# Rewrite Indefinite Integral in Terms of Elliptic Coordinates

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.

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

vela
Staff Emeritus
Homework Helper
Think Jacobian.

I'm sorry, can you be more specific?

vela
Staff Emeritus
Homework Helper

I think I got it, hold on.

So I simply use $\iint\limits_R\ f(x,y) dx \ dy = \iint\limits_{R^*}f(x(u,v),y(u,v)) det(J) du \ dv$?

vela
Staff Emeritus
Homework Helper
Yup.

Okay thanks! I'm gonna 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?

vela
Staff Emeritus
Homework Helper
I don't get that.

Darn. I'll try agian.

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

vela
Staff Emeritus