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

1. May 11, 2012

vmr101

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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

2. May 12, 2012

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.

3. May 14, 2012

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.

4. May 14, 2012

HallsofIvy

Staff Emeritus
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 $\sqrt{x}$.

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

5. May 14, 2012

CBR600RR101

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.

6. May 14, 2012

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 doesnt just drop off.
Result still seems not quite right.

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Last edited: May 14, 2012
7. May 14, 2012

vmr101

Using the above integral with the question just as a Type 1 region, i also now get 7/30 for the mass.
But isnt it meant to be worked out using the union of the type1 and 2 regions (from part a) ? And shouldnt the ansers be the same?
Cheers.

8. May 14, 2012

CBR600RR101

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?

9. May 15, 2012