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Evaluating Double Integrals: Switching the Order of Integration

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Alw
#1
Oct27-07, 03:23 PM
P: 8
1. The problem statement, all variables and given/known data

Evaluate the double integral by changing the order of integration in the iterated integral and evalutating the resulting iterated integral.

2. Relevant equations

[tex]\int[/tex][tex]^{1}_{0}[/tex] [tex]\int[/tex][tex]^{1}_{x}[/tex] cos(x/y)dydx

3. The attempt at a solution

I know how to solve a double integral after i've switched the order of integration, i'm having trouble with the acutal switching part The book we are using has one example in it regarding this, and it isn't very clear. If anyone would mind walking me through how to switch the order of integration, that'd be great

Thanks in advance,
-Andy

edit: The text for the integrals didnt come out well, to make it more clear, its the integral from 0 - to - 1 and the integral
from x - to - 1
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Gib Z
#2
Oct28-07, 12:58 AM
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Why must you change the order? The integral is easily solvable as it is.
siddharth
#3
Oct28-07, 02:21 AM
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Quote Quote by Gib Z View Post
Why must you change the order? The integral is easily solvable as it is.
Is it? I can't see how.

I know how to solve a double integral after i've switched the order of integration, i'm having trouble with the acutal switching part The book we are using has one example in it regarding this, and it isn't very clear. If anyone would mind walking me through how to switch the order of integration, that'd be great
It always helps if you sketch the area over which you're integrating. Notice that the limits in x are from 0 to 1.

So, the area over which you're integrating is bounded in the x direction by the lines x=0 and x=1. Also, since the limits in y are from x to 1, the boundaries in the y direction are the lines y=x and y=1. Can you sketch the area now? From this, can you figure out how to switch the order of integration?

Gib Z
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Oct28-07, 02:27 AM
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Evaluating Double Integrals: Switching the Order of Integration

Quote Quote by siddharth View Post
Is it? I can't see how.
I read it wrong :( I seemed to read cos (y/x) >.<" Damn
Alw
#5
Oct28-07, 10:12 AM
P: 8
Ok, thanks! so if i'm not mistaken then, the new equation is:

[tex]\int[/tex][tex]^{1}_{0}[/tex] [tex]\int[/tex] [tex]^{y}_{0}[/tex] cos(x/y)dxdy ?
HallsofIvy
#6
Oct28-07, 12:16 PM
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Yes. In your original integral x ranged from 0 to 1 and, for each x, y ranged from x to 1. That is the triangle with vertices (0,0), (1,1) and (0, 1). In the opposite order, to cover that triangle, y must range from 0 to 1 and, for each y, x must rage from 0 to y.


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