Math 2443, Section 16.3
Review
These notes will supplement (not replace) the lectures based on Section 16.3
Section 16.3
(i) Double integrals over general regions: We defined double integrals over rectangles in the last section. How does one define the double integral of a function f
over a general region D? Choose an arbitrary rectangle R which contains D. Define
a new function F on R as follows.
F (x, y) = f (x, y), on D
= 0, outside D.
The function F is the same is f on D and zero outside D. Hence, it doesn’t matter
which rectangle we choose. If F is integrable over R, we can define, for a general
region D,
ZZ
ZZ
f (x, y)dA =
F (x, y)dA.
D
R
If the region D is “nice”, the integral on the right-hand side will exist.
(ii) Type I and II regions: If a given region D is of a specific type, it simplifies the
double integral considerably.
(a) A planar region DI is of Type I if it lies between the graphs of two continuous
functions of x. That is,
DI = {(x, y)| a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}
where g1 and g2 are continuous on [a, b].
RR
To find DI f (x, y)dA, we choose a rectangle R = [a, b] × [c, d] which contains
DI . Then,
Z bZ d
ZZ
ZZ
F (x, y)dydx
F (x, y)dA =
f (x, y)dA =
DI
a
R
c
Since F (x, y) = 0 for y ≥ g2 (x) as well as y ≤ g1 (x), and F (x, y) = f (x, y)
when g1 (x) ≤ y ≤ g2 (x). Thus,
Z
d
F (x, y)dy =
Z
g2 (x)
f (x, y)dydx.
g1 (x)
c
This gives us
ZZ
f (x, y)dA =
DI
Z bZ
a
1
g2 (x)
g1 (x)
f (x, y)dydx.
2
Review
(b) A planar region DII is of type II if it lies between the graphs of two continuous
functions of y. That is,
DII = {(x, y)| c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)}
where h1 and h2 are continuous on [c, d].
Similar to above, we get
ZZ
f (x, y)dA =
DII
Z
c
d Z h2 (y)
f (x, y)dxdy.
h1 (y)
Remark: Remember that in both the instances, the limits of the inner integral
are functions of one variable but the integration is done with the opposite with
respect to the other variable.
0.1 Example: Evaluate the double integral
ZZ
x cos y dA, D bounded by y = 0, y = x2 , x = 1.
D
Solution: We can write D as
D = {(x, y)| 0 ≤ x ≤ 1, 0 ≤ y ≤ x2 }.
I have not drawn the region but you must do so. Clearly, D is Type I region and
thus
Z 1
Z 1 Z x2
ZZ
2
(x sin y]x0 dx
x cos y dydx =
x cos ydA =
0
0
0
D
Z 1
1 1
1
=
x sin x2 dx =
− cos x2 0 = (1 − cos 1)
2
2
0
0.2 Example: Evaluate the double integral
ZZ
y 2 exy dA, D = {(x, y)| 0 ≤ y ≤ 4, 0 ≤ x ≤ y}.
D
Solution: The region D is a Type II region, hence
y
Z 4 2
Z 4Z y
ZZ
y xy
y 2 exy dx dy =
y 2 exy dA =
dy
e
y
0
0
0
D
0
4
Z 4
1 y2 y 2
1 16
y2
(ye − y)dy =
=
e − 17 .
e −
=
2
2 0 2
0
0.3 Example: Find the volume of the solid under the surface z = xy and above
the triangle with vertices (1, 1), (4, 1) and (1, 2).
Solution: The triangle D with the given points as vertices can be seen to be a Type
I region as follows. In the triangle, we see that x lies between x = 1 and x = 4 while
3
Math 2443, Section 16.3
the y-component lies between y = 1 and the line joining the points (1, 2) and (4, 1),
which has the equation y = − 13 x + 73 . We thus write
D = {(x, y)| 1 ≤ x ≤ 4, 1 ≤ y ≤ −
x 7
+ }.
3 3
Again, I have not drawn the region but you have to do so. Note that the function
f (x, y) = xy is always non-negative on D. Hence, the volume V of the solid under
the surface z = xy and above the given triangle D is
V =
ZZ
xy dA =
Z
1
D
4 Z − x3 + 37
xy dydx
1
− x3 + 37
!
x 7 2
− x dx
x − +
=
3 3
1
1
1
4
Z 4
1
1 x4 14x3
3
2
2
=
x − 14x + 40x dx =
−
+ 20x
18 1
18 4
3
1
1 1
14
31
=
(256 − 1) − (64 − 1) + 20(16 − 1) = .
18 4
3
8
Z
4
xy 2
2
1
dx =
2
Z
4
0.4 Example: Find the volume of the tetrahedron bounded by the coordinate
planes and the plane 3x + 2y + z = 6.
Solution: The coordinate planes are given by x = 0, y = 0 and z = 0. Let’s find
the vertices of the tetrahedron first. These will be the intersections of three of the
four planes given. The intersection of x = 0, y = 0 and 3x + 2y + z = 6 is (0, 0, 6).
