# Rewrite Indefinite Integral in Terms of Elliptic Coordinates

wifi
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.

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

Staff Emeritus
Homework Helper
Think Jacobian.

wifi
I'm sorry, can you be more specific?

Staff Emeritus
Homework Helper
wifi
I think I got it, hold on.

wifi
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 ##?

Staff Emeritus
Homework Helper
Yup.

wifi
Okay thanks! I'm gonna LaTeX my work so you can see.

wifi
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?

Staff Emeritus
Homework Helper
I don't get that.

wifi
Darn. I'll try agian.

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

Staff Emeritus