Transcript Chapter 17

Chapter 17
Section 17.3
The Evaluation of Double Integrals by
Repeated Integrals
Double Integrals
A double integral of a function of
2 variables 𝑤 𝑥, 𝑦 requires a
region Ω in the xy-plane and an
area unit 𝑑𝐴. Area units for 2
variables come in two different
types.
The Double Integral:
y
𝑤 𝑥, 𝑦 𝑑𝐴
Ω
Ω
x
Where 𝑑𝐴 =
𝑑𝑥 𝑑𝑦
𝑑𝑦 𝑑𝑥
The area unit that is used determine 2 things that are required to evaluate the integral.
1. The limits that are placed on each integral.
2. The order that you integrate each variable in from the inside integral to the outside.
The 𝑑𝑦 𝑑𝑥 Area unit for 𝑑𝐴
In this case the region Ω is
bounded by top and bottom
curves 𝑓 𝑥 and 𝑔 𝑥 and left
and right end points a and b.
y
The 𝑑𝑥 𝑑𝑦 Area unit for 𝑑𝐴
In this case the region Ω is
bounded by left and right curves
ℎ 𝑦 and 𝑘 𝑦 and top and
bottom end points c and d.
y
𝑓 𝑥
Ω
a
𝑤 𝑥, 𝑦 𝑑𝐴 =
𝑏
𝑔 𝑥
b
x
Ω
𝑤 𝑥, 𝑦 𝑑𝑦 𝑑𝑥
𝑎
𝑔 𝑥
𝑤 𝑥, 𝑦 𝑑𝐴 =
d
ℎ 𝑦
c
𝑓 𝑥
Ω
𝑘 𝑦
𝑑
x
Ω
𝑘 𝑦
𝑤 𝑥, 𝑦 𝑑𝑥 𝑑𝑦
𝑐
ℎ 𝑦
Example
Set up and evaluate double integrals using both types of area units to
evaluate the integral to the right. Where Ω is the region bounded by 𝑦 = 𝑥 2
and 𝑦 = 2𝑥.
Begin by graphing the region Ω.
12𝑥𝑦 2 𝑑𝐴
Ω
For other integral solve each equation for x.
y
y
Find where curves
intersect by solving.
2𝑥 = 𝑥 2
0 = 𝑥 2 − 2𝑥
0=𝑥 𝑥−2
𝑥 = 0 and 𝑥 = 2
𝑦 = 2𝑥
𝑦 = 𝑥2
x
2
0
1st Integral
12𝑥𝑦 2 𝑑𝐴
2
=
0
Ω
2
=
0
2
=
4𝑥𝑦 3
32𝑥 4
2𝑥
𝑥
−
𝑑𝑥
2
4𝑥 7 𝑑𝑥
0
5 − 1𝑥 8 2
= 32
𝑥
5
2
0
1024
= 5 − 128 = 384
5
2𝑥
𝑥2
12𝑥𝑦 2 𝑑𝑦𝑑𝑥
Treat y as variable.
Plug in functions
for y variable.
Evaluate like an
ordinary integral.
4
Find projection on the
y-axis by plugging in.
𝑦 = 2𝑥
𝑥 = 𝑦2
If 𝑥 = 0 then 𝑦 = 0
𝑦 = 𝑥2
𝑥= 𝑦
0
If 𝑥 = 2 then 𝑦 = 4
x
2
0
1st Integral
12𝑥𝑦 2 𝑑𝐴 =
Ω
4
0
4
=
0
𝑦
𝑦
2
12𝑥𝑦 2 𝑑𝑥𝑑𝑦
𝑑𝑦
Treat x as variable.
6𝑦 3 − 32𝑦 4 𝑑𝑦
Plug in functions
for x variable.
0
4
=
6𝑥 2 𝑦 2
𝑦
𝑦
2
3 5
= 32𝑦 4 − 10
𝑦
4
0
384
= 384 − 1535
5 = 5
Evaluate like an
ordinary integral.
