Area between two curves

  • Thread starter Saitama
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  • #1
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Homework Statement


Consider curves ##C_1: (y-x)=(x+y-\sqrt{2})^2## and ##C_2: (x+y-\sqrt{2})=(y-x)^2##, then the area between ##C_1## and ##C_2## is
A)1/2
B)1/3
C)1/4
D)None


Homework Equations





The Attempt at a Solution


Finding out the points of intersection would be a lot difficult here. And even if I find them, integration would be dirty. This is a question from my test paper and I suspect that it has an easy solution but I am unable to figure that out. :confused:

Any help is appreciated. Thanks!
 
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Answers and Replies

  • #2
Dick
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Homework Statement


Consider curves ##C_1: (y-x)=(x+y-\sqrt{2})^2## and ##C_2: (x+y-\sqrt{2})=(y-x)^2##, then the area between ##C_1## and ##C_2## is
A)1/2
B)1/3
C)1/4
D)None


Homework Equations




The Attempt at a Solution


Finding out the points of intersection would be a lot difficult here. And even if I find them, integration would be dirty. This is a question from my test paper and I suspect that it has an easy solution but I am unable to figure that out. :confused:

Any help is appreciated. Thanks!
Did you think about trying a change of variables?
 
  • #3
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92
Did you think about trying a change of variables?
Nope. How would I do that here? Substitute ##y-x## with ##t##?
 
  • #4
Dick
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Nope. How would I do that here? Substitute ##y-x## with ##t##?
Sure. Call x+y+sqrt(2)=s, x-y=t. Find the area is s,t coordinates. Don't forget the Jacobian factor.
 
  • #5
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92
....Jacobian factor.
Sorry, never heard of that before. :uhh:

Hmm...using the substitution, the question is similar to finding area between ##y=x^2## and ##y^2=x##. The area between them is 1/3. This is the answer given in the answer key. Thank you Dick! :smile:
 

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