# Rewrite Indefinite Integral in Terms of Elliptic Coordinates

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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.

I should note that the task is to merely rewrite the integral, not to evaluate it.

Think Jacobian.

I'm sorry, can you be more specific?

I think I got it, hold on.

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 ##?

Yup.

Okay thanks! I'm going to LaTeX my work so you can see.

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?

I don't get that.

Darn. I'll try agian.

I think I messed up a negative. Now I'm getting ## J= (asinh(u)cos(v))^2+(acosh(u)sin(v))^2 ##

I'm not at my computer right now, but it looks like what I got.

Awesome! Thank you so much!