The order that you integrate a double integral in will not matter (i.e. give the same answer) as
long as you have the correct corresponding limits of integration:
𝑑
𝑘 𝑦
𝑏
𝑤 𝑥, 𝑦 𝑑𝑥 𝑑𝑦 =
𝑐
ℎ 𝑦
𝑓 𝑥
𝑤 𝑥, 𝑦 𝑑𝐴 =
𝑤 𝑥, 𝑦 𝑑𝑦 𝑑𝑥
𝑎
Ω
𝑔 𝑥
To determine which is better to use graph the region Ω and see if it has a single function that
forms it top, bottom, right and left boundaries. If a boundary is not a single function the
double will need to be split into more than one double integral.
y
y
y
Ω2
Ω1
Ω3
x
x
x
Use either:
𝑑𝑥 𝑑𝑦 or 𝑑𝑦 𝑑𝑥
Use:
top, bottom, left
and right are all
one function
top and bottom are
different functions
Use:
𝑑𝑥 𝑑𝑦
𝑑𝑦 𝑑𝑥
left and right are
different functions
Example
Set up a single double integral and evaluate for the integral to the right
where Ω is the region bounded by: 𝑦 = 𝑥, 𝑦 = −𝑥, and 𝑦 = 1.
Start by graphing Ω.
𝑦=1
0.8
𝑥 = −𝑦
0.6
𝑦= 𝑥
0.4
𝑥 = 𝑦2
Ω
This region does not always have the
same function as its bottom boundary.
Sometimes it is a line and some times a
parabola. To have one double integral we
will need to use 𝑑𝑥 𝑑𝑦.
1.0
𝑦 = −𝑥
24𝑥 2 𝑦𝑑𝐴
0.2
To do this solve to get x as a function of y.
1.0
0.5
24𝑥 2 𝑦𝑑𝐴
0.5
1
𝑦2
=
0
Ω
24𝑥 2 𝑦𝑑𝑥
1
𝑑𝑦 =
−𝑦
1
=
8𝑦 7
0
−
−8𝑦 4 𝑑𝑦
0
=
1.0
𝑦8
1
=
0
+
1
8 5
5𝑦 0
= 1 + 85 = 13
5
8𝑥 3 𝑦
𝑦2
−𝑦
𝑑𝑦
8𝑦 7 + 8𝑦 4 𝑑𝑦
Example
Set up double integrals and evaluate them for the integral to the right
24𝑥 2 𝑦𝑑𝑦 𝑑𝑥
where Ω is the region bounded by: 𝑦 = 𝑥, 𝑦 = −𝑥, and 𝑦 = 1.
Ω
𝑦=1
1.0
To be able to do this in this order we
0.8
must split this into two regions Ω1
Ω2
Ω1
and Ω2 which always have the same
0.6
𝑦 = −𝑥
𝑦= 𝑥
top and bottom curves.
0.4
0.2
1.0
0.5
0.5
24𝑥 2 𝑦𝑑𝑦𝑑𝑥
0
=
24𝑥 2 𝑦𝑑𝑦
𝑑𝑥
−1 −𝑥
Ω1
0
=
12𝑥 2 𝑦 2
−1
=
1
1.0
4𝑥 3
−
1
−𝑥
0
12 5
5 𝑥 −1
0
𝑑𝑥 =
24𝑥 2 𝑦𝑑𝑦𝑑𝑥 =
0
Ω2
1
12𝑥 2 − 12𝑥 4 𝑑𝑥
=
= 0 − (−4 +
=
8
5
2 2
12𝑥 𝑦
=
4𝑥 3
𝑥
−
3𝑥 4
8
1
0
1
24𝑥 2 𝑦𝑑𝑦 𝑑𝑥
𝑥
1
1
0
−1
12
5)
1
𝑑𝑥 =
12𝑥 2 − 12𝑥 3 𝑑𝑥
0
=4−3=1
Adding together the values of these two double integrals we get 5 + 1 =
13
5
which is the same.
