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Rewrite Indefinite Integral in Terms of Elliptic Coordinates

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  • #1
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Problem:

Rewrite the indefinite integral ## \iint\limits_R\, (x+y) dx \ dy ## in terms of elliptic coordinates ##(u,v)##, where ## x=acosh(u)cos(v) ## and ## y=asinh(u)sin(v) ##.


Attempt at a Solution:

So would it be something like,

## \iint\limits_R\, (x+y) dx \ dy = \iint\limits_R\, [acosh(u)cos(v)+asinh(u)sin(v)] dx \ dy ##. I need to calculate ##dx## and ##dy## in terms of ##u## and ##v##, then substitute them in the integral, correct?

The only thing that comes to mind is the multivariable chain rule: ## \frac{dx}{dt} = \frac{\partial x}{\partial u} \frac{du}{dt}+\frac{\partial y}{\partial v} \frac{dv}{dt} ## or ## \frac{dy}{dt} = \frac{\partial y}{\partial u} \frac{du}{dt}+\frac{\partial y}{\partial v} \frac{dv}{dt} ##

Any help is appreciated. Thanks in advance.
 

Answers and Replies

  • #2
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I should note that the task is to merely rewrite the integral, not to evaluate it.
 
  • #3
vela
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Think Jacobian.
 
  • #4
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I'm sorry, can you be more specific?
 
  • #6
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I think I got it, hold on.
 
  • #7
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So I simply use ## \iint\limits_R\ f(x,y) dx \ dy = \iint\limits_{R^*}f(x(u,v),y(u,v)) det(J) du \ dv ##?
 
  • #8
vela
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Yup.
 
  • #9
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Okay thanks! I'm gonna LaTeX my work so you can see.
 
  • #10
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For the Jacobian I got ## J = (a^2sinh^2(u)cos^2(v)-a^2cosh(u)sinh(u)cos(v)sin(v))du \ dv ##. Is that correct?
 
  • #11
vela
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I don't get that.
 
  • #12
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Darn. I'll try agian.
 
  • #13
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I think I messed up a negative. Now I'm getting ## J= (asinh(u)cos(v))^2+(acosh(u)sin(v))^2 ##
 
  • #14
vela
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I'm not at my computer right now, but it looks like what I got.
 
  • #15
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Awesome! Thank you so much!
 

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