Similarly, the other three vertices are (2, 0, 0), (0, 3, 0) and (0, 0, 0). Now, the given
tetrahedron is a solid that lies above the triangle D in the xy plane that has vertices
(0, 0), (2, 0) and (0, 3). Please draw this triangle and the tetrahedron, even though
I have not done so. The line joining (0, 3) and (2, 0) is given by y = − 23 x + 3. We
can write
3
D = {(x, y)| 0 ≤ x ≤ 2, 0 ≤ y ≤ − + 3},
2
which is a Type I region. Now, the volume V of the tetrahedron is the double
integral of the function 6 − 3x − 2y over D.
V =
=
Z
Z
(6 − 3x − 2y)dA =
2
0
6y − 3xy − y 2
Z
0
2 Z − 32 x+3
− 3 x+3
2
0
0
(6 − 3x − 2y)dy dx
dx
2 !
3
3
3
dx
6 − x + 3 − 3x − x + 3 − − x + 3
=
2
2
2
0
Z 2
9 2
9 2
−9x + 18 + x − 9x − x + 9x − 9 dx
=
2
4
0
2
Z 2
9 2
3 3 9 2
=
x − 9x + 9 dx =
x − x + 9x = 6.
4
4
2
0
0
Z
2
4
Review
0.5 Example: Find the volume of the solid enclosed by the cylinders z = x2 , y = x2
and the planes z = 0, y = 4.
Solution: In these type of problems, it is not necessary to draw the given solid.
You need to find the region in the xy plane above which the solid is located. In this
case, we know that the solid lies above the region D bounded by the line y = 4 and
the parabolay = x2 in the xy plane. In this region, x lies between −2 and 2. Hence
D = {(x, y)| − 2 ≤ x ≤ 2, x2 ≤ y ≤ 4},
which is a Type I region. Hence, the volume of the solid is the double integral over
D of the function f (x, y) = x2 .
V =
=
ZZ
Z
2
x2 dA =
=
2
−2
x2 y
−2
Z
4
Z
4 3 1 5
x − x
3
5
−2
x2 dydx
x2
dx =
x2
2
4
Z
2
−2
=
(4x2 − x4 )dx
128
.
15
(iii) Reversing the order of integration: For double integrals on general regions, we
cannot simply switch the order of integration because the Fubini Theorem doesn’t
directly apply on such regions. Nonetheless, it is possible that a given double integral
is difficult to solve and we’d like to reverse the order of integration. For this, we first
find the planar region D on which the double integral is being evaluated. Next, if
D is originally a Type I region, we can switch the order of integration by expressing
it as a Type II region and evaluating the integral.
0.6 Example: Evaluate the integral by reversing the order of integration.
Z 1Z 3
2
ex dx dy.
0
3y
Solution: We can see that the inner integral cannot be solved by elementary methods. In order to reverse the order of integration we first find the region on which
the integral is being evaluated. Looking at the limits on x and y, we can express
this region D as
D = {(x, y)| 0 ≤ y ≤ 1, 3 ≤ x ≤ 3y}.
This is a Type II region. It is given by the triangle in the xy plane which is bounded
by the lines y = x3 , y = 0 and x = 3. I have not drawn the region but you have to
do so. Hence,
ZZ
Z 1Z 3
2
2
ex dA.
ex dx dy =
0
3y
D
Now, to switch the order of integration, we express D as a Type I region. To do
this we must express point (x, y) on D such that x lies between two real numbers
5
Math 2443, Section 16.3
and y is between two continuous functions of x. It is clear that 0 ≤ x ≤ 3 and also
0 ≤ y ≤ x3 . That is,
D = {(x, y)|0 ≤ x ≤ 3, 0 ≤ y ≤
x
}.
3
Hence,
Z
0
1Z 3
x2
e dx dy =
3y
=
=
ZZ
Z
0
1
3
x2
e dA =
D
3
Z
0
Z
0
x
ye
9
i
2 x/3
0
3 Z x/3
2
ex dydx
0
dx =
Z
3
0
1 x2
xe dx
3
1 u
1 9
e − 1).
e du =
2
6
where we have used the substitution u = x2 in the last step.
0.7 Example: Evaluate the integral by reversing the order of integration.
Z 4Z 2
1
dydx.
√ y3 + 1
0
x
Solution: The double integral is taken over the region
√
D = {(x, y)|0 ≤ x ≤ 4, x ≤ y ≤ 2},
√
which is a Type I region and is represented by the region above the curve y = x,
below the line y = 2 and between x = 0 and x = 4. Now, we can express this region
as a Type II region as follows
D = {(x, y)| 0 ≤ y ≤ 2, 0 ≤ x ≤ y 2 }.
Therefore,
Z
0
4Z 2
√
x
Z 2 Z y2
1
1
dA
=
dxdy
3
3
y +1
0
0
D y +1
y2
Z 2
Z 2
y2
x
dy
=
dy
=
3
y3 + 1 0
0 y +1
0
Z 9
11
1
=
du = ln 9.
3
1 3u
1
dydx =
3
y +1
ZZ
where we have used the substitution u = y 3 + 1 in the last step.
(iv) If a region D is the union of two regions D1 and D2 . That is, D = D1 ∪ D2 then the
double integral over D is simply the sum of the double integrals over D1 and D2 .
ZZ
ZZ
ZZ
f (x, y)dA.
f (x, y)dA +
f (x, y)dA =
D
D1
D2
Sometimes you won’t be able to express a given region as Type I or Type II. I such
a case, see if you can write it as the union of two or more such regions.