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Let’s try another example with a different substitution.

Evaluating a triple integral with a change of variables

Evaluate the triple integral

0 3 0 4 y / 2 ( y / 2 ) + 1 ( x + z 3 ) d x d y d z

in x y z -space by using the transformation

u = ( 2 x y ) / 2 , v = y / 2 , and w = z / 3 .

Then integrate over an appropriate region in u v w -space .

As before, some kind of sketch of the region G in x y z -space over which we have to perform the integration can help identify the region D in u v w -space ( [link] ). Clearly G in x y z -space is bounded by the planes x = y / 2 , x = ( y / 2 ) + 1 , y = 0 , y = 4 , z = 0 , and z = 4 . We also know that we have to use u = ( 2 x y ) / 2 , v = y / 2 , and w = z / 3 for the transformations. We need to solve for x , y , and z . Here we find that x = u + v , y = 2 v , and z = 3 w .

Using elementary algebra, we can find the corresponding surfaces for the region G and the limits of integration in u v w -space . It is convenient to list these equations in a table.

Equations in x y z for the region D Corresponding equations in u v w for the region G Limits for the integration in u v w
x = y / 2 u + v = 2 v / 2 = v u = 0
x = y / 2 u + v = ( 2 v / 2 ) + 1 = v + 1 u = 1
y = 0 2 v = 0 v = 0
y = 4 2 v = 4 v = 2
z = 0 3 w = 0 w = 0
z = 3 3 w = 3 w = 1
On the left-hand side of this figure, there is a box G with sides 1, 2, and 1 along the u, v, and w axes, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u + v, y = 2v, and z = 3w. On the right-hand side of this figure there is a region D in xyz space that is a rotated box with sides 1, 4, and 3 along the x, y, and z axes. The rear plane is marked x = y/2 or y = 2x. The front plane is marked x = y/2 + 1 or y = 2x minus 2.
The region G in u v w -space is transformed to region D in x y z -space .

Now we can calculate the Jacobian for the transformation:

J ( u , v , w ) = | x u x v x w y u y v y w z u z v z w | = | 1 1 0 0 2 0 0 0 3 | = 6 .

The function to be integrated becomes

f ( x , y , z ) = x + z 3 = u + v + 3 w 3 = u + v + w .

We are now ready to put everything together and complete the problem.

0 3 0 4 y / 2 ( y / 2 ) + 1 ( x + z 3 ) d x d y d z = 0 1 0 2 0 1 ( u + v + w ) | J ( u , v , w ) | d u d v d w = 0 1 0 2 0 1 ( u + v + w ) | 6 | d u d v d w = 6 0 1 0 2 0 1 ( u + v + w ) d u d v d w = 6 0 1 0 2 [ u 2 2 + v u + w u ] 0 1 d v d w = 6 0 1 0 2 ( 1 2 + v + w ) d v d w = 6 0 1 [ 1 2 v + v 2 2 + w v ] 0 2 d w = 6 0 1 ( 3 + 2 w ) d w = 6 [ 3 w + w 2 ] 0 1 = 24.
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Let D be the region in x y z -space defined by 1 x 2 , 0 x y 2 , and 0 z 1 .

Evaluate D ( x 2 y + 3 x y z ) d x d y d z by using the transformation u = x , v = x y , and w = 3 z .

0 3 0 2 1 2 ( v 3 + v w 3 u ) d u d v d w = 2 + ln 8

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Key concepts

  • A transformation T is a function that transforms a region G in one plane (space) into a region R in another plane (space) by a change of variables.
  • A transformation T : G R defined as T ( u , v ) = ( x , y ) ( or T ( u , v , w ) = ( x , y , z ) ) is said to be a one-to-one transformation if no two points map to the same image point.
  • If f is continuous on R , then R f ( x , y ) d A = S f ( g ( u , v ) , h ( u , v ) ) | ( x , y ) ( u , v ) | d u d v .
  • If F is continuous on R , then
    R F ( x , y , z ) d V = G F ( g ( u , v , w ) , h ( u , v , w ) , k ( u , v , w ) ) | ( x , y , z ) ( u , v , w ) | d u d v d w = G H ( u , v , w ) | J ( u , v , w ) | d u d v d w .

In the following exercises, the function T : S R , T ( u , v ) = ( x , y ) on the region S = { ( u , v ) | 0 u 1 , 0 v 1 } bounded by the unit square is given, where R R 2 is the image of S under T .

  1. Justify that the function T is a C 1 transformation.
  2. Find the images of the vertices of the unit square S through the function T .
  3. Determine the image R of the unit square S and graph it.

x = u 2 , y = v 3

a. T ( u , v ) = ( g ( u , v ) , h ( u , v ) ) , x = g ( u , v ) = u 2 and y = h ( u , v ) = v 3 . The functions g and h are continuous and differentiable, and the partial derivatives g u ( u , v ) = 1 2 , g v ( u , v ) = 0 , h u ( u , v ) = 0 and h v ( u , v ) = 1 3 are continuous on S ; b. T ( 0 , 0 ) = ( 0 , 0 ) , T ( 1 , 0 ) = ( 1 2 , 0 ) , T ( 0 , 1 ) = ( 0 , 1 3 ) , and T ( 1 , 1 ) = ( 1 2 , 1 3 ) ; c. R is the rectangle of vertices ( 0 , 0 ) , ( 1 2 , 0 ) , ( 1 2 , 1 3 ) , and ( 0 , 1 3 ) in the x y -plane; the following figure.
A rectangle with one corner at the origin, horizontal length 0.5, and vertical height 0.34.

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x = 2 u v , y = u + 2 v

a. T ( u , v ) = ( g ( u , v ) , h ( u , v ) ) , x = g ( u , v ) = 2 u v , and y = h ( u , v ) = u + 2 v . The functions g and h are continuous and differentiable, and the partial derivatives g u ( u , v ) = 2 , g v ( u , v ) = −1 , h u ( u , v ) = 1 , and h v ( u , v ) = 2 are continuous on S ; b. T ( 0 , 0 ) = ( 0 , 0 ) , T ( 1 , 0 ) = ( 2 , 1 ) , T ( 0 , 1 ) = ( −1 , 2 ) , and T ( 1 , 1 ) = ( 1 , 3 ) ; c. R is the parallelogram of vertices ( 0 , 0 ) , ( 2 , 1 ) , ( 1 , 3 ) , and ( −1 , 2 ) in the x y -plane; see the following figure.
A square of side length square root of 5 with one corner at the origin and another at (2, 1).

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Practice Key Terms 4

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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