integrate f(x,y) = 1/y over R, where R is the region 0 < x < y, 0 <y < 1, x + y > .5

Aug 2010
4
0
Tried using y=0..1/4, x=1/2-y..1 and y=1/4..1, x=y..1 as integration limits and also tried the reverse order but there is always problem with integrating 1/y up to the x axis. Any help please? Thanks.
 

Ackbeet

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If you carefully examine R, I think you'll find that the x-axis doesn't even touch it. If you look here, you can see that R is the region enclosed by the triangle. So, using this information, can you write down the integral?
 
Aug 2010
4
0
I thought R was the region between the line y = x and the x axis (by the first 2 conditions) and then to the right of the line y = .5-x. Not sure what I am misreading if it isn't. ..
 

Ackbeet

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I think you need to be absolutely sure which side of a line is in the allowed region. To do that, pick a point that is obviously on one side of a line, and plug in the coordinates of that point into the inequality, and see if it holds. Does this help?
 

Ackbeet

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You got it now?
 
Apr 2010
384
153
Canada
Tried using y=0..1/4, x=1/2-y..1 and y=1/4..1, x=y..1 as integration limits and also tried the reverse order but there is always problem with integrating 1/y up to the x axis. Any help please? Thanks.


\(\displaystyle y = x \) and \(\displaystyle x + y = \frac{1}{2} \) intersect at \(\displaystyle ( \frac{1}{4} , \frac{1}{4} ) \)

Thus,

\(\displaystyle I = \int_{ \frac{1}{4} }^{1} \frac{1}{y} dy \int_{ \frac{1}{2} - y }^{ \frac{1}{4} } dx + \int_{ \frac{1}{4} }^{1} \frac{1}{y} dy \int_{ \frac{1}{4} }^{ y } dx \)
 
Aug 2010
4
0
Yeah i got it thanks a lot. I kept skipping that 0 < x < y .. used the wrong region. Embarrassing..lol
 

Ackbeet

MHF Hall of Honor
Jun 2010
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You're welcome. Hey, the experts mess up all the time. I wouldn't worry about it overmuch. Have a good one!