Sometimes reversing the order of integration will give a much easier double integral to
evaluate. This means you will need to graph the region Ω from the limits of the integral.
Example
Reverse the order of integration for the double integral to
the right and evaluate it.
1
0
1
𝑥2
𝑥3
𝑥4
+
𝑦2
1.0
Begin by graphing the region Ω is bounded by the
curves 𝑦 = 1 and 𝑦 = 𝑥 2 for x between 0 and 1.
0.8
𝑦 = 𝑥2
0.6
0.4
Solve each equation for x in order to be able to find
the new limits of integration.
1
0
1
𝑥2
𝑥3
𝑥4 + 𝑦2
Let 𝑢 = 𝑥 4 + 𝑦 2
so, 𝑑𝑢 = 4𝑥 3 𝑑𝑥
1
𝑦
𝑥3
𝑑𝑦 𝑑𝑥 =
0
𝑥4 + 𝑦2
0
1
=
0
1
=
0
1
=
0
2𝑦 2
𝑦2
𝑢
2
1
4 𝑢
𝑑𝑥 𝑑𝑦
𝑑𝑢 𝑑𝑦
0.2
1
=
0
1
=
0
=
𝑦2
𝑥4 + 𝑦2
2
0
=
0.4
0.6
0.8
1.0
2𝑦 𝑦
− 𝑑𝑦
2
2
2−1
𝑦 𝑑𝑦
2
2−1
2
𝑦
𝑑𝑦
𝑥= 𝑦
0.2
2𝑦 2
𝑑𝑦
𝑑𝑦 𝑑𝑥
2−1
2
1
𝑦 𝑑𝑦
0
𝑦2
2
1
=
0
2−1
4
Geometric Interpretation of Integrals
If 𝑦 = 𝑓 𝑥 where 𝑓 𝑥 > 0 (i.e. the
curve is above the x-axis), then the
value of the integral is the area under
the curve.
For the double integral of a
positive surface 𝑧 = 𝑓 𝑥, 𝑦
this is the volume under the
surface 𝑧 = 𝑓 𝑥, 𝑦 and
above the region Ω in the xyplane.
5
Area under curve 𝑦 = 𝑓 𝑥 and
above x-axis between a and b:
4
3
𝑏
2
𝑓 𝑥 𝑑𝑥
1
𝑎
1
1
2
3
4
5
1
z
Volume under the surface 𝑧 =
𝑓 𝑥, 𝑦 and above the region Ω in
the xy-plane:
𝑓 𝑥, 𝑦 𝑑𝐴
y
Ω
Ω
x
y
To find the volume under a surface 𝑧 = 𝑓 𝑥, 𝑦 you need to
draw the graph of how the volume you are interested in
finding will project into the xy-plane and this becomes your
region of integration Ω.
Ω
x
Example
Set up and evaluate a double integral to find the volume of the solid in the 1st octant bounded
by the three planes 𝑥 = 0, 𝑦 = 0, and 𝑧 = 0 along with the parabolic cylinder 𝑦 = 4 − 𝑥 2 and
z
the plane 𝑧 = 8 − 2𝑥
y
𝑧 = 8 − 2𝑥
Begin by drawing the
solid and its projection
in the xy-plane.
𝑦 = 4 − 𝑥2
y
Ω
4−𝑥 2
8 − 2𝑥 𝑑𝐴 =
8 − 2𝑥 𝑑𝑦 𝑑𝑥
0 0
2
Ω
=
8𝑦 − 2𝑥𝑦
0
2
=
4−𝑥 2
0
𝑑𝑥
8 4 − 𝑥 2 − 2𝑥 4 − 𝑥 2 𝑑𝑥
0
2
32 − 8𝑥 2 − 8𝑥 + 2𝑥 3 𝑑𝑥
=
0
= 32𝑥 −
8 3
3𝑥
𝑦 = 4 − 𝑥2
Ω
2
x
2
Projection in
the xy-plane.
