Hello, I think this thread belongs to Differential Geometry.

I want to compute Jacobian determinant and area differential of the following transformation. $T(u,v)=\langle u^3-3uv^2,3u^2v-v^3\rangle$

Answer:- Jacobian Determinant $\frac{\partial{(x,y)}}{\partial{(u,v)}}= \frac{\partial{x}}{\partial{u}}\frac{\partial{y}}{ \partial{v}}-\frac{\partial{x}}{\partial{v}}\frac{\partial{y}}{ \partial{u}}$

$=(3u^2-3v^2)(3u^2-3v^2)-(-6uv)(6uv)$ =$9u^4+18u^2v^2+9v^4$

and its area differential is $(9u^4+18u^2v^2+9v^4)du*dv$

So the Jacobian deteminant =36 at (u,v)=(1,1). The approximate area of the image of rectangle $[1,1.4]\times[1,1.2]=36*0.4*0.2=2.88.$Is this correct? I don't know the correct answer because answers are not given.