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