−
4𝑥 2
+
2
1 4
2𝑥 0
64
104
= 64 − 64
3 − 16 + 8 = 56 − 3 = 3
x
Integrals as Limits
Area under curve 𝑦 = 𝑓 𝑥 and
A 1 variable integral is the limit of a
above x-axis between a and b:
𝑛
Riemann Sum. Where 𝑑𝑥 is the limit of
𝑏
the ∆𝑥𝑖 and the integral is the limit of
𝑓 𝑥 𝑑𝑥 = lim
𝑓 𝑥𝑖∗ ∆𝑥𝑖
𝑛→∞
the sum of the heights of the rectangles. 𝑎
𝑖=1
𝑓 𝑥𝑖∗
a
∆𝑥𝑖∗
b
z
For the double integral we cut the
region Ω into a grid and look at
what happens when we take the
limit as both n and m go to
infinity, i.e. the mesh of the grid
gets smaller and smaller.
𝑓 𝑥𝑖∗ , 𝑦𝑗∗
Ω
y
∆𝑦𝑗
x
∆𝑥𝑖
y
Volume under the surface 𝑧 = 𝑓 𝑥, 𝑦 and above the region
Ω in the xy-plane:
𝑚
𝑛
𝑓 𝑥𝑖∗ , 𝑦𝑗∗ ∆𝑥𝑖 ∆𝑦𝑗
𝑓 𝑥, 𝑦 𝑑𝐴 = lim lim
Ω
𝑛→∞ 𝑚→∞
Ω
𝑖=1 𝑗=1
x
5
The area between two curves 𝑦
= 𝑓 𝑥 and 𝑦 = 𝑔 𝑥 can be
found by integrating the
difference of the functions.
Area between curves:
𝑦=𝑓 𝑥
4
3
𝑏
2
𝑓 𝑥 − 𝑔 𝑥 𝑑𝑥
1
𝑎
𝑦=𝑔 𝑥
a
b
1
z
𝑧 = 𝑓 𝑥, 𝑦
The volume between two
surfaces 𝑧 = 𝑓 𝑥, 𝑦 and 𝑧 =
𝑔 𝑥, 𝑦 can be found by the
double integral over the
projected region in the xyplane.
Volume between the
surfaces over Ω:
𝑓 𝑥, 𝑦 − 𝑔 𝑥, 𝑦 𝑑𝐴
𝑧 = 𝑔 𝑥, 𝑦
y
Ω
Ω
x
Again the region Ω is the projection in the xy-plane of where the
top and bottom surface come together.
Example
Find the volume of the solid that is bounded on the right by the plane 𝑦 = 4, on the left by the
parabolic cylinder 𝑦 = 𝑥 2 , below by the plane 𝑧 = 𝑦, and above by the parabolic cylinder 𝑧 =
𝑦 2 + 3 by setting up and evaluating a double integral.
First draw the solid along with the projection into the xy-plane.
The volume is the double integral:
2
𝑧=𝑦 +3
y
z
𝑡𝑜𝑝 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 − 𝑏𝑜𝑡𝑡𝑜𝑚 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑑𝐴
4
𝑦=4
Ω
Ω
3
𝑧=𝑦
𝑦 = 𝑥2
𝑦=𝑥
𝑦 2 + 3 − 𝑦 𝑑𝐴
=
2
2
4
y
-2
2
=
x
Ω
4
−2 𝑥 2
𝑦 2 + 3 − 𝑦 𝑑𝑦 𝑑𝑥
x
2
=
−2
=
=
1 3
3𝑦
+ 3𝑦 −
4
1 2
2𝑦 𝑥 2
2
𝑑𝑥 =
76
1 6
−
3
3𝑥
−2
−
3𝑥 2
+
1 4
2𝑥 𝑑𝑥
=
76
3𝑥
−
1 7
21𝑥
−
𝑥3
+
2
1 5
10𝑥 −2
152 128
16
−152 −128
−16
304 256
32
−
−8+
−
−
−8+
=
−
− 16 +
3
21
5
3
21
5
3
21
5
10640 − 1280 − 1680 + 672 8352 2784
=
=
105
105
35