Introduction to changing variables in double integrals - Math Insight
Math Insight
頁1/4
home
Introduction to changing variables in double
integrals
threads
index
Suggested background
Introduction to double integrals
Similar Pages
Imagine that you had to compute the double integral
∬ g(x,y)dA
D
(1)
where g(x,y) = x2 + y 2 and D is the disk of radius 7 centered at the origin.
Area calculation for changing variables
in double integrals
Double integral change of variable
examples
Double integrals as iterated integrals
Double integral examples
Double integrals as volume
Triple integral change of variables
story
Length, area, and volume factors
The integrals of multivariable calculus
Determinants and linear
transformations
An introduction to ordinary differential
equations
More similar pages
Threads
Math 2374 (Part 17)
Previous page: Finding a potential
function for three-dimensional
conservative vector fields
Next page: Area calculation for
changing variables in double
integrals
In terms of the standard rectangular coordinates x and y , the disk is given by
−7 ≤ x ≤ 7
−−−−−−
−−−−−−
−√49 − x2 ≤ y ≤ √49 − x2 .
Multivariable calculus (Changing
variables)
We could start to calculate the integral in terms of x and y as
∬ g(x,y)dA = ∫
D
7
−7
∫
√49−x 2
−√49−x 2
Previous page: Triple integral examples
Next page: Area calculation for
changing variables in double
integrals
(x2 + y 2 ) dy dx = a mess.
It turns out that this integral would be a lot easier if we could change variables to polar coordinates.
In polar coordinates, the disk is the region we'll call D∗ defined by 0 ≤ r ≤ 7 and 0 ≤ θ ≤ 2π .
Hence the region of integration is simpler to describe using polar coordinates.
Moreover, the integrand x2 + y 2 is simple in polar coordinates because x2 + y 2 = r2 . Using polar
coordinates, our lives will be a lot easier because it seems that all we need to do is integrate r2 over
the region D∗ defined by 0 ≤ r ≤ 7 and 0 ≤ θ ≤ 2π .
Unfortunately, it's not quite that easy. We need to account for one more consequence of changing
variables, which is how changing variables changes area. You may recall that dA stands for the area
of a little bit of the region D . In rectangular coordinates, we replaced dA by dx dy (or dy dx ). We
need to determine what dA becomes when we change variables. As you will see, in polar coordinates,
dA does not becomes dr dθ .
The relationship between rectangular (x,y) and polar (r,θ) coordinates is given by x = rcosθ ,
y = rsinθ . To see how area gets changed, let's write the change of variables as the function
(x,y) = T(r,θ) = (r cosθ,rsin θ).
(2)
The function T(r,θ) gives rectangular coordinates in terms of polar coordinates. The below applet
shows how T maps the region D∗ into the region D
http://mathinsight.org/double_integral_change_variables_introduction
2011/12/6
Introduction to changing variables in double integrals - Math Insight
頁2/4
See also
Double integral change of variable
examples
Examples of changing the order of
integration in double integrals
Triple integral change of variables story
Go Deeper
Area calculation for changing variables in
double integrals
Length, area, and volume factors
Introduction to a surface integral of a
scalar-valued function
This is a Java applet created with LiveGraphics3D. It looks like you do not have Java installed.
You can click here to get Java.
A map from a rectangle to a disk. The map (x,y) = T(r,θ) = (r cos θ,r sinθ) from polar
coordinates (r,θ) to rectangular coordinates (x,y) maps the rectangle D∗ described by
0 ≤ r ≤ 7 , 0 ≤ θ ≤ 2π into the disk D of radius 7 described by x2 + y 2 ≤ 72 . You can change
r and θ by dragging with the mouse the red point in D∗ . The yellow point moves to show the
point T(r,θ) in D . You cannot move the yellow point directly with the mouse.
Do you understand why the disk looks like a rectangle in polar coordinates (i.e., why the region D∗
on the left is a rectangle)? Remember, the disk is described by 0 ≤ r ≤ 7 and 0 ≤ θ ≤ 2π , which is
a rectangle when plotted in therθ -plane.
We can say that T(r,θ) parametrizes D for (r,θ) in D∗ . This uses the same language that we used
whenparametrizing a curve. We'll use it again when we talk about parametrizing surfaces.
