Double integral new coordinate system calculation

  • Thread starter cambo86
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  • #1
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Homework Statement


This is a 2 part question. I'm fine with the first part but the 2nd part I'm struggling with.

The first part asks us to calculate the double integral,
[itex]\int\int[/itex]Dx2dA
for, D = {(x,y)|0≤ x ≤1, x≤ y ≤1}
For this part I got an answer of 1/4.

For the 2nd part we introduce a new coordinate system for D,
x = 1-st, 0≤ s ≤1
y = s, 0≤ t ≤1

The Attempt at a Solution


[itex]\int\int[/itex]Dx2dA
= [itex]\int_0^1\int_0^1[/itex](1-st)2dt ds
= [itex]\int_0^1\int_0^1[/itex](1-2st+s2t2)dt ds
= [itex]\int_0^1[/itex]t-st2+[itex]\frac{1}{3}[/itex]s2t3 ds, from t=0 to t=1
= [itex]\int_0^1[/itex]1-s+[itex]\frac{1}{3}[/itex]s2 ds
= s-[itex]\frac{1}{2}[/itex]s2+[itex]\frac{1}{9}[/itex]s3, from s=0 to s=1
= 1-[itex]\frac{1}{2}[/itex]+[itex]\frac{1}{9}[/itex]
= [itex]\frac{11}{18}[/itex]

I feel like,
dA [itex]\neq[/itex] dt ds
I'm not sure what it equals though. I thought I could use polar coordinates but I don't have a constant radius.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


This is a 2 part question. I'm fine with the first part but the 2nd part I'm struggling with.

The first part asks us to calculate the double integral,
[itex]\int\int[/itex]Dx2dA
for, D = {(x,y)|0≤ x ≤1, x≤ y ≤1}
For this part I got an answer of 1/4.
Okay, so you got this wrong. In more detail this is
[tex]\int_{x= 0}^1\int_{y= x}^1 x^2dydx= \int_0^1 (1- x)x^2 dx[/tex]
What do you get for that?

For the 2nd part we introduce a new coordinate system for D,
x = 1-st, 0≤ s ≤1
y = s, 0≤ t ≤1

The Attempt at a Solution


[itex]\int\int[/itex]Dx2dA
= [itex]\int_0^1\int_0^1[/itex](1-st)2dt ds
You can't just replace "dydx" with "dtds". Since you are now measuring distance differently, the area will be measured differently. Use the Jacobian:
[tex]\left|\begin{array}{cc}\frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t}\end{array}\right|[/tex][tex]= \left|\begin{array}{cc}-s & -t \\ 0 & 1 \end{array}\right|= -s[/tex]

So you have to replace "dxdt" with -s dsdt as well as replacing [itex]x^2[/itex] with [itex](1- st)^2[/itex].

= [itex]\int_0^1\int_0^1[/itex](1-2st+s2t2)dt ds
= [itex]\int_0^1[/itex]t-st2+[itex]\frac{1}{3}[/itex]s2t3 ds, from t=0 to t=1
= [itex]\int_0^1[/itex]1-s+[itex]\frac{1}{3}[/itex]s2 ds
= s-[itex]\frac{1}{2}[/itex]s2+[itex]\frac{1}{9}[/itex]s3, from s=0 to s=1
= 1-[itex]\frac{1}{2}[/itex]+[itex]\frac{1}{9}[/itex]
= [itex]\frac{11}{18}[/itex]

I feel like,
dA [itex]\neq[/itex] dt ds
Yes, that is the difficulty. Use the Jacobian, as I said before.

I'm not sure what it equals though. I thought I could use polar coordinates but I don't have a constant radius.
 
  • #3
25
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Sorry, the lower bound on y for the first part should have been 1-x but I should be able to work through the rest of the reply and get an answer. Thank you for the detailed reply.
 

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