Double Integral Setup for Finding Area with Given Bounds

[TheBestMilk]

Homework Statement

I have the bounds, 0≤y$_{1}$≤2, 0≤y$_{2}$≤1, and 2y$_{2}$≤y$_{1}$.

I now have a line u=y$_{1}$-y$_{2}$ and I'm trying to find the area such that y$_{2}$≥y$_{1}$-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:

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

$\int^{2u}_{0}$$\int^{y_{1}/2}_{y_{1}-u}dy_{2}dy_{1}$ = u$^{2}$ (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!

 

[vrmuth]

Homework Statement

I have the bounds, 0≤y$_{1}$≤2, 0≤y$_{1}$≤1, and 2y$_{2}$≤y$_{1}$.

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

what is the limit for y2 ?

 

[TheBestMilk]

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!

 

[vrmuth]

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 $\int^{u}_{0}$$\int^{\frac{y1}{2}}_{0}$dy$_{2}$dy$_{1}$ + $\int^{2u}_{u}$$\int^{\frac{y_{1}}{2}}_{y_{1}-u}$dy$_{2}$dy$_{1}$

 

[TheBestMilk]

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!

 

[vrmuth]

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 $\frac{-u^{2}}{2}$, have you worked it out? , if yes pls show me

 

[TheBestMilk]

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:

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

Thanks again for the help setting it up. I'm pretty sure I got the integral right.

 

[vrmuth]

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:

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

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 $\int^{2u}_{u} (-\frac{1}{2}y_{1} +u)dy_{1}$ , Am i right ?

 

[TheBestMilk]

Yep, you're right. Forgot to carry that negative sign through.