Compute $iint (x+y),dx, dy$ with circle constraint $x^{2}+y^{2}=x+y$

There are 3 possible answers I can see, depending on what you meant.

1: What you said—a 2D integral over a 1D set. In this case the answer is 0, no working required.

2: A line integral over the curve $C:x^2+y^2=x+y$. We can parametrise this curve as $(x,y)=r(theta)=left(frac{1}{sqrt{2}}costheta+frac{1}{2},frac{1}{sqrt{2}}sintheta+frac{1}{2}right)$ for $thetain[0,2pi]$. A quick check shows that $|r'(theta)|=frac{1}{sqrt{2}}$. So then the integral is $$int_Cf(x,y)ds = int_0^{2pi}f(r(theta))|r'(theta)|dtheta = int_0^{2pi}left(frac{1}{2}(costheta+sintheta)+frac{1}{sqrt{2}}right)dtheta,$$ which I'm sure you can work out.

3: A genuine double integral over the disc $D:x^2+y^2leq x+y$. We can parametrise this in (I believe) a clearer way than has already been given: $(x,y)=left(rcostheta+frac{1}{2},rsintheta+frac{1}{2}right)$, for $thetain[0,2pi]$ and $rin[0,1/sqrt{2}]$. Recall that the area element in polar coordinates is $dxdy=rdrdtheta$. Thus the integral is: $$iint_D(x+y)dxdy=int_0^{2pi}int_0^{1/sqrt{2}}left(r^2(costheta+sintheta)+rright)drdtheta.$$ This splits into two terms and each term is simply the product of 2 1-variable integrals, so I'm sure you can work this out as well.

Assuming, that we integrate on $x^2+y^2leqslant x+y$, as noted by @Tom Sharpe, we can obtain all necessary limits from $0 leqslant r = sin phi + cos phi$

For example, let's obtain limits for $phi$:

we have $sin phi + cos phi = sqrt{2} cos left(frac{pi}{4} - phi right)$. Now most simple is draw graphic of this function and see where is it positive - we will see $[-frac{pi}{4},frac{3pi}{4}] $ as one possible solution, which agrees with geometrical view from $Oxy$ plane, that circle is in one side of $y=-x$ line. Obviously any other solution, from periodic nature of trigonometric function, is acceptable. Another way is take formal definition of $cos$ and find mentioned segment from it. Result will be same. $$intlimits_{-frac{pi}{4}}^{frac{3pi}{4}}intlimits_{0}^{sin phi + cos phi}r^2(sin phi + cos phi)d phi dr$$ I would like to say that the path indicated by @Alexey Burdin is closer to my heart, although, the decisive factor here, may be, is question what better to optimize - the limits of the integral or the integrand.

We are asked to evaluate the integral, $I$, of the function $f(x,y)=x+y$ over the disk defined by the boundary circle $x^2+y^2= x+y$. We can express $I$ in Cartesian coordinates as

$$I=int_{1/2-1/sqrt2}^{1/2+1/sqrt2} int_{1/2-sqrt{1/2-(y-1/2)^2}}^{1/2+sqrt{1/2-(y-1/2)^2}} (x+y),dx,dy$$

If we make a brute force transformation to polar coordinates, $(r,phi)$, then the locus of points on the boundary of the disk are given by $r=cos(phi)+sin(phi)$ with $phiin [-pi/4,3pi/4]$ serving as a parameter. Then, we have

$$begin{align} I&=int_{-pi/4}^{3pi/4} int_0^{cos(phi)+sin(phi)}(rcos(phi)+rsin(phi)),r,dr,dphi\\ end{align}$$

Can you finish now?