Let $E = (0,1) \times (0,1) \in \mathbb{R}^2$. Let the function $\Phi : E \to \mathbb{R}^2$ be given by $$ \Phi(x,y) = (e^{x+y}, e^{x-y}), \qquad \text{with } (x,y) \in E. $$ Determine the area of $\Phi(E)$.
What I have tried so far:
Let $\Phi(u,v) = (x,y) = (e^{u+v}, e^{u-v})$, and let $f(x,y) = f(e^{u+v}, e^{u-v})$.
We know the change of variables formula $$ \iint_{\Phi(E)} f(x,y) dA = \iint_E f(e^{u+v}, e^{u-v}) |\det(D\Phi)| \, dA. $$ With $D\Phi$ the Jacobian of $\Phi$, so $D\Phi = \begin{pmatrix} e^{u+v} & e^{u+v} \\ e^{u-v} & -e^{u-v} \end{pmatrix}$. Then calculating the determinant gives $\det(D\Phi) = -2 e^{2u}$. Substituting that in the formula gives
$$ \iint_{\Phi(E)} f(x,y) dA = \iint_E f(e^{u+v}, e^{u-v}) 2e^{2u} \, dA. $$
But I don't know how to go further with this. Do I have to define $f$ more specific than I did? And if so, how?
Or is this not the way to go about solving this problem?
Because the comments helped me getting to an answer, and thus no answer is needed anymore, this is what was in the comments.
We noted that the area of $\Phi(E)$ is
$$ \iint_{\Phi(E)} 1\ dA = \iint_E 1 |\det(D\Phi)|\ dA\,. $$
Then, using that we know $\det(D\Phi) = -2e^{2x}$, we can use this and the boundaries given for $E$ to calculate the area. Thus follows
$$\begin{align} \iint_{\Phi(E)} 1\ dA &= \iint_E 2e^{2x}\ dA \\ &= \int_0^1 \int_0^1 2e^{2x}\ dydx \\ &= \int_0^1 \left[2ye^{2x}\right]_0^1\ dx \\ &= \int_0^1 2e^{2x}\ dx \\ &= \left[e^{2x} \right]_0^1 \\ &= e^2 -1 \,. \end{align}$$