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Change of Variables

Back in Calculus I we had the substitution rule that told us that,

∫ a f ( g ( x ) ) g ′ ( x ) dx = ∫c f ( u ) du
b

d

where u = g ( x )

In essence this is taking an integral in terms of x’s and changing it into terms of u’s
...
In fact we’ve already done this to a certain
extent when we converted double integrals to polar coordinates and when we converted triple
integrals to cylindrical or spherical coordinates
...
If you recall, in each of those cases we
commented that we would justify the formulas for dA and dV eventually
...

While often the reason for changing variables is to get us an integral that we can do with the new
variables another reason for changing variables is to convert the region into a nicer region to work
with
...

That is not always the case however
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First we need a little notation out of the way
...
Also we will typically start out with a region, R, in xy-coordinates
and transform it into a region in uv-coordinates
...

2
y
u
2
= 1 and the transformation is x = , y = 3v
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[Solution]
2
2

Solution
2
y
u
2
(a) R is the ellipse x +
= 1 and the transformation is x = , y = 3v
...


⎛ u ⎞ ( 3v )
=1
⎜ ⎟ +
36
⎝2⎠
u 2 9v 2
+
=1
4 36
u 2 + v2 = 4
2

2

So, we started out with an ellipse and after the transformation we had a disk of radius 2
...

2
2
As with the first part we’ll need to plug the transformation into the equation, however, in this case
we will need to do it three times, once for each equation
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So, we have a triangle
...
We will apply the
transformation to each edge of the triangle and see where we get
...
Plugging in the transformation gives,

1
1
(u − v ) = − (u + v ) + 4
2
2
u − v = −u − v + 8
2u = 8
u 4
The first boundary transforms very nicely into a much simpler equation
...


Finally, let’s transform y = x3 − 34
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Let’s take a look at the new region that we get under the transformation
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[Return to Problems]

Note that we can’t always expect to transform a specific type of region (a triangle for example)
into the same kind of region
...

Notice that in each of the above examples we took a two dimensional region that would have
been somewhat difficult to integrate over and converted it into a region that would be much nicer
in integrate over
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In addition to converting the integrand into something simpler it will
often also transform the region into one that is much easier to deal with
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We will start with double integrals
...

Here is the definition of the Jacobian
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Here is how to compute the determinant
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Change of Variables for a Double Integral
Suppose that we want to integrate f ( x, y ) over the region R
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Also note that we are taking the absolute value of the Jacobian
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All that we need to do is use the formula above for dA
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Now, let’s do a couple of integrals
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Solution
First, let’s sketch the region R and determine equations for each of the sides
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e
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While we could do this integral in terms of x and y it would involve two integrals and so would be
some work
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We’ll do this by plugging the transformation
into each of the equations above
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2u − 3v = 2u + 3v
6v = 0
v=0
Transforming y = − x is similar
...


2u − 3v = − ( 2u + 3v ) + 5
4u = 5
5
u=
4

Finally, let’s transform y = x − 5
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The integral is then,

5
0≤v≤
6

∂ ( x, y ) 2 3
=
= −6 − 6 = −12
∂ ( u, v ) 2 −3

5
4

and v =

5
6

and so

5
6

⌠
x
+
y
dA
=
⎮
∫∫R
⌡0

5
6

=⌠
⎮
⌡0

5
4
0

∫ ( 2u + 3v ) + ( 2u − 3v ) −12 du dv
∫

5
4
0

5
6

48u du dv

=⌠
⎮ 24u
⌡0

5
2 4
0

dv

5
6

75
⌠
=⎮
dv
⌡0 2
5
6

75
= v
2 0
125
=
4
Example 4 Evaluate

2
2
2
2
x
−
xy
+
y
dA
where
R
is
the
ellipse
given
by
x
−
xy
+
y
= 2 and
∫∫
R

using the transformation x = 2 u −

2
3

v, y = 2u+

2
3

v
...


2 = x 2 − xy + y 2
2

⎛
2 ⎞ ⎛
2 ⎞⎛
= ⎜⎜ 2 u −
v ⎟⎟ − ⎜⎜ 2 u −
v ⎟⎟ ⎜⎜ 2 u +
3 ⎠ ⎝
3 ⎠⎝
⎝
4
2 2 ⎛ 2 2 2⎞
2
= 2u −
uv + v − ⎜ 2u − v ⎟ + 2u 2 +
3
3 ⎠
3
⎝

2 ⎞ ⎛
2 ⎞
v ⎟⎟ + ⎜⎜ 2 u +
v ⎟⎟
3 ⎠ ⎝
3 ⎠
4
2 2
uv + v
3
3

2

= 2u 2 + 2v 2
Or, upon dividing by 2 we see that the equation describing R transforms into

u 2 + v2 = 1
or the unit circle
...

Note as well that we’ve shown that the function that we’re integrating is

x − xy + y = 2 ( u + v
2

2

2

2

)

in terms of u and v so we won’t have to redo that work when the time to do the integral comes
around
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2

∂ ( x, y )
=
∂ ( u, v )

2

2
−
3

2
2
4
=
+
=
3
3
3
2
3

The integral is then,

4
⌠⌠
2
2
du dv
2 (u + v )
∫∫R x − xy + y dA = ⎮⎮
3
⌡⌡
2

2

S

Before proceeding a word of caution is in order
...
These equations are only valid on the
boundary of the region and we are looking at all the point interior to the boundary as well and for
those points neither of these equations will be true!
At this point we’ll note that this integral will be much easier in terms of polar coordinates and so
to finish the integral out will convert to polar coordinates
...
In this case we will again start with a region R and use
the transformation x = g ( u, v, w ) , y = h ( u, v, w ) , and z = k ( u, v, w ) to transform the region
into the new region S
...
Here is the definition of the Jacobian for this kind of transformation
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We saw how to
evaluate these when we looked at cross products back in Calculus II
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The integral under this transformation is,

∫∫∫
R

⌠⌠⌠
∂ ( x, y , z )
f ( x, y, z ) dV = ⎮⎮⎮ f ( g ( u , v, w ) , h ( u , v, w ) , k ( u , v, w ) )
du dv dw
∂ ( u , v, w )
⌡⌡⌡
S

As with double integrals we can look at just the differentials and note that we must have

∂ ( x, y , z )
dV =
du dv dw
∂ ( u , v, w )

We’re not going to do any integrals here, but let’s verify the formula for dV for spherical
coordinates
...

Solution
Here the transformation is just the standard conversion formulas
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We will leave it to you to check the formula for dV for cylindrical coordinates if you’d like to
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