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UPDATE: Apparently a bounded region with 3 sides can expand to a bounded region with 4 sides after a transformation. I was not aware of this. Though it seems like a reasonable thing to happen.

I've been working on this problem for about 6 hours.

$$ I = \int^1_0dx\int^x_0\frac{(x+y)e^\frac{x+y}{x-y}}{x^2}dy$$

with the change of variables:

$ u = y/x$

$ v = x+y$

I calculated the jacobian determinant to be:

$|J| = \frac{v}{1+u^2}$

alot of stuff cancels nicely since,

$x^2 = \frac{v^2}{(1+u)^2}$

then I just becomes a simple integral over the new region Q

$$ I = \int\int_Qe^vdvdu$$

To get this region I noted that the integral region in x,y is a triangle which is the intersection of the following lines:

$x=1$

$y=0$

$x=y$

which correspond to the triangular region Q of u-v plane given by the intersection of:

$ u-v = -1$

$ v = 0 $

$ u = 1 $

However pluging these boundaries into the integral gives:

$$ I = \int^2_0e^vdv\int^1_{v-1}du$$

This clearly evaluates to $I = e^2 -3$ whereas the correct answer from the back of the textbook is $ I = e^2-e-1$

this is kind of drivin' me nuts, would appreciate any insight as to what i might be doing wrong.

share|improve this question
    
is my region Q too big or something? –  Timtam Aug 4 '12 at 9:04
    
Note that you can get displayed equations using double dollar signs instead of single dollar signs. It makes fractions and integrals easier to read (and also centres the equations). –  joriki Aug 4 '12 at 9:04
    
let me try that –  Timtam Aug 4 '12 at 9:06
    
By the way, the integration region is not the intersection of three lines but the region bounded by those three lines. If you wanted to describe it as an intersection, it would have to be the intersection of three half-planes bounded by those three lines. –  joriki Aug 4 '12 at 9:12
    
i'm having issue finding the bounded region in uv –  Timtam Aug 4 '12 at 9:14

1 Answer 1

$v=0$ should be $u=0$. ${}{}{}$

share|improve this answer
    
i thin you are right but why? –  Timtam Aug 4 '12 at 9:10
    
@Timtam: Because that's what you get when you plug $y=0$ into $u=y/x$. Why should it be $v=0$? –  joriki Aug 4 '12 at 9:11
    
good point hmm... –  Timtam Aug 4 '12 at 9:11
    
this does not form a closed region in uv what do i do? sorry to bother you like this but i'm confused. –  Timtam Aug 4 '12 at 9:13
    
@Timtam: Why do you need a closed region? Just integrate over that region, closed or not. –  joriki Aug 4 '12 at 9:16

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