Math Camp Notes: Integration II

Double and Triple Integration

In order to evaluate the double integral Z b Z d
f (x, y)dxdy
a c
it is helpful to rewrite the integral as !
Z b Z d
f (x, y)dx dy
a c

and evaluate the inside integral rst. Then the outside integral can be evaluated to obtain the solution.
Example:
Evaulate (x + y)dxdy .
R2R4
1 2
Z 2Z 4 Z 2 Z 4  Z 2   Z 2
1
(x + y)dxdy = (x + y)dx dy = ( x2 + xy)|42 dy = (6 + 2y) dy = 6y + y 2 |21 = 9
1 2 1 2 1 2 1

Triple integrals (and higher order integrals) are evaluated in the same manner as doubles; we rst integrate
with respect to one variable and work our way back to lower order integrals.
! !
Z b Z d Z f Z b Z d Z f
f (x, y, z)dxdydz = f (x, y)dx dy dz
a c e a c e

Order of Integration
The limits of integration will only be scalar if we are integrating over a rectangle (or its equivalent in higher
dimensions). In this case, the simplicity of the order of integration does not matter.
Example:
Recall the double integral (x + y)dxdy above which we evaluated with respect to x rst. Now evaluate
R2R4
1 2
with respect to y rst.
Z 4 Z 2 Z 4 Z 2  Z 4  
1 2 2
(x + y)dydx = (x + y)dy dx = ( y + xy)|1 dy =
2 1 2 1 2 2
Z 4 
3 3 1
= + x dy = y + y 2 |42 = 3 + 6 = 9
2 2 2 2
However, we are integrating with respect to variables that are also in the limits of integration, then dierent
orders of integration inuence the diculty of the problem. To see this, consider the integral of f (x, y)
over the region R = {1 ≤ y ≤ 2, y ≤ x ≤ y 2 }. By drawing the area of integration, we can easily see
that integrating with respect
√ to x rst is much easier.√Notice that the region R is equivalent to the region
S = {1 ≤ x ≤ 2, x ≤ y ≤ x} ∪ {2 ≤ x ≤ 4, 2 ≤ y ≤ x}. If I integrate with respect to y rst, I must split
the integral into two parts, since the bounds of integration in the x direction change at x = 2.
Example:
Evaluate xy 2 dydx with respect to y rst:
R 2 R 2x
1 x
Z 2 Z 2x Z 2 Z 2
2 1 3 2x 7 4 7 52 7 5 217
xy dydx = ( xy |x )dx = x dx = x |1 = (2 − 15 ) =
1 x 1 3 1 3 15 15 15
Now evaluate with respect to x rst:
Z 2 Z 2x Z 2 Z y Z 4 Z 2 Z 2 Z 4
2 2 2 1 2 2y 1 2 22
xy dydx = xy dxdy + xy dxdy = x y |1 dy + x y | y dy =
1 x 1 1 2 y
2 1 2 2 2 2

1
 2 4
y4
Z Z
1 1 1 1 2 1
= ( y 4 − y 2 )dy + (2y 2 − )dy = ( y 5 − y 3 )|21 + ( y 3 − y 5 )|42 =
1 2 2 2 8 10 6 3 40
31 7 112 992 217
= − + − =
10 6 3 40 15

Transformations
Consider the integral ab cd f (x, y). As we have seen previously, integrals are sometimes easier to evaluate
R R

when we substitute a variable u in for a function of x and y . Consider the following integral:
Z 3 Z x−2
1
2 0 (x + y)(x − y)

We can guess that making a substitution u = x + y and v = x − y would make the integration a lot
simpler. However, there is more than meets the eye in making this transformation. Not only will the limits
of integration change, but we must also change the function itself beyond just making the substitution.
In general, say we are given f (x, y), a region of integration, and two transformation functions u(x, y) and
v(x, y). In order to integrate using the transformation, we rst solve the system of transformation functions
for x(u, v) and y(u, v). Then me must compute the Jacobian matrix
∂x ∂x
 
∂u ∂v
∂y ∂y
∂u ∂v

and nd the absolute value of its determinant. We will cover determinants later, but the absolute value of
the determinant of this specic matrix is ∂u ∂x ∂y
∂v · ∂u = |J|. We then go back to the original equation
· ∂v − ∂x ∂y

f (x, y), plug in x(u, v) and y(u, v) to get a new equation g(u, v ) and multiply it by the Jacobian to get our
new integrand Z Z
g(u, v) |J| dudv.