To look at how T(r,θ) changes area, we can chop up the region D∗ into small rectangles of width
∆r and height ∆θ . The function T(r,θ) maps each of these small rectangles into a “curvy rectangle”
in D , as shown below.
http://mathinsight.org/double_integral_change_variables_introduction
2011/12/6
Introduction to changing variables in double integrals - Math Insight
頁3/4
This is a Java applet created with LiveGraphics3D. It looks like you do not have Java installed.
You can click here to get Java.
A map from a rectangle to a disk transforms area. The map (x,y) = T(r,θ) = (r cosθ,rsin θ)
from polar coordinates (r,θ) to rectangular coordinates (x,y) maps each small rectangle in D∗
to a “curvy rectangle” in D . Although the small rectangles in D∗ are the same size, the
corresponding “curvy rectangles” vary greatly in size. Depending on the coordinates (r,θ) , the
map T(r,θ) shrinks or expands the area by different amounts. You can change r and θ by
dragging with the mouse the red point in D∗ . The yellow point moves to show the point T(r,θ)
in D . You cannot move the yellow point directly with the mouse.
The area of each small rectangle in D∗ is ∆r∆θ . But we don't care about area in D∗ . The dA in the
integral of equation (1) is based on area in D not area in D∗ . So we need to estimate the area of each
“curvy rectangle” in D , which we'll denote by ∆A .
We can calculate the area of the “curvy rectangle”by approximating it as a parallelogram with sides
∂T
∆θ . The area of a parallelogram is the magnitude of the cross product
∆r and ∂T
∂r
∂θ
∂T
∥ × ∂T ∥ ∥ ∆r∆θ of the two vectors spanning the parallelogram. Plus, since we are in two∥ ∂r
∂θ
dimensions, we write the area more simply by a 2 × 2 determinant.After some simplification, the
area of the“curvy rectangle” reduces to the expression
∆A ≈ |detDT(r,θ)|∆r∆θ,
where DT is derivative matrix of the map T .
Note that the D in DT(r,θ) is not the same D as the region D of integration. Because we need to
take the absolute value of the determinant, we typically use the notation“det” to denote determinant to
avoid confusion (see discussion at end of the page on matrices and determinants).
For T given by equation (2) , you can calculate that |detDT(r,θ)| = r so that the area of each
“curvy rectangle” is r∆r∆θ . This agrees with the above picture, since the “curvy rectangles” were
larger whenr was larger.
The important point is that T stretches or shrinks D∗ when it maps D∗ onto D . Consequently, when
we convert from area inD∗ to area in D , we need to multiply by the area expansion factor
|detDT(r,θ)| . The area expansion factor captures how the “curvy rectangles” change size as you
move the point around in the above applet.
We now put everything back together. We started off trying to integrate the function
g(x,y) = x2 + y 2 over the region D . If we use (x,y) = T(r,θ) to change variables, we can instead
integrate the function g(T(r,θ)) = r2 over the regionD∗ . However, we need to include the area
expansion factor |detDT(r,θ)| = r in dA to account for the stretching by T . We can replace dA
with r dr dθ . We end up with the formula
∬ g(x,y)dA = ∬
D∗
D
which for our example is
∬ (x2 + y 2 )dA = ∫
D
2π
0
g(T(r,θ))|detDT(r,θ)|dr dθ,
∫
0
7
r2 r dr dθ = ∫
0
2π
∫
0
7
r3 dr dθ.
You can compute that this integral is 74 π/2 much easier using this form than you could using the
original integral of equation (1) .
For a general change of variables, we tend to use the variables u andv (rather than r and θ ). In this
case, if we change variables by (x,y) = T(u,v) , our integral is
∬ g(x,y)dA = ∬
D
D∗
g(T(u,v))|detDT(u,v)|du dv,
where D is a region in the xy -plane that is parametrized by(x,y) = T(u,v) for (u,v) in the region
D∗ .
