Evaluating the integral by changing the order

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Homework Statement


We are told to change the order of the given double integral. It is as follows:-
[tex]\int^{k}_{o}[/tex][tex]\int^{k+\sqrt{k^{2}-y^{2}}}_{-k+\sqrt{k^{2}-y^{2}}}[/tex] f(x,y)dxdy


Homework Equations


The answer that I get after sketching the limits and finding the region is different from the answer given in my book. I would like to ascertain if its me who is wrong or the book(since the book has quite a lot of mistakes).


The Attempt at a Solution


First I sketched the given limits, and I got the upper half of the two given circles since y=0 is also a limit. Then I got a region bounded by the limits. Since it is kinda hard to describe, I am just writing the answer that I am getting after changing the order. It is as follows:-
[tex]\int^{0}_{-k}[/tex][tex]\int^{k}_{\sqrt{k^{2}-(x+k)^{2}}}[/tex] f(x,y)dydx + [tex]\int^{k}_{0}[/tex][tex]\int^{k}_{\sqrt{k^{2}-(x-k)^{2}}}[/tex] f(x,y)dydx

But the answer given in my textbook has three parts, thereby indicating that the region I got is wrong.
Any help pointing me in the right direction would be helpful.
 

Answers and Replies

  • #2
gabbagabbahey
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Hmmm....what was the books answer?
 
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  • #3
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Hmmm....what was the books answer?

The book has found some weird regions, and the whole region is divided into three parts, the first of which coincides with the first part of my answer. The whole answer is as follows:-

[tex]\int^{0}_{-k}[/tex][tex]\int^{k}_{\sqrt{k^{2}-(x+k)^{2}}}[/tex] f(x,y) dydx + [tex]\int^{k}_{0}[/tex][tex]\int^{k}_{0}[/tex] f(x,y) dydx + [tex]\int^{2k}_{k}[/tex][tex]\int^{\sqrt{k^{2}-(x-k)^{2}}}_{0}[/tex] f(x,y) dydx.


I have no idea how the above answer was formulated. My experience with double integrals and common logic should dictate my original answer, but of course, I may be wrong. Any help?

Thanks in advance.
 
  • #4
gabbagabbahey
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I agree with the textbook.

The region you've used is bounded by the upper-right part of the circle [itex](x+k)^2+y^2=k^2[/itex], and the upper-left part of the circle [itex](x-k)^2+y^2=k^2[/itex] (as well as [itex]y=0[/itex] and [itex]y=k[/itex]). However, the original integral covers a region bounded by the upper-right part of the circle [itex](x+k)^2+y^2=k^2[/itex], and the upper-right part of the circle [itex](x-k)^2+y^2=k^2[/itex] (as well as [itex]y=0[/itex] and [itex]y=k[/itex]).
 
  • #6
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I agree with the textbook.

The region you've used is bounded by the upper-right part of the circle [itex](x+k)^2+y^2=k^2[/itex], and the upper-left part of the circle [itex](x-k)^2+y^2=k^2[/itex] (as well as [itex]y=0[/itex] and [itex]y=k[/itex]). However, the original integral covers a region bounded by the upper-right part of the circle [itex](x+k)^2+y^2=k^2[/itex], and the upper-right part of the circle [itex](x-k)^2+y^2=k^2[/itex] (as well as [itex]y=0[/itex] and [itex]y=k[/itex]).

I dont get it completely...

The question given just describes a double integral, and the region. Sketching the region gives me the upper half of the two circles. Now the region that is bounded by the two curves is the region in between the two circles, right? I dont get how the upper right part of the circle [itex](x-k)^2+y^2=k^2[/itex] is supposed to count, since the limits do not mention anything like that. Also, the original integral just mentions the two circles and y=k.
Kindly elaborate it a little bit.

One more question:-
If the books answer is correct, then on the basis of symmetry, I can have the upper left of the circle [itex](x-k)^2+y^2=k^2[/itex] and the upper right and left of the circle [itex](x+k)^2+y^2=k^2[/itex], right(by changing the appropriate limits in the answer, of course)?
 
  • #7
gabbagabbahey
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I dont get it completely...

I dont get how the upper right part of the circle [itex](x-k)^2+y^2=k^2[/itex] is supposed to count, since the limits do not mention anything like that. Also, the original integral just mentions the two circles and y=k.
Kindly elaborate it a little bit.

Let's take a look at the limits of your original integral; specifically, what does the upper [itex]x[/itex] limit, [itex]k+\sqrt{k^2-y^2}[/itex], vary from/to as [itex]y[/itex] ranges from zero to [itex]k[/itex]? What part of the circle does that curve describe?

One more question:-
If the books answer is correct, then on the basis of symmetry, I can have the upper left of the circle [itex](x-k)^2+y^2=k^2[/itex] and the upper right and left of the circle [itex](x+k)^2+y^2=k^2[/itex], right(by changing the appropriate limits in the answer, of course)?

That's only true if [itex]f(x,y)=f(-x,y)[/itex] (i.e. if [itex]f[/itex] is also symmetric about the y-axis)
 
  • #8
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Let's take a look at the limits of your original integral; specifically, what does the upper [itex]x[/itex] limit, [itex]k+\sqrt{k^2-y^2}[/itex], vary from/to as [itex]y[/itex] ranges from zero to [itex]k[/itex]? What part of the circle does that curve describe?

OK.....I think I get it now. Thanks a lot for your help, for without it I would have never known where I was going wrong.
But the question was tricky. If I didnt see the answer in the book, I would still be believing that my original answer was correct!!!

Once again thanks for all your efforts. Am glad to be a part of the Physics Forums community.
 
  • #9
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Oh yeah....one last thing.

Do I have to mark this thread as "solved" or something like that? If yes, then from where? I couldnt find an option(maybe I didnt look too well?).
 
  • #10
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God I hate it when the surface isn't the sphere at infinity, and the contents haven't decayed exponentially...
 

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