Change of variables in double integrals
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1. The document discusses change of variables for double integrals, introducing the Jacobian determinant which relates the differentials of the original and transformed variables. 2. It provides an example of using a change of variables (u=x-y, v=x+y) to evaluate an integral over a parallelogram region. 3. Polar coordinates are also discussed as a common change of variables technique for double integrals, with an example evaluating an integral over a circular region in polar coordinates.
![1 Change of variables in double integrals
Review of the idea of substitution
Consider the integral
2
x cos(x2 )dx.
0
To evaluate this integral we use the u-substitution
u = x2 .
This substitution send the interval [0, 2] onto the interval [0, 4]. Since
du = 2xdx (1)
the integral becomes
1 4 1
cos udu = sin 4.
2 0 2
We want to perform similar subsitutions for multiple integrals.
Jacobians
Let
x = g(u, v) and y = h(u, v) (2)
be a transformation of the plane. Then the Jacobian of this transformation
is defined by
∂x ∂x
∂(x, y) ∂x ∂y ∂x ∂y
= ∂u ∂y =
∂y
∂v − (3)
∂(u, v) ∂u ∂v ∂u ∂v ∂v ∂u
Theorem
Let
x = g(u, v) and y = h(u, v)
be a transformation of the plane that is one to one from a region S in the
(u, v)-plane to a region R in the (x, y)-plane. If g and h have continuous
partial derivatives such that the Jacobian is never zero, then
∂(x, y)
f (x, y)dxdy = f [g(u, v), h(u, v)]| |dudv (4)
R S ∂(u, v)
Here | . . . | means the absolute value.
1](https://image.slidesharecdn.com/changeofvariablesindoubleintegrals-121123013142-phpapp01/85/Change-of-variables-in-double-integrals-1-320.jpg)
![Remark A useful fact is that
∂(x, y) 1
| | = ∂(u,v) (5)
∂(u, v) | ∂(x,y) |
Example
Use an appropriate change of variables to evaluate
(x − y)2 dxdy (6)
R
where R is the parallelogram with vertices (0, 0), (1, 1), (2, 0) and (1, −1).
(excercise: draw the domain R).
Solution:
We find that the equations of the four lines that make the parallelogram
are
x−y =0 x−y =2 x+y =0 x+y =2 (7)
The equations (7) suggest the change of variables
u=x−y v =x+y (8)
Solving (8) for x and y gives
u+v v−u
x= y= (9)
2 2
The Jacobian is
∂x ∂x 1 1
∂(x, y) ∂u ∂v 2 2
1 1 1
= = = + = (10)
∂(u, v) ∂y
∂u
∂y
∂v
−1
2
1
2 4 4 2
The region S in the (u, v) is the square 0 < u < 2, 0 < v < 2. Since
x − y = u, the integral becomes
2 2 1 2 u3 2 2 4 8
u2 dudv = [ ] dv = dv =
0 0 2 0 6 0 0 3 3
Polar coordinates
We know describe examples in which double integrals can be evaluated
by changing to polar coordinates. Recall that polar coordinates are defined
by
x = r cos θ y = r sin θ (11)
2](https://image.slidesharecdn.com/changeofvariablesindoubleintegrals-121123013142-phpapp01/85/Change-of-variables-in-double-integrals-2-320.jpg)
Change of variables in double integrals
- 1. 1 Change of variables in double integrals Review of the idea of substitution Consider the integral 2 x cos(x2 )dx. 0 To evaluate this integral we use the u-substitution u = x2 . This substitution send the interval [0, 2] onto the interval [0, 4]. Since du = 2xdx (1) the integral becomes 1 4 1 cos udu = sin 4. 2 0 2 We want to perform similar subsitutions for multiple integrals. Jacobians Let x = g(u, v) and y = h(u, v) (2) be a transformation of the plane. Then the Jacobian of this transformation is defined by ∂x ∂x ∂(x, y) ∂x ∂y ∂x ∂y = ∂u ∂y = ∂y ∂v − (3) ∂(u, v) ∂u ∂v ∂u ∂v ∂v ∂u Theorem Let x = g(u, v) and y = h(u, v) be a transformation of the plane that is one to one from a region S in the (u, v)-plane to a region R in the (x, y)-plane. If g and h have continuous partial derivatives such that the Jacobian is never zero, then ∂(x, y) f (x, y)dxdy = f [g(u, v), h(u, v)]| |dudv (4) R S ∂(u, v) Here | . . . | means the absolute value. 1
- 2. Remark A useful fact is that ∂(x, y) 1 | | = ∂(u,v) (5) ∂(u, v) | ∂(x,y) | Example Use an appropriate change of variables to evaluate (x − y)2 dxdy (6) R where R is the parallelogram with vertices (0, 0), (1, 1), (2, 0) and (1, −1). (excercise: draw the domain R). Solution: We find that the equations of the four lines that make the parallelogram are x−y =0 x−y =2 x+y =0 x+y =2 (7) The equations (7) suggest the change of variables u=x−y v =x+y (8) Solving (8) for x and y gives u+v v−u x= y= (9) 2 2 The Jacobian is ∂x ∂x 1 1 ∂(x, y) ∂u ∂v 2 2 1 1 1 = = = + = (10) ∂(u, v) ∂y ∂u ∂y ∂v −1 2 1 2 4 4 2 The region S in the (u, v) is the square 0 < u < 2, 0 < v < 2. Since x − y = u, the integral becomes 2 2 1 2 u3 2 2 4 8 u2 dudv = [ ] dv = dv = 0 0 2 0 6 0 0 3 3 Polar coordinates We know describe examples in which double integrals can be evaluated by changing to polar coordinates. Recall that polar coordinates are defined by x = r cos θ y = r sin θ (11) 2
- 3. The Jacobian of the transformation (11) is ∂x ∂x ∂(x, y) cos θ −r sin θ = ∂r ∂y ∂θ ∂y = = r cos2 θ + r sin2 θ = r (12) ∂(r, θ) ∂r ∂θ sin θ r cos θ Example Let R be the disc of radius 2 centered at the origin. Calculate sin(x2 + y 2 )dxdy (13) R Using the polar coordinates (11) we rewrite (13) as 2π ∂(x, y) sin(r2 )| |drdθ (14) 0 ∂(r, θ) Substituting (12) into (14) we obtain 2π 2 r sin r2 drdθ (15) 0 0 Using the subsitutuion t = r2 we have 2π 4 1 sin t dtdθ = π(1 − cos 4). 0 0 2 3