
# Jacobians

### The Idea of Substitution

Consider the integral

$\int _0^2 x\, \text{cos}\, (x^2) \; dx$

To evaluate this integral we use the u-substitution

$u = x^2$

This substitution sends the interval [0,2] onto the interval [0,4].  We can see that there is stretching of the interval. The stretching is not uniform. In fact, the first part [0,0.5] is actually contracted. This is the reason why we need to find $$du$$.

$\dfrac{du}{dx} = 2x$

or

$\dfrac{dx}{du} = \dfrac{1}{2x}$

This is the factor that needs to be multiplied in when we perform the substitution. Notice for small positive values of $$x$$, this factor is greater than 1 and for large values of $$x$$, the factor is smaller than 1. This is how the stretching and contracting is accounted for.

### Jacobians

We have seen that when we convert to polar coordinates, we use

$dy\,dx = r\,dr\,dq$

With a geometrical argument, we showed why the "extra $$r$$" is included. Taking the analogy from the one variable case, the transformation to polar coordinates produces stretching and contracting. The "extra r" takes care of this stretching and contracting. The goal for this section is to be able to find the "extra factor" for a more general transformation. We call this "extra factor" the Jacobian of the transformation. We can find it by taking the determinant of the two by two matrix of partial derivatives.

Definition: Jacobian

Let

$$x = g(u,v)$$  and  $$y = h(u,v)$$

be a transformation of the plane. Then the Jacobian of this transformation is

Example 1

Find the Jacobian of the polar coordinates transformation

$$x(r,q) = r \cos q$$   and       $$y(r,q) = r \sin q$$

SOLUTION

We have

This is comforting since it agrees with the extra factor in integration.

### Double Integration and the Jacobian

Theorem: Integration and Coordinate Transformations

Let

given by

$$x = g(u,v)$$  and $$y = h(u,v)$$

be a transformation on the plane that is one to one from a region $$S$$ to a region $$R$$. If $$g$$ and $$h$$ have continuous partial derivatives such that the Jacobian is never zero, then

Remark:  A useful fact is that the Jacobian of the inverse transformation is the reciprocal of the Jacobian of the original transformation.

This is a consequence of the fact that the determinant of the inverse of a matrix $$A$$ is the reciprocal of the determinant of $$A$$.

Proof

As usual, we cut $$S$$ up into tiny rectangles so that the image under $$T$$ of each rectangle is a parallelogram.

We need to find the area of the parallelogram. Considering differentials, we have

$T(u + \Delta u,v) @ T(u,v) + (x_u\Delta u,y_u\Delta u)$

$T(u,v + \Delta v) @ T(u,v) + (x_v\Delta v,y_v\Delta v)$

Thus the two vectors that make the parallelogram are

$\vec{P} = g_u \Delta D u \hat{i} + h_u \Delta {D} u \hat{j}$

$\vec{Q} = g_v \Delta v \hat{i} + h_v \Delta v \hat{j}$

To find the area of this parallelogram we just cross the two vectors.

and the extra factor is revealed.

Example 1

Use an appropriate change of variables to find the volume of the region below

$z = (x - y)^2$

above the x-axis, over the parallelogram with vertices (0,0), (1,1), (2,0), and (1,-1)

SOLUTION

We find the equations of the four lines that make the parallelogram to be

y  =  x        y  =  x - 2        y  =  -x        y  =  -x + 2

or

x - y  =  0        x - y  =  2        x + y  =  0        x + y  =  2

The region is given by

0  <  x - y  <  2        and        0  <  x + y  < 2

This leads us to the inverse transformation

u(x,y)  =  x - y        v(x,y)  =  x + y

The Jacobian of the inverse transformation is

Since the Jacobian is the reciprocal of the inverse Jacobian we get

The region is given by

$$0 < u < 2$$  and  $$0 < v < 2$$

and the function is given by

$z = u^2$

Putting this all together, we get the double integral

### Jacobians and Triple Integrals

For transformations from $$R^3$$ to $$R^3$$, we define the Jacobian in a similar way

Example 3

Find the Jacobian for the spherical coordinate transformation

$x = r\, cos\,\theta \; sin\,\phi \;\;\;\;\; y = r\, sin\, \theta\; sin\, \phi \;\;\;\;\; z = r\, cos\, \phi$

SOLUTION

We take partial derivatives and compute

### Contributors

20:47, 15 Dec 2013