set the rst variable of integration to zero (so, if you integrated with respect to z rst, make
z = 0)
to
create a plane. It might even be helpful to draw a new graph in a normal 2-variable Cartesian plane.
Now stab in the direction of the middle variable of integration. Find the entrance and exit wounds.
4. Now you are left with one variable to nd the limits of integration for. The bounds of the remaining
variable is the projection of the region on that axis.
3 Don't
worry. I'm not really that violent.
4
How to switch the bounds of a triple integral (and how to graph a 3D function given the bounds):
1. Since the middle and outer integrals' bounds are the projection of the 3D region, plot this 2D projection
rst. Be careful of this major fact: If you have
˝
dx dy dz ,
you'd graph a
yz
projection rst. If one
D
of the bounds for, let's say,
y
is something like
equals and substitute that in for
y=z
then you can't graph that directly! See what
z
y = z.
2. From here, the bounds for the middle integral can be found as usual. The bounds for the outer integral
are the projection of the 2D graph on the axis of the variable that the outer integral is with respect to
(make sure that the outer integral has constants for the bounds)
3. Now, extrude this projection in the dimension of the inner integral variable. The bounds for the inner
integral can then be found as normal.
4. The key is to realize which bounds are functions of which variables and to adjust them accordingly.
5
Example:
Set up a triple integral to nd the volume of the given solid region in the rst octant bounded by the plane
12x + 16y + 12z = 48
and the coordinate planes.
Given graph of the boundary:
Step 1: Pick an order of integration. I will pick
dz dy dx
Step 2: Imagine stabbing this boundary in the
z
plane solved for
Step 3: Set
z,
z=0
which is
4y
4−x− .
3
arbitrarily. Any order can be used.
z =0
4y
0≤z ≤4−x−
3
axis. It enters at
Therefore,
and exits at the equation of the
to create a new Cartesian plane. Re-plotting the boundary might be helpful. The graph
is shown below.
Step 4: Find the boundaries for
point.
y.
The hypothetical knife would enter at
There are two easy ways to do this.
One is to simply solve for
y
y = 0. Now
z = 0.
when
you need the exit
This would make
3
y = − x + 3. Another way to do this, which is equally as easy, is to recognize that the slope of a line is
4
∆y
3
the
, which is − . This, in conjunction with the knowledge that the y intercept is (0, 3) can yield the
∆x
4
3
3
equation of the line, which is − x + 3. Therefore, 0 ≤ y ≤ − x + 3.
4
4
Step 5: Find the remaining boundary for x. The projection of the previous boundary on the x axis yields
0 ≤ x ≤ 4.
Step 6: Use this information to piece together a triple integral. It would thus be:
ˆ
4
ˆ
− 3x
4 +3
ˆ
4−x− 4y
3
dx dy dz
0
0
0
6
13.5, Part I - Triple Integrals in Cylindrical Coordinates
The triple integral of
f
over
D
in cylindrical coordinates is:
˚
ˆ
β
ˆ
h(θ)
ˆ
H(r cos(θ),r sin(θ))
f (r, θ, z) dz r dr dθ
f (x, y, z) dV =
D
G(r cos(θ),r sin(θ))
g(θ)
α
The following are important conversion rules between rectangular and cylindrical coordinates:
1.
x = r cos(θ)
2.
y = r sin(θ)
3.
z=z
Example:
Set up, but do not evaluate,
´ 2 ´ √4−x2 ´√
2
√
2
0
− 4−x
x2 +y 2
(x2 + y 2 ) dz dy dx
Step 1: Plot the boundary region. The tip of the cone is at
Step 2: Find the boundaries of
z
in cylindrical coordinates.
(0, 0, 0)
and opens from
z=0
to
z=2
by converting to cylindrical coordinates. The boundaries become
Also note that the integrand itself becomes
Step 3: Collapse the cone so that
z=0
r ≤ z ≤ 2.
r2 .
and plot it (indicated by shaded region)
Remember to check the boundaries given
Step 4: Find the boundaries for
r
and
θ.
With this graph, it is clear that
from a smaller angle to a higher angle for the boundary, so
− π2 ≤ θ ≤
0 ≤ r ≤ 2.
The angle must sweep
π
2.
Step 5: Set up the triple integral. It would become, with the boundaries already determined:
ˆ
π
2
−π
2
ˆ
2
ˆ
2
r2 r dz dr dθ
0
r
13.5, Part II - Triple Integrals in Spherical Coordinates
There are three new coordinates to recognize in spherical coordinates:
1.
ρ
is the distance from the origin to a point, P
2.
φ
is the angle between the positive z-axis and an arbitrary line
3.
θ
is the same angle as in cylindrical coordinates and measures rotation about the z-axis relative to the
OP
that goes from
0
to
π4
positive x-axis
4φ = 0
is a line from the origin going upward vertically and
φ=π
7
is a line from the origin going downward vertically
The triple integral of
f
D,
over a region,
in spherical coordinates is:
˚
ˆ
β
ˆ
ˆ
b
h(φ,θ)
f (ρ, φ, θ)ρ2 sin φ dρ dφ dθ
f (ρ, φ, θ) dV =
D
g(φ,θ)
a
α
The following are important conversion rules between rectangular and spherical coordinates:
1.
x = ρ sin φ cos θ
2.
y = ρ sin φ sin θ
3.
Also
z = ρ cos φ
√
note: ρ =
r2 + z 2
ρ2 = x2 + y 2 + z 2 = r2 + z 2
and
Example:
˝
Set up, but do not evaluate, the following integral in spherical coordinates:
e−(4x
2
3
+4y 2 +4z 2 ) 2
dV
where
D
D
is a sphere of radius
6.
Step 1: Convert known functions to spherical coordinates. The triple integral becomes
˝
e−4r
2
23
dV
D
Step 2: Find the bounds for
ρ.
Step 3: Find the bounds for
φ. The bounds for this will be from 0 to π . The
θ take care of that part of the sweeping action
2π
is because the bounds of
Step 4: Find the bounds for
θ.
This will extend from
0
to the sphere, which is
6
reason that it is not from
0
The bounds for this will be one full revolution from
to
0
to
2π
Step 5: Set up the triple integral. It will be:
ˆ
2π
ˆ
π
ˆ
6
3
e−8ρ ρ2 sin φ dρ dφ dθ
0
0
0
Example:
Set up, but do not evaluate, an expression for the volume of the smaller region cut from the solid sphere
ρ ≤ 14
by the plane
z = 7.
ρ.
Step 1: Find the bounds for
z = ρ cos φ,
Step 2: Find the bounds for
7
cos φ
An equation for a plane is given by
so the lower bound for
equal to
14
φ.
Therefore, the lower bound is
is
6 sec φ,
a sec φ
14.
because
z=7
is the plane and
and the upper bound is
The lower limit is
and solving for
Step 3: Find the bounds for
ρ
φ = 0,
and the upper limit of
π
3
can be found by setting
φ.
θ. These are simply those
0 and upper bound is 2π .
of a circle in order to create one full revolution.
Step 4: Write out the triple integral.
ˆ
2π
ˆ
π/3
ˆ
14
ρ2 sin φ dρ dφ dθ
0
0
7/ cos φ
8
9