Evaluate double integral ln((x-y)/(x+y)) dy dx where the double integral region is the triangle with vertices (1,0),(4,3), (4,1). Hint: use a transformation with the Jacobian.

This exact transformation is discussed on pp 3-5 of

http://math.etsu.edu/multicalc/prealpha/Chap4/Chap4-4/printversion.pdf

To evaluate the given double integral using a change of variables, we can perform a transformation using a new set of variables, u and v, in terms of the original variables, x and y.

The transformation we will use is:
u = x - y
v = x + y

To find the new limits of integration, we need to transform the vertices of the triangle in the uv-plane.

Given the vertices of the triangle in the xy-plane: (1,0), (4,3), and (4,1), we can substitute these values into the transformation equations to find the corresponding vertices in the uv-plane.

For (1,0), we have:
u = 1 - 0 = 1
v = 1 + 0 = 1

For (4,3), we have:
u = 4 - 3 = 1
v = 4 + 3 = 7

For (4,1), we have:
u = 4 - 1 = 3
v = 4 + 1 = 5

Therefore, the new limits of integration in the uv-plane are as follows:

1 ≤ u ≤ 3
1 ≤ v ≤ 7

Now, we need to find the Jacobian of the transformation. The Jacobian is calculated as the determinant of the matrix of partial derivatives of the transformation:

Jacobian(J) = ∂(x,y) / ∂(u,v) = | ∂x/∂u ∂x/∂v |
| ∂y/∂u ∂y/∂v |

To find the partial derivatives, we solve the transformation equations for x and y in terms of u and v:

x = (u + v) / 2
y = (v - u) / 2

Taking the partial derivatives:

∂x/∂u = 1/2
∂x/∂v = 1/2
∂y/∂u = -1/2
∂y/∂v = 1/2

Now, we can substitute these values into the Jacobian:

J = | 1/2 1/2 |
|-1/2 1/2 |

Taking the determinant, we get:
J = (1/2)(1/2) - (1/2)(-1/2) = 1/2

Now, we can convert the original integral into the new variables:

∫∫ ln((x-y)/(x+y)) dy dx = ∫∫ ln(u/v) * (1/2) dv du

With the new limits of integration: 1 ≤ u ≤ 3 and 1 ≤ v ≤ 7.

The final step is to evaluate the transformed integral in the uv-plane:

∫∫ ln(u/v) * (1/2) dv du from u=1 to u=3 and v=1 to v=7.

This can be evaluated using numerical methods or by finding an antiderivative with respect to v and then integrating with respect to u.