Finding limits of double integral

In summary, the problem is to solve an integral using a coordinate transformation. After attempting to find the limits of integration through a graphical approach, it is determined that the limits for v should be [-∞,0] instead of [0,∞]. However, considering the points given, it is concluded that the limits for v should be [0,-∞]. The correct approach is to view both the XY plane and the UV plane as complex planes, which reveals that the 1st quadrant maps to the upper half plane (v≥0). Therefore, the limits of integration should be v range [0,∞) and u range (-∞,+∞).
  • #1
unscientific
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Homework Statement



The problem is to solve the integral. First I did coordinate transformation by finding jacobian = (1/4)(x2 + y2).

The problem is, I do not know the limits of integration after transformation...I tried using a graphical approach: by considering 2 cases: y>x and y<x and adding both integrals.

For y < x it is straightforward, the limits of u: [0,∞] and v: [0,∞].

However, for y > x: limits of u: [0,∞] which is easy as x,y>0 so v>0.

But the limits for v i obtained: v: [0,-∞].

When I add both integrals they sum to zero.

My suspicion is that the limits of v for y > x should be [-∞,0] and not [0,∞]. But this doesn't make sense as considering the following points:

1. y > x
2. y, x are forever increasing

Combining both, it implies " x and y are increasing with y increasing faster than x". This implies v goes from 0 to -∞..

Homework Equations


The Attempt at a Solution



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  • #2
As x,y >= 0, v = 2xy >= 0 in both cases.
The integral is symmetric with respect to y>x and y<x, you can restrict your integral to one of those cases and multiply with 2 afterwards.
I don't think you need those cases at all.
 
  • #3
mfb said:
As x,y >= 0, v = 2xy >= 0 in both cases.
The integral is symmetric with respect to y>x and y<x, you can restrict your integral to one of those cases and multiply with 2 afterwards.
I don't think you need those cases at all.

I know that's possible as 1/(1+u2) is an even function, but I don't like using this approach as I feel that I won't always be this lucky.

I'm interested in finding the limits of du and dv.
 
  • #4
If you write z = x + iy = r e then z2 = u+iv = r2e2iθ. Thus the transformation from the XY plane to the UV plane fans out the first quadrant, doubling all the polar angles, to cover the first two quadrants.
 
  • #5
haruspex said:
If you write z = x + iy = r e then z2 = u+iv = r2e2iθ. Thus the transformation from the XY plane to the UV plane fans out the first quadrant, doubling all the polar angles, to cover the first two quadrants.

I'm not sure what you mean...Usually to find out the limits we consider the boundary curves and work out its shape in the R' region... using the geometric approach.

But in this case I'm not sure what geometric approach should be taken.
 
  • #6
What does the u-axis correspond to in the xy-plane?
 
  • #7
vela said:
What does the u-axis correspond to in the xy-plane?

u = x2 - y2 = (x+y)(x-y)

I would love to say it's the distance from origin squared (x2 + y2) but sadly it's not...

I have no idea.
 
  • #8
Try again. Note that the u-axis is v=0, just like the x-axis is y=0.
 
  • #9
unscientific said:
I'm not sure what you mean...Usually to find out the limits we consider the boundary curves and work out its shape in the R' region... using the geometric approach.

But in this case I'm not sure what geometric approach should be taken.
I'm saying that if you think of both the XY plane and the UV plane as being complex planes, the geometry of the mapping becomes clear. (x+iy)2 = u+iv. A radial line from the origin, polar angle theta say, maps to a radial line from the origin at angle 2 theta. (Radial distances become squared.) Thus the 1st quadrant maps to the upper half plane (v≥0).
 
  • #10
vela said:
Try again. Note that the u-axis is v=0, just like the x-axis is y=0.

First, we have to assume that u and v are orthogonal...let's say it is.

For v = 0, u = x2 or -y2

For u = 0, x = y and v = 2x2 or 2y2

doesn't this make the limits of v[0,∞] and u[0,-∞] or u[0,∞] which are the cases i mentioned earlier...Summing them would give a result of 0.
 
  • #11
I think your problem is that you're trying to map x and y ranges individually, in some sense, to u and v ranges. That's the only way I can understand how you get ranges like "u[0,-∞] or u[0,∞]". It doesn't work like that. You need to determine the set of points (u,v) that are in the region, then express that as ranges for integration purposes. The region is v >= 0, which you can express as v range [0,∞) and u range (-∞,+∞).
 

Related to Finding limits of double integral

1. What are double integrals and why do we need to find their limits?

Double integrals are a type of mathematical operation used to find the area under a two-dimensional surface. The limits of a double integral refer to the boundaries of the region over which the integration is performed. Finding the limits is necessary in order to accurately calculate the area under the surface.

2. How do we determine the limits of a double integral?

The limits of a double integral are determined by the boundaries of the region over which the integration is performed. This region can be defined by the given function and/or by the shape of the surface. In some cases, the limits may also be determined by the specific problem or scenario being studied.

3. Are there any specific techniques or methods for finding the limits of a double integral?

Yes, there are several techniques that can be used to find the limits of a double integral. These include using geometric reasoning, converting the integral to polar coordinates, and using symmetry to simplify the limits. It is important to carefully analyze the given function and problem in order to determine the most appropriate method for finding the limits.

4. Can the limits of a double integral be negative?

Yes, the limits of a double integral can be negative. This can occur when the region over which the integration is performed extends into the negative x or y-axis. It is important to carefully consider the orientation of the region and the given function in order to determine the correct limits for the integral.

5. What are the practical applications of finding limits of double integrals?

Finding limits of double integrals has various practical applications, such as in physics, engineering, and economics. It can be used to calculate the area of a two-dimensional surface, to find the center of mass of an object, and to determine the volume of a three-dimensional object. It is a fundamental tool in many fields of science and mathematics.

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