find the area of its image under the map $f(x,y)=(e^{x+y},e^{x-y})$ $D$ be the open unit square in first quadrant, I need to find the area of its image under the map $f(x,y)=(e^{x+y},e^{x-y})$
will it be $e^2$? It is just a guess! 
Thank you!
 A: All points @Samarat pointed are true. I am adding additional point which is, when we use transformations of multiple integrals, we need to consider the Jacobian of $x$ and $y$ with respect to $u$ and $v$. See this. In fact we have the following integrals instead:
$$\iint_{D_t}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|dudv$$ in which $$\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=|x_uy_v-x_vy_u|$$
A: Let $D$ be the unit square. Note that $$\frac{\partial(f_1,f_2)}{\partial(x,y)}=e^{x-y}e^{x+y}-(-e^{x-y}e^{x+y})=2e^{2x}$$
thus your function has nonvanishing Jacobian. Moreover your function is injective (check this). Of course it is $\mathscr C^1$ so we can use the change of variables formula for integrals. This says that if $D$ is the unit square and $f$ you diffeomorphism, $$\int_{f(D)}1dxdy=\int_{D}1\circ f(x,y)|J_f(x,y)|dxdy$$
But of course $1\circ f(x,y)=1$ so we get that $$\int_{f(D)}1dxdy=\int_{D}|J_f(x,y)|dxdy$$
Thus the area is $$\int\limits_0^1 {\int\limits_0^1 {2{e^{2x}}dx} dy}  = {e^2} - 1$$
A: If $$u=e^{x+y}\quad v=e^{x-y}$$ then the transformed area in the $u-v$ plane will be bounded by the curves $$uv=1,\quad uv=e^2, \quad u=v,\quad u=e^2v$$ which I denote by the region $D_t$ Then your integration becomes $$\iint_{D_t}du\ dv$$