We've obviously skipped quite a few details on this introductory page in an effort to just give the big
picture. In particular, we've glossed over how we obtained the expression for the area expansion
factor. You can read how weobtain that formula. You can also see some examples of changing
variables, including more details on the disk example.
http://mathinsight.org/double_integral_change_variables_introduction
2011/12/6
Introduction to changing variables in double integrals - Math Insight
頁4/4
Stretching by maps
The area expansion factor for changing variables in double integrals is an example of
accounting for the stretching of a map, in this case, the function T . We encounter such
factors frequently in multivariable and vector calculus.
The simplest example of such an expansion factor is the ∥c′ (t)∥ we obtain when calculating
thearc lengthor a line integralover a curve parametrized by c : R → R2 . We could call this
a length expansion factor. Just like for the T : R2 → R2 of changing variables, the
expansion factor for parametrized curves involves the magnitude of some expression
involving the derivative matrix of the map.
For the map T : R3 → R3 used to change variables in triple integrals, the volume
expansion factor|detDT(u,v,w)| is essentially the same as for double integrals.
Lastly, imagine you took the blue disk D in the above applet and lifted it out of the plane so
that it was a surface floating in three-dimensional space. Then, our mapping T becomes the
function Φ : R2 → R3 thatparametrizes a surface. We can repeat the calculationfor the area
expansion factor virtually without change to obtain the area expansion factor forsurface
areaor surface integrals. The only difference that results from being in three dimensions is
that you cannot change the cross productfor the parallelogram area to a 2 × 2 determinant.
∥ .
∥ ∂Φ × ∂Φ ∥ Hence, the area expansion factor for parametrized surfaces is the cross-product∥ ∂u
∂v
Cite this as
Nykamp DQ, “Introduction to changing variables in double integrals.” From Math Insight. http://mathinsight.org/double_integral_change_variables_introduction
Keywords: integral, double integral, change variables
Introduction to changing variables in double integrals by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 3.0 License.
For permissions beyond the scope of this license, please contact us.
http://mathinsight.org/double_integral_change_variables_introduction
2011/12/6
Double integral examples - Math Insight
頁1/3
Math Insight
home
Double integral examples
Suggested background
To illustrate computing double integrals asiterated integrals, we start with the
simplest example of a double integral over a rectangle and then move on an
integral over a triangle.
Example 1
Compute the integral
∬ xy 2 dA
D
where D is the rectangle 0 ≤ x ≤ 2 , 0 ≤ y ≤ 1 .
Solution: We will compute the double integrals as theiterated integral
∫ (∫
1
0
2
0
xy 2 dx)dy.
We first integrate with respect to x inside the parentheses. Similar to the
procedure withpartial derivatives, we must treat y as a constant during this
integration step. Since the integral of cx iscx2 /2 , we calculate
∫ (∫
1
0
0
2
x=2
x2 ∣ xy 2 dx)dy = ∫ ( y 2 ∣ )dy
2 ∣ x=0
0
1
=∫ (
1
0
=∫
0
1
22 2 02 2
y − y )dy
2
2
2y 2 dy.
Note that in the second line above, we wrote the limits as x = 2 andx = 0 so it is
unambiguous that x is the variable we just integrated.
threads
index
Introduction to double integrals
Double integrals as iterated integrals
Similar Pages
Double integrals as volume
Examples of changing the order of
integration in double integrals
Double integrals as area
Double integrals where one integration
order is easier
Introduction to changing variables in
double integrals
An introduction to ordinary differential
equations
Introduction to triple integrals
Green's theorem examples
Triple integral examples
Subtleties of differentiability in higher
dimensions
More similar pages
Threads
Math 2374 (Part 9)
Previous page: Double integrals as
iterated integrals
Next page: Double integrals as volume
Multivariable calculus (Double
integrals)
Previous page: Double integrals as
iterated integrals
Next page: Double integrals as volume
To finish, we need to compute the integral with respect to y , which is simple.
