3 The Jacobian in 3 dimensions

When changing the coordinate system of a triple integral

I = V f ( x , y , z ) d V

we need to extend the above definition of the Jacobian to 3 dimensions.

Key Point 12

Jacobian in Three Variables

For given transformations x = x ( u , v , w ) , y = y ( u , v , w ) and z = z ( u , v , w ) the Jacobian is

J ( u , v , w ) = x u x v x w y u y v y w z u z v z w

The same pattern persists as in the 2-dimensional case (see Key Point 10). Across a row of the determinant the numerators are the same and down a column the denominators are the same.

The volume element d V = d x d y d z becomes d V = J ( u , v , w ) d u d v d w . As before the limits and integrand must also be transformed.

Example 26

Use spherical coordinates to find the volume of a sphere of radius R .

Figure 32

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Solution

The change of coordinates from Cartesian to spherical polar coordinates is given by the transformation equations

x = r cos θ sin ϕ y = r sin θ sin ϕ z = r cos ϕ

We now need the nine partial derivatives

x r = cos θ sin ϕ x θ = r sin θ sin ϕ x ϕ = r cos θ cos ϕ y r = sin θ sin ϕ y θ = r cos θ sin ϕ y ϕ = r sin θ cos ϕ z r = cos ϕ z θ = 0 z ϕ = r sin ϕ

Hence we have

J ( r , θ , ϕ ) = cos θ sin ϕ r sin θ sin ϕ r cos θ cos ϕ sin θ sin ϕ r cos θ sin ϕ r sin θ cos ϕ cos ϕ 0 r sin ϕ

J ( r , θ , ϕ ) = cos ϕ r sin θ sin ϕ r cos θ cos ϕ r cos θ sin ϕ r sin θ cos ϕ + 0 r sin ϕ cos θ sin ϕ r sin θ sin ϕ sin θ sin ϕ r cos θ sin ϕ

Check that this gives J ( r , θ , ϕ ) = r 2 sin ϕ . Notice that J ( r , θ , ϕ ) 0 for 0 ϕ π , so J ( r , θ , ϕ ) = r 2 sin ϕ . The limits are found as follows. The variable ϕ is related to ‘latitude’ with ϕ = 0 representing the ‘North Pole’ with ϕ = π 2 representing the equator and ϕ = π representing the ‘South Pole’.

The variable θ is related to ‘longitude’ with values of 0 to 2 π covering every point for each value of ϕ . Thus limits on ϕ are 0 to π and limits on θ are 0 to 2 π . The limits on r are r = 0 (centre) to r = R (surface).

To find the volume of the sphere we then integrate the volume element d V = r 2 sin ϕ d r d θ d ϕ between these limits.

 Volume = 0 π 0 2 π 0 R r 2 sin ϕ d r d θ d ϕ = 0 π 0 2 π 1 3 R 3 sin ϕ d θ d ϕ = 0 π 2 π 3 R 3 sin ϕ d ϕ = 4 3 π R 3
Example 27

Find the volume integral of the function f ( x , y , z ) = x y over the parallelepiped with the vertices of the base at

( x , y , z ) = ( 0 , 0 , 0 ) , ( 2 , 0 , 0 ) , ( 3 , 1 , 0 ) and ( 1 , 1 , 0 )

and the vertices of the upper face at

( x , y , z ) = ( 0 , 1 , 2 ) , ( 2 , 1 , 2 ) , ( 3 , 2 , 2 ) and ( 1 , 2 , 2 ) .

Figure 33

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Solution

This will be a difficult integral to derive limits for in terms of x , y and z . However, it can be noted that the base is described by z = 0 while the upper face is described by z = 2 . Similarly, the front face is described by 2 y z = 0 with the back face being described by 2 y z = 2 . Finally the left face satisfies 2 x 2 y + z = 0 while the right face satisfies 2 x 2 y + z = 4 .

The above suggests a change of variable with the new variables satisfying u = 2 x 2 y + z , v = 2 y z and w = z and the limits on u being 0 to 4 , the limits on v being 0 to 2 and the limits on w being 0 to 2 .

Inverting the relationship between u , v , w and x , y and z , gives

x = 1 2 ( u + v ) y = 1 2 ( v + w ) z = w

The Jacobian is given by

J ( u , v , w ) = x u x v x w y u y v y w z u z v z w = 1 2 1 2 0 0 1 2 1 2 0 0 1 = 1 4

Note that the function f ( x , y , z ) = x y equals 1 2 ( u + v ) 1 2 ( v + w ) = 1 2 ( u w ) . Thus the integral is

w = 0 2 v = 0 2 u = 0 4 1 2 ( u w ) 1 4 d u d v d w = w = 0 2 v = 0 2 u = 0 4 1 8 ( u w ) d u d v d w = w = 0 2 v = 0 2 1 16 u 2 1 8 u w 0 4 d v d w = w = 0 2 v = 0 2 1 1 2 w d v d w = w = 0 2 v v w 2 0 2 d w = w = 0 2 2 w d w = 2 w 1 2 w 2 0 2 = 4 4 2 0 = 2
Task!

Find the Jacobian for the following transformation:

x = 2 u + 3 v w , y = v 5 w , z = u + 4 w

Evaluating the partial derivatives,

x u = 2 , x v = 3 , x w = 1 ,

y u = 0 , y v = 1 , y w = 5 ,

z u = 1 , z v = 0 , z w = 4

so the Jacobian is

x u x v x w y u y v y w z u z v z w = 2 3 1 0 1 5 1 0 4 = 2 1 5 0 4 + 1 3 1 1 5 = 2 × 4 + 1 × ( 14 ) = 6

where expansion of the determinant has taken place down the first column.