Define the Domain of a lamina as union of Type 1 and Type 2 regions?

[vmr101]

Homework Statement

Consider a lamina (two dimensional plate) with edges given by the lines y = sqrt(x) and
y = -x + 2x^(2), for which the density is given by P(x; y) = x.
(a) Define the domain of the lamina as the union of a Type 1 region and a Type 2 regions.
b) Calculate the mass.
Just looking for other peoples ideas for the domain, I have worked it out but am not sure if it is correct.
Thanks.

Homework Equations

The Attempt at a Solution

D = D(1) U D(2)
D(1) = 0≤ x ≤1/2 , -x+2x^(2)≤ y ≤0
D(2) = 0≤ y ≤1 , 1/4 (1 + sqrt(1+8y)≤ x ≤y^2

From this I got the mass equals 521/75.
I also tried calculating the mass from the domain just as a single typeI but got a mass of37/30.
Any help is appreciated. Thanks

 

[LCKurtz]

Without seeing the actual integrals you worked out, it's hard to help you. Did you do D(2) as a double integral? I don't think either of your answers are correct. Aren't your x limits backwards on D(2)? Show us your integrals.

 

[CBR600RR101]

I actually need some help with this question as well - when I did it I integrated X bound by the Type II region.

But my Type II region was, 0 ≤ y ≤ 1, y^2 ≤ x ≤ 1/4 (√(1+8y)+1)
I ended up getting a really small answer though which makes me think I did something wrong.
With all of that said, I can work out the integrals myself, I am more curious about what my bounds should be. I would even show my working out for my integration, If I knew how to represent all of that - but I can't even show my integral because I don't know how.

So to sum up my question, what are the bounds meant to be? Do we just use Type I or Type II to integrate P(x,y)?

Other bounds that have come to mind are:
0 ≤ x ≤ 1
√x ≤ y ≤ 2x^2-x

If I just knew the bounds this question would be going so much more better, I just need to be sure of what I am working with.

 

[HallsofIvy]

I actually need some help with this question as well - when I did it I integrated X bound by the Type II region.

But my Type II region was, 0 ≤ y ≤ 1, y^2 ≤ x ≤ 1/4 (√(1+8y)+1)
I ended up getting a really small answer though which makes me think I did something wrong.
With all of that said, I can work out the integrals myself, I am more curious about what my bounds should be. I would even show my working out for my integration, If I knew how to represent all of that - but I can't even show my integral because I don't know how.

So to sum up my question, what are the bounds meant to be? Do we just use Type I or Type II to integrate P(x,y)?

Other bounds that have come to mind are:
0 ≤ x ≤ 1
√x ≤ y ≤ 2x^2-x

You have this reversed. If x= 1/4, for example, \sqrt{x}= 1/2 while 2x^2- x= 1/8- 1/4= -1/8. For x between 0 and 1, 2x^2- x is less than \sqrt{x}.

If I just knew the bounds this question would be going so much more better, I just need to be sure of what I am working with.

\int_{x=0}^1\int_{y= 2x^2- x}^\sqrt{x} x dydx

 

[CBR600RR101]

You have this reversed. If x= 1/4, for example, \sqrt{x}= 1/2 while 2x^2- x= 1/8- 1/4= -1/8. For x between 0 and 1, 2x^2- x is less than \sqrt{x}.


\int_{x=0}^1\int_{y= 2x^2- x}^\sqrt{x} x dydx

Wow. I was just working on the question again, and I realized that I gave in the wrong information here so I came here to edit my previous post to fix it up - but you had already pointed it out. Thanks though! I ended up getting an answer of 7/30. I think I did it correctly, even though it is a small number.

 

[vmr101]

I found out the domain should actualy be:
D = D(1) U D(2)
D(1) = 0≤ x ≤1/2 , -x+2x^(2)≤ y ≤0
D(2) = 0≤ y ≤1 , y^2 ≤ x ≤ 1/4 (1 + sqrt(1+8y)

From this my answer is now: 313/5120 which is really small too ≈0.0611
I am not sure if i made a small mistake anywhere. Working attached as pdf.
I might have the wrong integral for the upper limit of the D(2) integral.
*Edit:made adjustment to second last line, forgot to put 0 into lower limit - as it doesn't just drop off.
Result still seems not quite right.

 

[vmr101]

! I ended up getting an answer of 7/30. I think I did it correctly, even though it is a small number.

Using the above integral with the question just as a Type 1 region, i also now get 7/30 for the mass.
But isn't it meant to be worked out using the union of the type1 and 2 regions (from part a) ? And shouldn't the ansers be the same?
Cheers.

 

[CBR600RR101]

Using the above integral with the question just as a Type 1 region, i also now get 7/30 for the mass.
But isn't it meant to be worked out using the union of the type1 and 2 regions (from part a) ? And shouldn't the ansers be the same?
Cheers.

I don't think so...
I thought Question 1a)
Was basically asking for the two domains, so say...
0≤ x ≤1/2, y^2 ≤ x ≤ 1/4 (1 + sqrt(1+8y)

But question 1b) you would have to work out the mass using one of the two regions. Can anyone confirm?

 

[vmr101]

Just had this answer from Simon.
"You can calculate the mass by splitting it up two regions, or if you can do it with one region, that is fine. No penalty either way. "
So its all good to use a single domain.
Cheers.