# Evaluate double integral using transformations

1. Jan 7, 2012

### sharks

The problem statement, all variables and given/known data

The attempt at a solution
I plotted the graph x-y:

$y=\frac{1}{2}(u+v)$ and $x=\frac{1}{2}(u-v)$

So, after finding the Jacobian, the double integral becomes: $\int\int \frac{1}{1+u^2}\,.\frac{1}{2}dvdu$

Now, to find the transformed limits, i plot another graph u-v:

From the graph above, the shaded area is what i need to find. Since the area is symmetrical about the v-axis, i decided to evaluate the lower half of the right-hand side and then multiply by 4.

So, the limits for the lower right-half (meaning the section found under the line v=a) is:
For u fixed, v varies from v=u to v=a
u varies from u=0 to u=a

The double integral becomes:
$$4\int^a_0 \int^a_u \frac{1}{1+u^2}\,.dvdu$$
The answer to that integral above is ultimately incorrect, hence why i have no idea where i messed up. The answer i got is:
$$2a\arctan a - ln(1+a^2)$$

Last edited: Jan 7, 2012
2. Jan 7, 2012

### Dick

How can your variable transformation turn a triangle into a quadrilateral? Which point in the triangle is going to map to (2a,0)?

3. Jan 7, 2012

### sharks

OK, i have fixed the graph of u-v:

Since the area of region is symmetrical about the u-axis, taking limits of upper triangular section and then multiply by 2.

The limits:
For u fixed, v varies from v=0 to v=u
u varies from u=0 to u=a

$$2\int^a_0 \int^u_0 \frac{1}{1+u^2}\,.\frac{1}{2}.dvdu$$

The answer is $\frac{1}{2}ln (1+a^2)$

4. Jan 8, 2012

### ehild

What did you got for the first integral with respect to v? Take care, the integrand does not depend on v!

ehild

5. Jan 9, 2012

### sharks

For the first integral w.r.t.v. i got (after putting in the limits for v and then u):
$$\frac{u}{1+u^2}\,.\frac{1}{2}du=\frac{1}{2}ln (1+a^2)$$

Last edited: Jan 9, 2012
6. Jan 9, 2012