Calc Change of Var: Non-Graphical Limits of Integration

[Krypton]

How to calculate, non-graphically, the new limits of integration when change of variables r done to a double integral?

 

[Hootenanny]

How to calculate, non-graphically, the new limits of integration when change of variables r done to a double integral?

Perhaps if you posted a specific example, we could help you out?

 

[Krypton]

Plz wait 4 a day., its midnight in INDIA n ma brain had almost slept. I 'ill b vhappy if u could solv ma prob... Thanks in advance!

 

[gamesguru]

Wow...what kind of english do they learn in India?

To answer the question, it always helps to draw a graph but if you post a specific example we'll show you how to use some algebra to find the new limits.

 

[Krypton]

"inte[x=a to b] [y=c to d] f(x,y) dxdy = inte[r=sqrt(a^2+c^2) to r =sqrt(b^2+d^2)] [ß=arctan(c/a) to arctan(d/b)] f(rcosß,rsinß) rdrdß "
is this expression currect ?

 

[Hootenanny]

Edit: I see that I was mistaken.

 

[HallsofIvy]

\int_{x=a}^b\int_{y= c^d} f(x,y)dxdy
\int_{r=\sqrt{a^2+ c^2}}^{\sqrt{b^2+ d^2}}\int_{\theta= arctan(c/a)}^{arctan(d/b)} f(rcos(\theta}, rsin(\theta)) r dr d\theta

No those are not the same. The first is over a rectangle in the plane while the second is over a portion of an annulus- that is, over the region between two circles bounded by two angles. Those will always be what you get if your limits of integration in rectangular and polar coordinates are all constants.

In general, the limits of integration for the "inside" integral will depend on the other variable. If you really had to put that first integral into polar coordinates (not a good idea because of the lack of symmetry) you would probably do best to break it into four areas: draw lines from the origin to the four vertices and do a separate integral on each of those areas in order to avoid the "discontinuity" at the corners. For example, the first integral might be with \theta between arctan(c/b) and arctan(c/a) to get the region between those two lower vertices (I am assuming that d-c> b- a. Otherwise you will "hit" the vertex (b,d) before (a,c).) Now on each line between those, r must go from the lower line, y= c, to the vertical line x= b. In polar coordinates, that is r sin(\theta)= c to r cos(\theta)= b. That means that r varies from c/sin(\theta) to b/cos(\theta).
The first of the four integrals would be
\int_{\theta= arctan(d/b)}^{arctan(c/a)}\int_{r= c/sin(\theta)}^{b/cos(\theta)}f(r cos(\theta),r sin(\theta)) r dr d\theta

How you change a double integral from one set of coordinates to another depends very strongly on the specific geometry of the situation.

 

[Krypton]

Thanx ! , both u geniees ...

 

[Krypton]

Plz show me how u arrived at the new limits '' "

 

[HallsofIvy]

Plz show me how u arrived at the new limits '' "

I DID:

integral might be with \theta between arctan(c/b) and arctan(c/a) to get the region between those two lower vertices (I am assuming that d-c> b- a. Otherwise you will "hit" the vertex (b,d) before (a,c).)

Those values are the angle of the straight line from the origin to each of the two points (b,c), (a,c).

Now on each line between those, r must go from the lower line, y= c, to the vertical line x= b. In polar coordinates, that is r sin(\theta)= c to r cos(\theta)= b. That means that r varies from c/sin(\theta) to b/cos(\theta).

 

[Krypton]

If r varies frm c/sinß to b/cosß ,
what are the values of ß in both the cases ?

 

[HallsofIvy]

It is a variable. I said before that taking constants for all limits of integration will give only very limited areas- in the case of polar coordinates it would be a part of an annulus. For more general areas- and in polar coordinates a rectangle is a "general area", only the limits of the "outer integral" must be constants. The limits of integration of the "inner integrals" may be functions of the variiables that still remain.