Finally, we must change the limits of integration. This is done simply by looking at the possible values x
and y took on, and then seeing what that would correspond to in u and v . For example, if x, y ∈ (0, 1) and
u = xy and v = xy , we can see that u ∈ (0, 1) and v ∈ (0, ∞). We then integrate over that region.
Example:
Transform using u = x + y and v = x − y .
R 3 R x−2 1
2 0 (x+y)(x−y) dydx

Solving the system for x and y we get x = u+v

2 , y= u−v
2 . The jacobian of this system is
1 1
 
2 2
1
1 =
2 − 21 2

Plugging this into the original equation, we have

Z Z Z Z
1 1 1
u−v · dudv = dudv
( u+v
2
u−v u+v
+ 2 )( 2 − 2 )
2 2uv

To nd the new limits of integration, we notice that there are three boundaries to the original integral:
x ≤ 3, y ≥ 0, y ≤ x − 2. Substituting in for and x and y in each of these three, we get:
u+v
≤3⇒u+v ≤6
2
u−v
≥0⇒u≥v
2
u−v u+v
≤ −2⇒v ≥2
2 2

2
Plotting these three, we see that the area of integration in u, v space is now an isolceles triangle with corners
at (2, 2), (3, 3), and (4, 2). We can then set up the orders of integration so the integral now reads
Z 3 Z 6−v
1
dudv.
2 v 2uv

This is hard to integrate, so we will leave it for the interested student to do at home.

Homework

Evaluate the following integrals:

R 2 R 4−x2
1. 0 0
xydydx
R√
2. 1−x2
R1
0 1−x
x2 ydydx

3. xdydx for the region bounded by y = 2x and y = 3 − x2

RR

4. 01 y1 x2 sin(xy)dxdy
R R

Sketch the following regions:

1. 1 ≤ x ≤ 2, 2 ≤ y ≤ 7
2. 0 ≤ x ≤ 2, x2
2 ≤y≤x
3. 1 ≤ x ≤ 2, x2 ≤ y ≤ x + 2
4. 0 ≤ x ≤ 1, x2 ≤ y ≤ x
Integrate the following:
1. f (x, y) = sin(x2 ) over the area 0 ≤ x ≤ 2, 0 ≤ y ≤ x
2 with respect to x rst
2. f (x, y) = sin(x2 ) over the area 0 ≤ x ≤ 2, 0 ≤ y ≤ x
2 with respect to y rst
In 1st semester econometrics, you will be asked to integrate probability density functions (or p.d.f.s) in order
to nd the probability that certain events will occur. Sometimes it is necessary to transform the random
variables. For example, if we are given the density functions for X1 and X2 , and Y1 and Y2 as functions of
X1 and X2 , we can then transform this system to solve for the p.d.f.'s of Y1 and Y2 . The following problems
are taken from p.d.f.'s in the rst semester econometrics textbook, section 3.7. For each of the following,
• Find X1 and X2 as functions of Y1 and Y2
• Find the determinant of the Jacobian
• Sketch the region of integration in terms of X1 and X2
• Sketch the region of integration in terms of Y1 and Y2
• Evaluate the new integral in terms of Y1 and Y2

1. f (X1 , X2 ) = e−X1 −X2 over the region X1 > 0, X2 > 0, where Y1 = X1

X1 +X2 and Y2 = X1 + X2 .
2. f (X1 , X2 ) = 8X1 X2 over the region 0 ≤ X1 ≤ X2 ≤ 1, where Y1 = X1
X2 and Y2 = X2 .
Hint: As a check, realize that all p.d.f.'s integrate to one. So the original integrals with respect to X1 and
X2 , as well as the transformed integral with respect to Y1 and Y2 , should integrate to one. Try it!