Double intergral limits

  • Thread starter DIrtyPio
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  • #1
DIrtyPio
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Hi, I am doing my homework, but the book does not have any solutions so I can not verify my results. i wouldn't be a problem if I were 100% sure that I'm doing it right, but since I had some problems calculateing the limits of integration I'll ask you folks to help me! SO here is my first problem: SSdxdy over DELTA. DELTA is characterised by these four equations: x*y=1; x*y=4; y=x; y=2x. I think that the integral should look like this S sqrt(2)/2 -> 2 ( S x -> 2x dy) dx . Am I right?
 

Answers and Replies

  • #2
36,711
8,712
Have you drawn a graph of the region over which you're integrating? You need to do this in order to find intersections of the two lines and two hyperbolas. By the way, finding the limits of integration can be the hardest part in some integration problems.

You didn't say what you're trying to find with your integral. Is it the area of the region Delta? That will determine what the integrand looks like.

You wrote this as your integral:
S sqrt(2)/2 -> 2 ( S x -> 2x dy) dx
This is pretty much gibberish. I get that you're using 'S' to indicate an integral, but most of the rest makes no sense to me. What is "S sqrt(2)/2" supposed to mean? Are you integrating the constant sqrt(2)/2? With respect to what variable? Where did this constant come from?

What does "->" represent? I don't see how "S sqrt(2)/2" could be equal to what you have on the other side.
 
  • #3
eok20
200
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Hi, I am doing my homework, but the book does not have any solutions so I can not verify my results. i wouldn't be a problem if I were 100% sure that I'm doing it right, but since I had some problems calculateing the limits of integration I'll ask you folks to help me! SO here is my first problem: SSdxdy over DELTA. DELTA is characterised by these four equations: x*y=1; x*y=4; y=x; y=2x. I think that the integral should look like this S sqrt(2)/2 -> 2 ( S x -> 2x dy) dx . Am I right?

I don't think it's quite right. y=x is not always the lower bound on y (sometimes 1/x is) and y=2x is not always the upper bound on y (sometimes y=4/x is). I think you might have to break this up into a few integrals to get all of the different regions.
 
  • #4
DIrtyPio
18
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Ok, I figured out how should I do this integral. I made a varialbe change x*y=u and x/y=v so I found my new integral and the limits like this are pretty simple. And in the previous method I shuld have used 3 integrals devideing the main integral to 3 smaller ones which are orthogonal with respect to x or y axis.
 

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