How do I approach this double integral? Let $R$ be the region inside $$x^2+y^2 = 1$$ but outside $$x^2+y^2 = 2y$$ with $x \ge 0 $ and $y \ge 0$
Let $$u = x^2 + y^2$$ and $$v = x^2+ y^2 - 2y$$
Compute $ \iint_R xe^y dxdy$ using this change of coordinates.
I am not sure how to find the new limits of integration for this problem. I tried using $u = 1$ and $v = 0$, but this gives a point, not a region, and saying "outside" $v = 0$ and "inside" $u = 1$ makes no sense.
I also don't see how to put the integrand in terms of $u$ and $v$.
I did find that the Jacobian is $\left(\frac{-1}{4x}\right)$.
Thank you for your help! I am studying for an exam and am struggling with change of variables.
 A: If I'm not mistaken the variabel substitution should go as follows:


*

*$u=x^2+y^2$

*$v=x^2+y^2-2y$
Using 1 in 2:
$v=u-2y$
$y=\frac{u-v}{2}$
and using $y$ in 1:
$x^2=u-y^2$
$x= (u-\frac{(u-v)^2}{4})^{1/2}$
If we now take the jacobian which is invariant thus the order of the partial derivatives doesn't matter in the matrix, by this I mean that:
$J= \begin{vmatrix}
  \frac{dy}{dv} & \frac{dy}{du} \\
  \frac{dx}{dv}& \frac{dx}{du} 
 \end{vmatrix}$ 
or
$J= \begin{vmatrix}
  \frac{dy}{du} & \frac{dy}{dv} \\
  \frac{dx}{du}& \frac{dx}{dv} 
 \end{vmatrix}$
Can be used, observe that you can also swap the rows. Now taking the partial derivatives:
$\frac{dy}{du}= \frac{1}{2} $
$\frac{dy}{dv}=-\frac{1}{2} $
$\frac{dx}{du}= \frac{1}{2}(u-\frac{(u-v)^2}{4})^{-1/2}(\frac{u-v}{2}) $
$\frac{dx}{dv}= \frac{1}{2}(u-\frac{(u-v)^2}{4})^{-1/2}(1+\frac{v-u}{2})$
Taking the determinant and factoring out the negative square root term:
$J=\frac{-1}{4}(u-\frac{(u-v)^2}{4})^{-1/2}(\frac{u-v}{2}+1+\frac{v-u}{2})=\frac{-1}{4}(u-\frac{(u-v)^2}{4})^{-1/2}$
The integral should then become:
$\iint_R xe^y dxdy = \iint_R x(u,v)e^{y(u,v)} |J| dvdu = \iint_R (u-\frac{(u-v)^2}{4})^{1/2}e^{\frac{u-v}{2}}\frac{1}{4}(u-\frac{(u-v)^2}{4})^{-1/2}dudv$
The square root terms cancel out and thus we get:
$\frac{1}{4}\iint_Re^{\frac{u-v}{2}}dvdu$
From here I don't really understand with R being outside and inside are you trying to say that R is the region enclosed by $x^2+y^2=1$, $x^2+y^2=2y$, $x\geq 0$ and $y\geq 0$ in the first quadrant? I would have asked via the comments but since I don't have enough rep I couldn't.
A: By completing the square on the second equation it's easy to see that the region of integration is the region inside the unit disc centered at $(0,0)$ but outside the unit disc centered at $(0,1)$. It's sort of triangular-shaped (draw it). We want to change from $(x,y)$ variables to $(u,v)$ variables.
First let's figure out to what region $R'$ in the $uv$-plane the region $R$ in the $xy$-plane corresponds. The region $R$ is bounded by three curves. The first is the portion of the $x$ axis from $x=0$ to $x=1$. This corresponds to the line $v=u$ between $(0,0)$ and $(1,1)$. The second is a portion of the unit circle centered at $(0,0)$. This is a $u=1$ constant curve, and it goes from $v=0$ to $v=1$ (check). The third is a portion of the unit circle centered at $(0,1)$. This is a $v=0$ constant curve, and it goes from $x=0$ to $x=1$ (check). Thus the region $R'$ in the $uv$-plane is the triangle bounded by $v=0$, $u=1$, and $v=u$.
Now we can transform the integral. I'll leave out the computation of the Jacobian since you've already done it. We can solve for $y$ in terms of $u$ and $v$ to get $y=(u-v)/2$. Thus the integral in $(u,v)$ coordinates is
$$\iint_{R'} xe^{(u-v)/2}\left|\frac{-1}{4x}\right|\,dA = \frac{1}{4}\iint_{R'} e^{(u-v)/2}\,dA.$$
Express the above integral as an itegrated integral over $R'$ to compute the answer.
