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Double Integral Setup

  • #1

Homework Statement


I have the bounds, 0≤y[itex]_{1}[/itex]≤2, 0≤y[itex]_{2}[/itex]≤1, and 2y[itex]_{2}[/itex]≤y[itex]_{1}[/itex].

I now have a line u=y[itex]_{1}[/itex]-y[itex]_{2}[/itex] and I'm trying to find the area such that y[itex]_{2}[/itex]≥y[itex]_{1}[/itex]-u.

The integral comes down to two parts, the first of which I'm stuck on (when 0≤y1≤1). I'm pretty sure I have one way setup correctly, when I take the integral of dy2 first and then dy1, but for some reason I cannot get the double integral of dy1dy2 to workout properly. This is what I have the setups as:

[itex]\int^{u}_{0}[/itex][itex]\int^{u+y_{2}}_{2y_{2}}dy_{1}dy_{2}[/itex] = u[itex]^{2}[/itex]/2 (This is the one I believe is correct)

[itex]\int^{2u}_{0}[/itex][itex]\int^{y_{1}/2}_{y_{1}-u}dy_{2}dy_{1}[/itex] = u[itex]^{2}[/itex] (This is the one I cannot get to match the first)

Any insight would be very much appreciated. I'm not sure what bounds I'm messing up, but I'm sure that's it.

Thanks!
 
Last edited:

Answers and Replies

  • #2
80
0

Homework Statement


I have the bounds, 0≤y[itex]_{1}[/itex]≤2, 0≤y[itex]_{1}[/itex]≤1, and 2y[itex]_{2}[/itex]≤y[itex]_{1}[/itex].

I now have a line u=y[itex]_{1}[/itex]-y[itex]_{2}[/itex] and I'm trying to find the area such that y[itex]_{2}[/itex]≥y[itex]_{1}[/itex]-u.
what is the limit for y2 ?
 
  • #3
Ah, it seems I wrote the original bounds incorrectly. Sorry about that.

The bounds are actually: 0≤y2≤1, 0≤y1≤2, and 2y2≤y1.

Thanks!
 
  • #4
80
0
Ah, it seems I wrote the original bounds incorrectly. Sorry about that.

The bounds are actually: 0≤y2≤1, 0≤y1≤2, and 2y2≤y1.

Thanks!
its [itex]\int^{u}_{0}[/itex][itex]\int^{\frac{y1}{2}}_{0}[/itex]dy[itex]_{2}[/itex]dy[itex]_{1}[/itex] + [itex]\int^{2u}_{u}[/itex][itex]\int^{\frac{y_{1}}{2}}_{y_{1}-u}[/itex]dy[itex]_{2}[/itex]dy[itex]_{1}[/itex]
 
  • #5
Thank you very much. That makes perfect sense to me. Next time I'll draw the graph better to scale, which I think was throwing off where the y2 = y1 - u line was above the xaxis.

Thanks!
 
  • #6
80
0
Thank you very much. That makes perfect sense to me. Next time I'll draw the graph better to scale, which I think was throwing off where the y2 = y1 - u line was above the xaxis.

Thanks!
you are welcome !
but though i figured out what was wrong and set up the limits correctly i'm getting [itex]\frac{-u^{2}}{2}[/itex], have you worked it out? , if yes pls show me
 
  • #7
Hey,

Sorry it's been a few days. I had a friend in from out of town and a test in another class I was studying for. This is what I managed to come up with:

[itex]\int^{u}_{0}\int^{y_{1}/2}_{0} dy_{2}dy_{1} + \int^{2u}_{u}\int^{y_{1}/2}_{y_{1}-u} dy_{2}dy_{1}[/itex]
= [itex]\int^{u}_{0} \frac{y_{1}}{2}dy_{1} + \int^{2u}_{u} (-\frac{1}{2}y_{1} -u)dy_{1}[/itex]
= [itex]\frac{1}{4}u^{2} + (-\frac{1}{4}(4u^{2}) + 2u^{2} + \frac{1}{4}u^{2} - u^{2})[/itex]
=[itex]\frac{1}{4}u^{2} + \frac{1}{4}u^{2}[/itex]
=[itex]\frac{u^{2}}{2}[/itex]

Thanks again for the help setting it up. I'm pretty sure I got the integral right.
 
  • #8
80
0
Hey,

Sorry it's been a few days. I had a friend in from out of town and a test in another class I was studying for. This is what I managed to come up with:

[itex]\int^{u}_{0}\int^{y_{1}/2}_{0} dy_{2}dy_{1} + \int^{2u}_{u}\int^{y_{1}/2}_{y_{1}-u} dy_{2}dy_{1}[/itex]
= [itex]\int^{u}_{0} \frac{y_{1}}{2}dy_{1} + \int^{2u}_{u} (-\frac{1}{2}y_{1} -u)dy_{1}[/itex]
= [itex]\frac{1}{4}u^{2} + (-\frac{1}{4}(4u^{2}) + 2u^{2} + \frac{1}{4}u^{2} - u^{2})[/itex]
=[itex]\frac{1}{4}u^{2} + \frac{1}{4}u^{2}[/itex]
=[itex]\frac{u^{2}}{2}[/itex]

Thanks again for the help setting it up. I'm pretty sure I got the integral right.
You are welcome , it's nice to help each other so same to you :) , I think the 2nd integral must be [itex] \int^{2u}_{u} (-\frac{1}{2}y_{1} +u)dy_{1}[/itex] , Am i right ?
 
  • #9
Yep, you're right. Forgot to carry that negative sign through.
 

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