Evaluate double integral using transformations

In summary: The only thing is that the limits are not correct. In summary, the conversation is about finding the limits for a double integral after transforming the variables and plotting the corresponding graphs. The final answer is \frac{1}{2}ln (1+a^2) and the correct limits for the integral are u varies from u=0 to u=a and v varies from v=0 to v=u.
  • #1
DryRun
Gold Member
838
4
Homework Statement
http://s2.ipicture.ru/uploads/20120107/vVVkUT7f.jpg

The attempt at a solution
I plotted the graph x-y:
http://s2.ipicture.ru/uploads/20120107/ja3V9aSV.jpg

[itex]y=\frac{1}{2}(u+v)[/itex] and [itex]x=\frac{1}{2}(u-v)[/itex]

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

Now, to find the transformed limits, i plot another graph u-v:
http://s2.ipicture.ru/uploads/20120107/RrA8UKQ1.jpg

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:
[tex]4\int^a_0 \int^a_u \frac{1}{1+u^2}\,.dvdu[/tex]
The answer to that integral above is ultimately incorrect, hence why i have no idea where i messed up. The answer i got is:
[tex]2a\arctan a - ln(1+a^2)[/tex]
 
Last edited:
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  • #2
How can your variable transformation turn a triangle into a quadrilateral? Which point in the triangle is going to map to (2a,0)?
 
  • #3
OK, i have fixed the graph of u-v:
http://s2.ipicture.ru/uploads/20120108/cvc1aJm9.jpg

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

[tex]2\int^a_0 \int^u_0 \frac{1}{1+u^2}\,.\frac{1}{2}.dvdu[/tex]

The answer is [itex]\frac{1}{2}ln (1+a^2)[/itex]
 
  • #4
What did you got for the first integral with respect to v? Take care, the integrand does not depend on v!

ehild
 
  • #5
For the first integral w.r.t.v. i got (after putting in the limits for v and then u):
[tex]\frac{u}{1+u^2}\,.\frac{1}{2}du=\frac{1}{2}ln (1+a^2)[/tex]
 
Last edited:
  • #6
sharks said:
For the first integral w.r.t.v. i got (after putting in the limits for v and then u):
[tex]\frac{u}{1+u^2}\,.\frac{1}{2}du=\frac{1}{2}ln (1+a^2)[/tex]

Your answer look ok to me.
 

1. What is a double integral?

A double integral is a type of integral in calculus that involves integrating a function of two variables over a region in a two-dimensional plane. It represents the sum of infinitesimal areas under a surface or over a region in the plane.

2. What does it mean to evaluate a double integral using transformations?

Evaluating a double integral using transformations involves changing the variables in the integral to simplify the integration process. This is done by using a transformation function to map the original region of integration onto a simpler region, making the integration easier to solve.

3. When should I use transformations to evaluate a double integral?

Transformations are useful when the original region of integration is complex or difficult to integrate over. By using a transformation, the integral can be rewritten in terms of new variables which can make the integration process simpler and more manageable.

4. What are some common transformations used for double integrals?

Some common transformations used for double integrals include polar coordinates, Cartesian coordinates, and parametric equations. These transformations can help to simplify the integral by reducing the number of variables or by transforming the region of integration into a more manageable shape.

5. How do I know if I have set up the correct transformation for a double integral?

To ensure that the transformation is correct, you can check that the new limits of integration match the boundaries of the transformed region. Additionally, you can also check that the Jacobian determinant of the transformation is not equal to zero, as this would result in an invalid transformation.

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