I need to use
$u=x+y,\ v=x^2+y^2$
to find $I=\int_D x+y\ dxdy$ where $D=\{ (x,y)\ :\ x^2+y^2=1, y\geq 0\}$.
I think the region for the $u-v$ plane is the rectangle $\{(u,v)\ :\ -1\leq u\leq \frac{1}{\sqrt{2}},0\leq v\leq 1\}$ but i'm having trouble finding the jacobian.
When finding the jacobian I used that $\frac{\partial y,x}{\partial u,v} =1/\frac{\partial u,v}{\partial x,y} $ and found that it's $\frac{1}{2\sqrt{2v-u^2}}$ though I'm having trouble computing the integral still when i put into the change of coordinate formula. And for the region I used a sketch.
Best Answer
Defining
$$ f :(x,y) \to (x+y,x^2+y^2)$$
$$ \nabla f(x,y)=\begin{pmatrix} 1 & 1\\ 2x & 2y \end{pmatrix}$$
$$J_f(x,y)=(1-2y)-(1-2x)=2(x-y) $$
So $f$ is a bijection from $D$ to $D'=[-1,\sqrt2]\times\{1\}- \{ x(u,v)=y (u,v)\}$
So once you have the expression of $$ x(u,v) \\ y(u,v) $$
Put in $J_f$,invert it and then :
Your integral is :
$$ I=\int_{D'}(x(u,v)+y(u,v))|J_f^{-1}(u,v)|dudv $$