Since x is gone, it's just a regular one-variable integral. We calculate that our
double integral is
∬ xy 2 dA = ∫
0
D
=
1
2y 2 dy
1
2y 3 ∣ 2(13 ) 2(03 ) 2
−
=
∣ =
3 ∣ 0
3
3
3
To double check, we can compute the integral in the other direction, integrating
first with respect to y and then with respect to x . The only trick is to remember
that when integrating with respect to y , we must think of x as a constant. Since
the integral of cy 2 is cy 3 /3 , we calculate
http://mathinsight.org/double_integral_examples
2011/12/6
Double integral examples - Math Insight
頁2/3
See also
Introduction to double integrals
2
1
Double
integrals
as iterated integrals
2
∬ xy dA =Examples
∫ (∫ x
dy)dx the order of
ofychanging
D
0
0
integration
in double integrals
y=1
2
y 3 ∣ = ∫ (x ∣ )dx
3 ∣ y=0
0
2
As it must, this iterated integral gives the2 same
answer.
x
=∫
dx
3
0
Example 2
=
2
x2 ∣ 4 2
∣ = = .
6 ∣ 0 6 3
Rectangular regions were easy because the limits (a ≤ x ≤ b and c ≤ y ≤ d )
were fixed. Here's an example where we integrate over the triangle defined by
0 ≤ x ≤ 2 , 0 ≤ y ≤ x/2 .
Use same function f(x,y) = xy 2 . Compute ∬D xy 2 dA when D is that triangle.
Solution: Note for the triangle defined by 0 ≤ x ≤ 2 and 0 ≤ y ≤ x/2 , the
limits on y depend on x . For a given value of x , y ranges from 0 to x/2 , as
illustrated above by the vertical dashed line from (x,0) to (x,x/2) .
In a double integral, the outer limits must
be
constant, but the inner limits can
2
x/2
depend on the outer
put y as the inner integration
∬variable.
xy 2 dAThis
= ∫means,
(∫ we must
xy 2 dy)dx
D
0
0
variables, as was done in the second way of computing Example 1. The only
y=x/2
difference from Example 1 is that the2 upper
of y is x/2 . The double integral
x limit
= ∫ ( y 3 ∣ )dx
∣ is
3
y=0
0
2
x x 3 x
= ∫ ( ( ) − 03 )dx
3 2
3
0
2
x4
dx
0 24
2
32
4
x5 ∣ =
=
∣ =
5 ⋅ 24 ∣ 0 5 ⋅ 24
15
=∫
Example 2'
Now compute the integral over the same triangle D , but make y be the outer
integration variable.
Solution: Now we need to give constant limits for y . As illustrated below, for
point in the triangle y ranges between between 0 and 1 . Then, for a given value
of y , x takes on values between 2y and 2 (as shown by the horizontal dashed line
http://mathinsight.org/double_integral_examples
2011/12/6
Double integral examples - Math Insight
頁3/3
between (2y,y) and (2,y) ). Hence, we can describe the triangle by0 ≤ y ≤ 1
and 2y ≤ x ≤ 2 .
Is it confusing that the limits of x are 2y ≤ x ≤ 2 rather than 0 ≤ x ≤ 2 (which
would more closely parallel the above Example 2)? If we let x range from 0 to 2y
, then the triangle would be the upper-left triangle in the above picture. We want
1
2
to compute the integral
over the region D , which is the lower-right triangle
∬ xy 2 dA = ∫ (∫ xy 2 dx)dy
shaded in red. In
2y (as we are now using in Example 2').
D this triangle,0 $y2y$
x=2
1
x2 y 2 ∣ ∣ =∫ (
)dy
The double integral is similar0to the 2first∣ x=2y
way of computing Example 1, with the
only difference being that the lower
limit(2y)
of x2 yis22y . The integral is
1
= ∫ (2y 2 −
)dy
2
0
1
= ∫ (2y 2 − 2y 4 )dy
0
= 2[
y3 y5
− ]
3
5
1
0
1 1
2
4
= 2( − − (0 − 0)) = 2 ⋅
= .
3 5
15 15
Thankfully, this does agree with the answer we obtained in Example 2.
To go from Example 2 to Example 2', we “changed the order of integration.” You
can see more examples of changing the order of integration in double integrals.
Cite this as
Nykamp DQ, “Double integral examples.” From Math Insight. http://mathinsight.org/double_integral_examples
Keywords: integral, double integral
Double integral examples by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 3.0 License. For
permissions beyond the scope of this license, please contact us.
http://mathinsight.org/double_integral_examples
2011/12/6